chapter 8 equations of lines and planes

Review of Prerequisite Skills,pp. 424–425

1. a.

b.

2. a. The points A, B, and C are collinear if andonly if the vectors and are collinear.

So and so and are collinear.b. The points J, K, and L are collinear if and only if the vectors and are collinear.

So and so and are collinear.c. The points A, B, and C are collinear if and only if the vectors and are collinear.

So and so and are collinear.d. The points R, S, and T are collinear if and only if the vectors and are collinear.

Since the ratios of the components are not equal,and are not collinear. So R, S, and T do not

lie on the same line.3. ABC is a right triangle if its sides,

and satisfy the Pythagorean theorem.

So

So

So

.Since ABC is a right triangle.4. The vectors and are perpendicular if

So if then 5. a. A vector, is perpendicular to if

and and are not both zero.

So if and then Sois perpendicular to

b. A vector, is perpendicular to ifand and are not both zero.


c. A vector, is perpendicular to ifand are not all zero.

5 27(t1) 2 4(t2) 1 0(t3)

c>

? (t1, t2, t3) 5 (27, 24, 0) ? (t1, t2, t3)

t3t2,t1,c>

? (t1, t2, t3) 5 0

c>

(t1, t2, t3),

b>

.(5, 6)

a>

? (t1, t2) 5 0.t2 5 6,t1 5 5

5 6(t1) 2 5(t2)

b>

? (t1, t2) 5 (6, 25) ? (t1, t2)

t2t1b>

? (t1, t2) 5 0

b>

(t1, t2),

a>

.(3, 1)

a>

? (t1, t2) 5 0.t2 5 1,t1 5 3

5 1(t1) 2 3(t2)

a>

? (t1, t2) 5 (1, 23) ? (t1, t2)

t2t1a>

? (t1, t2) 5 0

a>

(t1, t2),

u>

? v>

5 0.t 5 18,

5 t 2 18

5 2t 1 (21)t 1 3(26)

u>

? v>

5 (t, 21, 3) ? (2, t, 26)

u>

? v>

5 0.

v>

u>

@AB>

@ 2 1 @BC>

@ 2 5 @AC>

@ 2,5 "14

@BC>

@ 5 "32 1 (22)2 1 (21)2

5 (3, 22, 21)

BC>

5 (5, 3, 2) 2 (2, 5, 3)

5 "41.

@AC>

@ 5 "42 1 (23)2 1 42

5 (4, 23, 4)

AC>

5 (5, 3, 2) 2 (1, 6, 22)

5 "27.

@AB>

@ 5 "12 1 (21)2 1 52

5 (1, 21, 5)

AB>

5 (2, 5, 3) 2 (1, 6, 22)

@BC>

@ ,@AC

>

@ ,@AB>

@ ,

RT>

RS>

5 (1, 2, 3)

RT>

5 (2, 4, 0) 2 (1, 2, 23)

5 (3, 21, 6)

RS>

5 (4, 1, 3) 2 (1, 2, 23)

RT>

RS>

AC>

AB>

AC>

5 2AB>

,

5 2(3, 5, 21)

5 (6, 10, 22)

AC>

5 (7, 12, 21) 2 (1, 2, 1)

5 (3, 5, 21)

AB>

5 (4, 7, 0) 2 (1, 2, 1)

AC>

AB>

JL>

JK>

JK>

5 12 JL

>

,

51

2(8, 2)

5 (4, 1)

JL>

5 (0, 4) 2 (24, 3)

5 (8, 2)

JK>

5 (4, 5) 2 (24, 3)

JL>

JK>

AC>

AB>

AC>

5 23AB>

,

5 23(3, 5)

5 (29, 215)

AC>

5 (28, 218) 2 (1, 23)

5 (3, 5)

AB>

5 (4, 2) 2 (1, 23)

AC>

AB>

5 (13, 212, 241)

5 (10, 215, 220) 1 (3, 3, 221)

1 (3 3 1, 3 3 1, 3 3 (27))

5 (5 3 2, 5 3 (23), 5 3 (24))

5(2, 23, 24) 1 3(1, 1, 27)

5 (2, 29, 6)

1 2 (25))5 (3 2 1, 22 2 7,

(3, 22, 1) 2 (1, 7, 25)

CHAPTER 8Equations of Lines and Planes

8-1Calculus and Vectors Solutions Manual

So if and thenSo is perpendicular

to 6. The area of a parallelogram formed by two vectors is determined by the magnitude of the crossproduct of the vectors.

7. a.

So is a vector perpendicular toboth vectors.b.

So is a vector perpendicular to bothvectors.8.

9. a.

b.

c.

d.

10. a.

b.

c.

d.

11. a. The y-intercept occurs when

So the y-intercept is The slope is equal to b.

So the y-intercept is The slope is equal to c.

So So the y-intercept is The slope is equal to d.

So So the y–intercept is 3. The slope is equal to a. –d.

12. Any positive scalar multiple of a vector is acollinear vector in the same direction. Answers mayvary. For example:a.b. 3(25, 4, 3) 5 (215, 12, 9)

2(4, 7) 5 (8, 14)

20

64y = –2x –5

4x – 8y = 8

5x = 5y – 15

3x – 5y + 1 = 0

2

–2–4–6

–4–8 –6–10 –2 x

y

55 5 1.

5y 5 15

5(0) 5 5y 2 15

5x 5 5y 2 15

35.

2125 5 1

5.

25y 5 21

3(0) 2 5y 1 1 5 0

3x 2 5y 1 1 5 0

48 5 1

2.21.

5 21

y 58

28

4(0) 2 8y 5 8

4x 2 8y 5 8

22.25.

5 25

5 22(0) 2 5

y 5 22x 2 5

x 5 0.

5 (4, 25, 24)

p>

5 (4, 0, 24) 2 (0, 5, 0)

5 (22, 8, 25)

p>

5 (1, 2, 4) 2 (3, 26, 9)

5 (210, 214)

p>

5 (27, 26) 2 (3, 8)

5 (7, 3)

p>

5 (4, 8) 2 (23, 5)

5 (24, 5, 4)

p>

5 (0, 5, 0) 2 (4, 0, 24)

5 (2, 28, 5)

p>

5 (3, 26, 9) 2 (1, 2, 4)

5 (10, 14)

p>

5 (3, 8) 2 (27, 26)

5 (27, 23)

p>

5 (23, 5) 2 (4, 8)

y

z

x

A

B

C D

(0, 0, 23)

5 0

1 (0)(23)5 (22)(0) 1 (21)(0)

b>

? (222, 28, 213)

5 0

a>

? (0, 0, 23) 5 (21)(0) 1 (22)(0) 1 (0)(23)

5 (0, 0, 23)

2 (21)(0), (21)(21) 2 (22)(22))

a>

3 b>

5 ((22)(0) 2 (0)(21), (0)(22)

(222, 28, 213)

5 0

5 266 1 40 1 26

1 (22)(213)

b>

? (222, 28, 213) 5 (3)(222) 1 (25)(28)

5 0

5 244 2 8 1 52

1 (24)(213)

a>

? (222, 28, 213) 5 (2)(222) 1 (1)(28)

5 (222, 28, 213)

2 (2)(22), (2)(25) 2 (1)(3))

a>

3 b>

5 ((1)(2) 2 (24)(25), (24)(3)

5 "2802

5 "(229)2 1 352 1 (226)2

5 0 (229, 35, 226) 0 A 5 area of parallelogram

5 (229, 35, 226)

(4)(1) 2 (10)(3))

(9)(3) 2 (4)(2),5 ((10)(22) 2 (9)(1),

(4, 10, 9) 3 (3, 1, 22)

c>

.

(24, 7, 0)c>

? (t1, t2, t3) 5 0.

t3 5 0,t2 5 7,t1 5 24,

8-2 Chapter 8: Equations of Lines and Planes

c.

d.13. To simplify can be written in algebraic notation. So a.

b.

So

c.

d.

e. is merely the negative of So

f.

14. The dot product of two vectors yields a realnumber, while the cross product of two vectorsgives another vector.

8.1 Vector and ParametricEquations of a Line in pp. 433–434

1. Direction vectors for a line are unique only up toscalar multiplication. So since each of the givenvectors is just a scalar multiple of each is anacceptable direction vectors for the line.

2. a. Simply find x and y coordinates for three values of t. Three possible values are and At and

At and At and So (1, 5), and

are three points on the line.b. Find the t value when the y-coordinate is 15. Sosolve for t.

If the Sois a point on the line.

3. Answers may vary. For example:a. is a point on the line and is adirection vector for the line.b. is a point on the line and is adirection vector for the line.c. is a point on the line and (0, 2) is a directionvector for the line.d. is a point on the line and is adirection vector for the line.4. Answers may vary. For example: One possibleline has A(2, 1) as its origin point and as itsdirection vector, while another has as itsorigin point and as its direction vector.

So the first case is .

The second case is 5. a. Find the t value when the y-coordinate is 18.So solve for t.

If the So is apoint on the line.b. Answers may vary. For example: A directionalvector for the line is Since is apoint on the line, a possible vector equation is

.c. Answers may vary. For example: We may take

to find another point on the line. Soand Hence

is a point on the line. So another vectorequation is .6. Answers may vary. For example:a. Three different s values will yield three different points on the line. If thens 5 21,

tPRr>

5 (22, 4) 1 t(21, 2),

(22, 4)

y 5 4 1 2(0) 5 4.x 5 22 2 0 5 22

t 5 0

tPRr>

5 (29, 18) 1 t(21, 2),

R(29, 18)(21, 2).

R(29, 18)x 5 22 2 7 5 29.t 5 7,

t 5 7

2t 5 14

18 5 4 1 2t

sPR.q>

5 (23, 5) 1 s(5, 24),

BA>

5 (2, 1) 2 (23, 5) 5 (5, 24)

tPRr>

5 (2, 1) 1 t(25, 4),

AB>

5 (23, 5) 2 (2, 1) 5 (25, 4)

BA>

B(23, 5)

AB>

(25, 0)(0, 6)

(4, 1)

(2, 27)(1, 3)

(2, 1)(3, 4)

P(214, 15)

x 5 1 1 3(25) 5 214.t 5 25,

t 5 25

22t 5 10

15 5 5 2 2t

(4, 3)

(22, 7),y 5 5 2 2(1) 5 3.

x 5 1 1 3(1) 5 4t 5 1,y 5 5 2 2(0) 5 5.

x 5 1 1 3(0) 5 1t 5 0,y 5 5 2 2(21) 5 7.

x 5 1 1 3(21) 5 22t 5 21,t 5 1.

t 5 0,t 5 21,

(13,

16)

R2,

5 (55, 40, 2140)

2 (220)(24)) (12)(25)

(21)(24) 2 (12)(23),

2 (21)(25), (2u>

1 v>

) 3 (u>

2 2v>

) 5 ((220)(23)

5 (24, 25, 23)

5 (4, 29, 21) 2 (8, 24, 2)

u>

2 2v>

5 (4, 29, 21) 2 2(4, 22, 1)

5 (12, 220, 21)

5 (8, 218, 22) 1 (4, 22, 1)

2u>

1 v>

5 2(4, 29, 21) 1 (4, 22, 1)

5 (11, 8, 228)

v>

3 u>

5 2 (211, 28, 28)

u>

3 v>

.v>

3 u>

5 (211, 28, 28)

2 (4)(1), (4)(22) 2 (29)(4))

u>

3 v>

5 ((29)(1) 2 (21)(22), (21)(4)

5 77

1 (0)(22)

(u>

1 v>

) ? (u>

2 v>

) 5 (8)(0) 1 (211)(27)

5 (0, 27, 22)

u>

2 v>

5 (4, 29, 21) 2 (4, 22, 1)

5 (8, 211, 0)

u>

1 v>

5 (4, 29, 21) 1 (4, 22, 1)

5 233

5 216 2 18 1 1

2v>

? u>

5 (24)(4) 1 (2)(29) 1 (21)(21)

5 (24, 2, 21)

2v>

5 21(4, 22, 1)

5 33

5 16 1 18 2 1

u>

? v>

5 (4)(4) 1 (29)(22) 1 (21)(1)

v>

5 (4, 22, 1)

v>

4(25i>

1 8j>

1 2k>

) 5 220i>

1 32j>

1 8k>

1

2(2i

>

1 6j>

2 4k>

) 5 i>

1 3j>

2 2k>


If then and if then Hence

and are three points on the line.

b. is a line that passes through theorigin different from the line in part a.c. If then So

, is a line that passesthrough the origin with a direction vector of Hence this describes the same line as part a.7. One can multiply a direction vector by a constantto keep the same line, but multiplying the pointyields a different line.8. a.

b. is a possible direction vector for this lineand Q is a point on the line.

So a vector equation for the line is. The corresponding

parametric equation is .9. a.

b. is a possible direction vector for this lineand M is a point on the line.


parametric equation is .10. a. A line perpendicular to L would have adirection vector that is perpendicular to the direction vector of L. is perpendicular to if


So if and then So an equation for a line with as a directionvector and P(2, 0) as a line is

.b. The line intersects the y-axis when the x coordinateis zero. The x coordinate is zero, when or

The y coordinate at this point is orSo the line intersects the y-axis at the point

11. The line crosses the x-axis, when soor So the x coordinate at this

point is The line crossesthe y-axis, when so or

So the y coordinate at this point isSo the triangle formed by the

origin, A(6, 0), and B(0, 3) is a right triangle with abase of six units and a height of three units. So thearea is 12. First all the relevant vectors are found.

a.

b.

c.

13. a. Find the t values such that x and y coordinatessatisfy or similarly

So when or LetA be the point where So x coordinate of Ais and the y coordinate is Let B be the point where So x coordinateof B is and the y coordinate is

So A is and B is

b. and

hence the length of

or about 24.045 "578,

5 "(217)2 1 (217)2

AB 5 @AB>

@AB

>

5 (212, 25) 2 (5, 12) 5 (217, 217)

(212, 25)(5, 12)9 2 14 5 25.

2 2 14 5 212,

t 5 214.

9 1 3 5 12.2 1 3 5 5,

t 5 3.

t 5 214.t 5 3x2 1 y2 2 169 5 0,

5 (t 2 3)(t 1 14)

5 t2 1 11t 2 42

1 t2 2 169

5 4 1 4t 1 t2 1 81 1 18t x2 1 y2 2 169 5 (2 1 t)2 1 (9 1 t)2 2 169

x2 1 y2 2 169 5 0.x2 1 y2 5 169

AC>

5 (24, 6) 52

3(26, 9) 5

2

3 AD

>

AD>

5 (26, 9) 5 3(22, 3) 5 3AB>

AC>

5 (24, 6) 5 2(22, 3) 5 2AB>

5 (26, 9)

AD>

5 ((1, 2) 1 3(22, 3)) 2 ((1, 2) 1 0(22, 3))

5 (24, 6)

AC>

5 ((1, 2) 1 2(22, 3)) 2 ((1, 2) 1 0(22, 3))

5 (22, 3)

AB>

5 ((1, 2) 1 1(22, 3)) 2 ((1, 2) 1 0(22, 3))

12(3)(6) 5 9.

y 5 8 1 (25) 5 3.

s 5 25.

210 2 2s 5 0,x 5 0,

x 5 210 2 2(28) 5 6.

s 5 28.8 1 s 5 0,

y 5 0,

(0, 21.2).

y 5 21.2.

0 1 3tt 5 20.4.

2 1 5t 5 0,

tPRr>

5 (2, 0) 1 t(5, 23),

(5, 23)

(3, 5) ? (t1, t2) 5 0.t2 5 23,t1 5 5

(3, 5) ? (t1, t2) 5 3(t1) 1 5(t2)

t2t1(3, 5) ? (t1, t2) 5 0

(3, 5)(t1, t2)

tPRy 5 5,x 5 4 1 5t,tPRr

>

5 (4, 5) 1 t(5, 0),

MN>

5 (9, 5) 2 (4, 5) 5 (5, 0)

(4, 5)

MN>

–6 6 80

42

–2

N(9, 5)

M(4, 5)

68

10y

x

2 4–4 –2

tPRy 5 7 1 2t,x 5 0,

tPRr>

5 (0, 7) 1 t(0, 2),

QR>

5 (0, 9) 2 (0, 7) 5 (0, 2)

(0, 7)

QR>

–6 60

42

–2

Q(0, 7)R(0, 9)

68

10y

x

2 4–4 –2

(3, 4).

tPRr>

5 (9, 12) 1 t(3, 4),

(9, 12) 1 t(3, 4) 5 (0, 0).t 5 23,

tPRr>

5 t(1, 1),

(3, 4)(0, 0),

(23, 24),s(3, 4) 5 (3, 4).s 5 1,

s(3, 4) 5 (0, 0)s 5 0,s(3, 4) 5 (23, 24).


14. In the parametric form, the second equationbecomes . If t is solved

for in this equation, we obtain and

Setting these two expressions equal to each other,

the line is described by or by

simplifying, So the second

equation describes a line with a slope of . If y is

solved for in the first expression, we see thatis on the second line but not the

first. Hence both equations are lines with slope of with no point in common and must be parallel.

8.2 Cartesian Equation of a Line,pp. 443–444

1. a. is a direction vector parallel tothe line.b. For a vector perpendicular to the line, a suitable

has to be found, such that is a such a vector.c. If then so (0, 9) is a point on thegiven line.d. A direction vector was found in part a., so a vector equation for a parallel line passing through A is . The corresponding parametric equation is

.e. A direction vector was found in part b., so a vector equation for a perpendicular line passingthrough is .The corresponding parametric equation is

.2. a.-b.

c. Switching the components of the direction vector with the coordinates of the point on the lineproduces a different line.

3. a. A direction vector parallel to the line is and if then So is a point onthe line. So a vector equation for the line is

. The correspondingparametric equation is .b. A direction vector parallel to the line is (2, 3),and if then So (0, 5) is a point on theline. So a vector equation for the line is

. The correspondingparametric equation is .c. The equation describes a horizontal linein the xy-plane, so a direction vector parallel to this line is Also is a point on this line, so a vector equation for the line is

, which gives a parametric equation of .d. The equation describes a vertical line inthe xy-plane, so a direction vector parallel to thisline is Also is a point on this line, so avector equation for the line is

, which gives a parametric equation of .

4. If the two lines have direction vectors that arecollinear and share a point in common, then the twolines are coincident. In this example, both have

as a parallel direction vector and both haveas a point on the line. Hence the two lines

are coincident.5. a. The normal vectors for the lines are and which are collinear. Since in twodimensions, any two direction vector perpendicularto are collinear, the lines have collineardirection vectors. Hence the lines are parallel.b. The lines will be coincident if they share a common point. is a point in the first line. So the lines are coincident if and only if

or equivalently So only if are the lines coincident.6. Since the normal vector is the Cartesianequation of the line is for someconstant k. Since is a point on the graph,

So So the equation of the line is 7. So the slope of this line is equal to Hence the equation for the line satisfies

or by multiplying both sides by y 2 5

x 2 (23)5 21,

4 2 522 2 (23) 5 21.

4x 1 5y 2 21 5 0.

k 5 4 2 25 5 221.

4(21) 1 5(5) 1 k 5 0.

A(21, 5)

4x 1 5y 1 k 5 0,

(4, 5),

k 5 12,

k 5 12.4(0) 2 6(2) 1 k 5 0,

(0, 2)

(2, 23)

(4, 26),

(2, 23)

(24, 0)

(3, 2)

tPRy 5 t,x 5 4,tPR

r>

5 (4, 0) 1 t(0, 1),

(4, 0)(0, 1).

x 5 4

tPRy 5 21,x 5 t,tPRr

>

5 (0, 21) 1 t(1, 0),

(0, 21)(1, 0).

y 5 21

tPRy 5 5 1 3t,x 5 2t,tPRr

>

5 (0, 5) 1 t(2, 3),

y 5 5.x 5 0,


>

5 (0, 26) 1 t(8, 7),

(0, 26)y 5 26.x 5 0,

(8, 7),

–6 60

42

–2

68

10y

x

2 4–4 –2

rq

tPRy 5 1 1 6t,x 5 22 1 5t,

tPRr>

5 (22, 1) 1 t(5, 6),B(22, 1)

tPRy 5 9 2 5t,x 5 7 1 6t,

tPRr>

5 (7, 9) 1 t(6, 25),(7, 9)

y 5 9,x 5 0,

n>

5 (5, 6)m>

? n>

5 0.n>

m>

5 (6, 25)

23

(1, 6)y 5 23x 1 5.

23

y 2 6 5 23x 2 2

3.

x 2 1

65 y 2 6

4,

t 5y 2 6

4.t 5

x 2 1

6

tPRy 5 6 1 4t,x 5 1 1 6t,


Moving everythingthe left hand side yields or

which is the equation in Cartesianform.8. So the directional vector of the line is collinearwith the normal vector and so has slopeequal to Furthermore is a point on theline. Hence the equation for the line satisfies

or by multiplying both sides by

Moving everything to the left side yields or

which is the equation inCartesian form.9. a.

b. First solve for t in both coordinates. So

and Then set these two sides equal to

each other to obtain or simply

So or

10. The acute angle of the intersection between twovectors and is found by taking the inverse

cosine of the absolute value of

a.and So the acute angle is

b. and

So the acute angle is

c. The direction vector for the first line is anda direction vector for the second is

and


d. A direction vector for the second line is and


e.

and So the acute angle is

f. has a direction vector of and thedirection vector for the second line is

and


11. a.

b. The normal vectors are and (1, 2).

and


and the obtuse angle is

12. a. Let the coordinates of C be (x, y). They mustsatisfy the equation Rewrite this equation in Cartesian form. The slopeis . The equation is of the form

. Substitute into the equation to solve for b.

The equation of the line is .If C is the vertex of the right triangle, and must be perpendicular, meaning that their dot product must be 0.

So Substitute for y.24

3 x 2 2

224 2 5x 1 x2 1 8 2 6y 1 y2 5 0.

(23 2 x)(8 2 x) 1 (2 2 y)(4 2 y) 5 0

CA>

? CB>

5 (23 2 x)(8 2 x) 1 (2 2 y)(4 2 y)

CB>

5 (8 2 x, 4 2 y)

CA>

5 (23 2 x, 2 2 y)

CB>

CA>

y 5 243 x 2 2

22 5 b 6 5 8 1 b

6 5 24

3(26) 1 b

(26, 6)y 5 243 x 1 b

m 5 243

(x, y) 5 (26, 6) 1 t(3, 24).

180°245° 5 135°.

cos21( 5!10!5) 5 45°

0 (1, 2) 0 5 "5.

0 (1, 23) 0 5 "10,(1, 23) ? (1, 2) 5 25,

(1, 23)

4 60

642

–2 2–4–6 –2

x

y

cos21( 1!1!5) 8 63°.

0 (2, 1) 0 5 "5.

0 (0, 1) 0 5 "1,(0, 1) ? (2, 1) 5 1,

(2, 1).

(0, 1)x 5 3

cos21( 13!29!17

) 8 54°.

0 (24, 1) 0 5 "17.

0 (2, 25) 0 5 "29,(2, 25) ? (24, 1) 5 213,

cos21( 8!20!5) 8 37°.

0 (2, 1) 0 5 "5.

0 (2, 4) 0 5 "20,(2, 4) ? (2, 1) 5 8,

(2, 1).

cos21( 5!25!5) 8 63°.

0 (4, 23) 0 5 "25.

0 (2, 1) 0 5 "5,(2, 1) ? (4, 23) 5 5,

(4, 23).

(2, 1)

cos21( 29!41!37) 8 42°.

0 (1, 26) 0 5 "37.

0 (25, 4) 0 5 "41,(25, 4) ? (1, 26) 5 229,

cos21( 3!29!17) 8 82°.

0 (24, 21) 0 5 "17.

0 (2, 25) 0 5 "29,(2, 25) ? (24, 21) 5 23,

a>

? b>

@a>

@ @b>

@.

b>

a>

4x 2 y 2 14 5 0.

212 1 4x 5 y 1 224(3 2 x) 5 y 1 2.

3 2 x 5y 1 2

24,

t 5y 1 2

24.

t 5 3 2 x

4 60

642

–2–4–6

2–4–6 –2

x

y

2x 1 y 2 16 5 0,

y 2 2 1 2x 2 14 5 0,

y 2 2 5 22(x 2 7).

x 2 7,y 2 2

x 2 75 22,

P(7, 2)22.

(2, 24),

x 1 y 2 2 5 0,

y 2 5 1 x 1 3 5 0,

y 2 5 5 21(x 1 3).x 2 (23),


So When When But then C would have thesame coordinates as A. This would not produce aright triangle. So the coordinates of C are .b.

c.

Since the dot product of the vectors is 0, the vectorsare perpendicular, and 13. The sum of the interior angles of a quadrilateralis 360°. The normals make 90° angles with theirrespective lines at A and C. The angle of thequadrilateral at B is 180° . Let x represent themeasure of the interior angle of the quadrilateralat O.

Therefore, the angle between the normals is thesame as the angle between the lines.

14. The normal vector for the first line is and (1, k) for the second.

and

So and We

obtain after squaring both sides that

So or simplySolving by the quadratic equation

gives

8.3 Vector, Parametric, and SymmetricEquations of a Line in pp. 449–450

1. a. A point on this line is b. A point on this line is c. A point on this line is d. A point on this line is e. A point on this line is

f. A point on this line is 2. a. A direction vector is b. A direction vector is c. A direction vector is d. A direction vector is e. A direction vector is (0, 0, 2).

f. A direction vector is which if multipliedby the least common denominator, 4, yields a vectorof 3. a.is a direction vector, as well as

So, is one possible

vector equation is another.

b. The parametric equation corresponding with thefirst vector equation is

. The second parametric equation is

4. a.So is a direction vector for the equation,and so may be used as the direction vector.Hence , is a vectorequation for a line containing the points

and b. The corresponding parametric equation is

.c. Since two of the coordinates in the directionvector are zero, a symmetric equation cannot exist.

tPRz 5 24,y 5 5,x 5 21 1 t,

B(2, 5, 24).A(A21, 5, 24)

tPRr>

5 (21, 5, 4) 1 t(1, 0, 0),

(1, 0, 0)

(3, 0, 0)

AB>

5 (2, 5, 24) 2 (21, 5, 24) 5 (3, 0, 0).

sPR.z 5 5 2 s,y 5 23 1 5s,x 5 3 2 4s,

tPRz 5 4 1 t,y 5 2 2 5t,x 5 21 1 4t,

sPRq>

5 (3, 23, 5) 1 s(24, 5, 21),

tPRt(4, 25, 1),r>

5 (21, 2, 4) 1

(3, 23, 5) 5 (24, 5, 21).BA>

5 (21, 2, 4) 2

AB>

5 (3, 23, 5) 2 (21, 2, 4) 5 (4, 25, 1)

(2, 21, 2).

(12, 2

14,

12),

(21, 0, 2).

(3, 24, 21).

(2, 1, 21).

(21, 1, 9).

(13, 2

34,

25).

(3, 22, 21).

(22, 23, 1).

(22, 1, 3).

(1, 21, 3).

(23, 1, 8).

R3,

k 5 2 6 "3.

k2 2 4k 1 1 5 0.

2 2 4k 1 2k2 5 1 1 k2

1 2 2k 1 k2

2(1 1 k2)5 1

4.

cos (60°) 5 0.5.cos (60°) 51 2 k

!2!1 1 k2

0 (1, k) 0 5 "1 1 k2.0 (1, 21) 0 5 "2,

(1, 21) ? (1, k) 5 1 2 k,

(1, 21)

x 5 u

360° 2 u 1 x 5 360°

90° 1 90° 1 180° 2 u 1 x 5 360°

2 u

/ACB 5 90°.

CA>

? CB>

5 (23)(8) 1 (4)(6)

5 224 1 24

5 0

5 (8, 6)

CB>

5 (8 2 0, 4 2 (22))

5 (23, 4)

CA>

5 (23 2 0, 2 2 (22))

60

42

–2–4

C(0, –2)

A(–3, 2)

68

10y

x

2 4–2

B(8, 4)

(0, 22)

x 5 23, y 5 2.

x 5 0, y 5 22.

x 5 0 or x 5 23.

224 2 5x 1 x2 1 8 2 6y 1 y2 5 0

216 2 5x 1 x2 2 6a24

3x 2 2b 1 a2

4

3x 2 2b

2

5 0

216 2 5x 1 x2 1 8x 1 12 116

9x2 1

16

3x 1 4 5 0

25

9x2 1 3x 1

16

3x 5 0

25x2 1 75x 5 0

25x(x 1 3) 5 0


5. a. So is avector equation for the line and the correspondingparametric equation is

. So the symmetric equation is

b.is a direction vector for the line. So

is a vector equation for the line and the corresponding parametricequation is . So the

symmetric equation is

c.is a direction vector for the line. Since

is also a direction vector for this line. So

is a vector equationfor the line and the corresponding parametric equationis . So the


d.is a direction vector for the line. So

is a vector equationfor the line and the corresponding parametric equationis . Since two of thecoordinates in the direction vector are zero, there isno symmetric equation for this line.e. is adirection vector for the line. So

is a vector equation for the line and thecorresponding parametric equation is


f. The direction vector for the z-axis is soa line parallel to the z-axis has (0, 0, 1) as a directionvector. So is a vectorequation for the line and the corresponding parametricequation is . Since twoof the coordinates in the direction vector are zero,there is no symmetric equation for this line.

6. a. So the first line is given by

. If x, y, and z are

solved for in terms of t, the corresponding parametricequations is

. So the first line has a direction vector ofThe second line is given by

If x and y are solved

for in terms of s, and areobtained. So the parametric equation for the secondline is andso has a direction vector of b.

and So the angle between the

two lines is

7. The directional vector of the first line isSo is a

directional vector for the first line as well. Sinceis also the directional vector of the

second line, the lines are the same if the lines sharea point. is a point on the second line. Since

is a point on thefirst line as well. Hence the lines are the same.8. a. The line that passes through with adirectional vector of is given by theparametric equation is

. So the y coordinate is equal to only whenAt and

So is a pointon the line. So the y coordinate is equal to 5 onlywhen At and

So is apoint on the line.b. Since the point A occurs when and pointB occurs when the line segment connectingthe two points is precisely all the t values between

and 5. So the equation is

9. The direction vector for the first line isand for the second line is

The lines are perpendicular precisely whenSo

So ifthen and

the lines are perpendicular.10. a. Three different points occur at three differentvalues of t. At the corresponding point onthe line is At the corresponding point on the line ist 5 1,

(4, 22, 5) 2 (24, 26, 8) 5 (8, 4, 23).

t 5 21,

(k, 2, k 2 1) ? (22, 0, 1) 5 0,k 5 21,

1(k 2 1) 5 2k 2 1.5 22(k) 1 0(2) 1

(k, 2, k 2 1) ? (22, 0, 1)

(k, 2, k 2 1) ? (22, 0, 1) 5 0.

(22, 0, 1).(k, 2, k 2 1)

22 # t # 5.z 5 3 2 6t,y 5 t,x 5 23t,22

t 5 5,

t 5 22,

B(215, 5, 227)z 5 3 2 6(5) 5 227.

x 5 23(5) 5 215t 5 5,t 5 5.

A(6, 22, 15)z 5 3 2 6(22) 5 15.

x 5 23(22) 5 6t 5 22,t 5 22.

22tPRz 5 3 2 6t,y 5 t,x 5 23t,

(23, 1, 26)

(0, 0, 3)

(1, 1, 3)1 5 1 1 78 5 1 1 1

2 5 3 2 522 ,

(1, 1, 3)

(24, 21, 1)

(24, 21, 1)(8, 2, 22) 5 22(24, 21, 1).

cos21( 2!2!3

) 8 35.3°.

0 (1, 21, 1) 0 5 "3.

0 (1, 21, 0) 0 5 "20(1) 5 2.5 1(1) 2 1(21) 1

(1, 21, 1) ? (1, 21, 0)

(1, 21, 0).

sPR,z 5 5,y 5 11 2 s,x 5 27 1 s,

y 5 11 2 sx 5 27 1 s,

x 1 7

15

y 2 11

21(5s), z 5 5.

(1, 21, 1).

tPR

z 5 7 1 t,y 5 10 2 t,x 5 26 1 t,

x 1 6

15 y 2 10

215 z 2 7

1(5t)

tPRz 5 4 1 t,y 5 2,x 5 1,

tPRr>

5 (1, 2, 4) 1 t(0, 0, 1),

(0, 0, 1),

x24

5y3, z 5 0.

tPRz 5 0,y 5 3t,x 5 24t,

tPRr>

5 t(24, 3, 0),

XO>

5 (24, 3, 0) 2 (0, 0, 0) 5 (24, 3, 0)

tPRz 5 0,y 5 t,x 5 21,

tPRt(0, 1, 0),r>

5 (21, 0, 0) 1

DE>

5 (21, 1, 0) 2 (21, 0, 0) 5 (0, 1, 0)

x 5 22.y 2 3

15

z1,

tPRz 5 t,y 5 3 1 t,x 5 22,

tPRt(0, 1, 1),r>

5 (22, 3, 0) 1

(0, 1, 1)(0, 26, 26) 5 26(0, 1, 1),

MN>

5 (22, 4, 7) 2 (22, 22, 1) 5 (0, 26, 26)

x 5 21.y 2 1

15

z1,

tPRz 5 t,y 5 1 1 t,x 5 21,

tPRt(0, 1, 1),r>

5 (21, 1, 0) 1

AB>

5 (21, 2, 1) 2 (21, 1, 0) 5 (0, 1, 1)

x 1 1

35

y 2 2

225

z 2 1

1.

tPRz 5 1 1 t,y 5 2 2 2t,x 5 21 1 3t,

tPR,r>

5 (21, 2, 1) 1 t(3, 22, 1),



The pointat the origin is b. Three different points occur at three different values of s. At the corresponding point onthe line is when

and Atthe corresponding point on the line is when

andSo and (1, 1, 3) are

two points on the line. The point at the origin is

c. is actually equal to

for any . So we can

pick different t values to obtain different points onthe lines. At the corresponding point on theline is found by solving for x, y, and z, in the

equation So

and So is a point onthe line. At and solving for x, y, and z, in the

equation yields

andSo (2, 1, 4) is a point on the line.

Also the point at the origin is

d. is actually equal to

for any . So we

can pick different t values to obtain different pointson the lines. At the corresponding point onthe line is found by solving for x, y, and z, in the

equation So

andSo is a point

on the line. At and solving for x, y, and z, in

the equation yields

and So is a point on the

line. Also the point at the origin is 11. For part a. the corresponding parametricequation is

. The corresponding symmetric equation isx 2 4

245

y 1 2

265

z 2 5

8.

tPRz 5 5 1 8t,y 5 22 2 6t,x 5 4 2 4t,

(24, 2, 3).

(24, 5, 8)z 5 (1)5 1 3 5 8.

y 5 (1)(3) 1 2 5 5,x 5 24,

y 2 2

35

z 2 3

5(51),x 5 24,

t 5 1

(24, 21, 22)z 5 (21)5 1 3 5 22.

y 5 (21)(3) 1 2 5 21,x 5 24,

y 2 2

35 z 2 3

5(5 21).x 5 24,

t 5 21,

tPRy 2 2

35

z 2 3

5(5t),x 5 24,

y 2 2

35 z 2 3

5x 5 24,

(21, 2, 0).

z 5 (1)4 5 4.

y 5 (1)(21) 1 2 5 1,x 5 (1)3 2 1 5 2,

x 1 1

35

y 2 2

215

z4

(51),

t 5 1

(24, 3, 24)z 5 (21)4 5 24.

y 5 (21)(21) 1 2 5 3,x 5 (21)3 2 1 5 24,

x 1 1

35

y 2 2

215

z4

(5 21).

t 5 21,

tPRx 1 1

35 y 2 2

215 z

4( 5 t),

x 1 1

35 y 2 2

215 z

4

(24, 2, 9).

(29, 3, 15)z 5 9 2 6(1) 5 3.

y 5 2 2 (1) 5 1,x 5 24 1 5(1) 5 1,

s 5 1,

z 5 9 2 6(21) 5 15.y 5 2 2 (21) 5 3,

x 5 24 1 5(21) 5 29,

s 5 21,

(4, 22, 5).

(4, 22, 5) 1 (24, 26, 8) 5 (0, 28, 13). For part b. the corresponding vector equation isThe

corresponding symmetric equation is

For part c. the point at the origin is and thedirection vector is So the correspondingvector equation is ,and parametric equation

For part d. the point at the origin is and adirection vector is So the correspondingvector equation is ,and parametric equation

.12. The direction vector of the first line is

and the direction vector of the secondline is The cross product of these two vectors gives a vector that is perpendicular to bothdirection vectors.

So a line with a direction vector of isperpendicular to the two initial lines. A parametricequation of such a line passing through the point

is .

13. Since and ifthen

or equivalently

So if then or Also if or then So the only two points occur at and

At and or

At and or

14. Let andbe two such points

for some real numbers s and t. So isperpendicular to the lines and and so sinceL2,L1

P1P2

>

P2(22 1 3s, 27 1 2s, 2 2 3s)

P1(4 1 2t, 4 1 t, 23 2 t)(2, 1, 2).z 5 2,y 5 5 1 (24) 5 1,

x 5 10 1 2(24) 5 2,s 5 24,

(22, 21, 2).z 5 2,y 5 5 1 (26) 5 21,

x 5 10 1 2(26) 5 22,s 5 26,s 5 24.

s 5 26

x2 1 y2 1 z2 5 9.s 5 24,s 5 26

s 5 24.s 5 26x2 1 y2 1 z2 5 9,

5 5(s 1 6)(s 1 4).

5 5s2 1 50s 1 120

(10 1 2s)2 1 (5 1 s)2 1 (2)2 2 9

(2)2 2 9 5 0.

(10 1 2s)2 1 (5 1 s)2 1(2)2 5 9

(10 1 2s)2 1 (5 1 s)2 1x2 1 y2 1 z2 5 9,

z 5 2,y 5 5 1 s,x 5 10 1 2s,

tPRz 5 13t,y 5 25 1 25t,x 5 2 2 34t,(2, 25, 0)

(234, 25, 13)

5 (234, 25, 13)

5 (228 2 6, 9 1 16, 28 1 21)

5 ((27)4 2 (3)2, (3)3 2 (24)4, (24)2 2 (27)3)

(24, 27, 3) 3 (3, 2, 4)

(3, 2, 4).

(24, 27, 3)

tPRz 5 3 1 5t,y 5 2 1 3t,x 5 24,

tPRr>

5 (24, 2, 3) 1 t(0, 3, 5),

(0, 3, 5).

(24, 2, 3)

tPR.z 5 4t,y 5 2 2 t,x 5 21 1 3t,

tPRr>

5 (21, 2, 0) 1 t(3, 21, 4)

(3, 21, 4).

(21, 2, 0)

x 1 4

55

y 2 2

215

z 2 9

26.

sPR.r>

5 (24, 2, 9) 1 s(5, 21, 26),

the direction vectors for the lines are andrespectively, and

So

So

So

Yet

So or Since

or So At

and At

and So andare the points that work.

15. The direction vector for the first line is and the direction vector for the second line is

and Sothe angle between the two lines is

Chapter 8 Mid-Chapter Review,pp. 451–452

1. a. Any three different t values yield three differentpoints. At

At and at So and

are three points on the line.b. Pick any three s values. At

Atand at So

(2, 3), and (5, 1) are three points on the line.c. Pick three different x values and solve for y toobtain the three points. At

or So when

Similarly at orAt or

So three points on the line are and (1, 1).


for any . So we can pick different t values to obtain different pointson the lines. At the corresponding point on the line is found by solving for x, y, and z, in the

equation So and

So is a point onthe line. At and solving for x, y, and z, in the

equation yields and

So (4, 0, 6) is another point onthe line. Also the point at the origin is 2. a. The x-intercept occurs when so solvefor the t values when to find the point. At

so So

So the x-intercept is at

The y-intercept occurs when or

So at the y-intercept, SoSo the y-intercept is at (0, 6).

b. The x-intercept occurs when so solve forthe s values when to find the point. At

so So

So the x-intercept is at The y-intercept occurs when or So at the y-intercept, So So the y-intercept is at 3. The direction vector for the first line is and the direction vector for the second is (2, 1).

and

So the angle between the lines is

The acute angle between the lines is approximately 4. The direction vector for the x-axis is (1, 0) and thedirection vector for the y-axis is (0, 1). The directionvector of the line is

and So the angle the line makes with the x-axis is

(0, 1) 5 25,(4, 25) ?( 41!41) 8 51°.cos21

0 (1, 0) 0 5 "1 5 1.0 (4, 25) 0 5 "41,

(4, 25) ? (1, 0) 5 4,(4, 25).

180° 2 93.2° 5 86.8°.

cos21( 21!5!65) 8 93.2°.

0 (2, 1) 0 5 "5.

0 (24, 7) 0 5 "65,(24, 7) ? (2, 1) 5 21,

(24, 7)

(0, 23).

y 5 3 2 2(3) 5 23.t 5 3.

26 1 2s 5 0.x 5 0,

(2143 , 0).

x 5 26 1 2(23) 5 214

3 .s 5 23.3 2 2s 5 0,

y 5 0,y 5 0,

y 5 0,

y 5 1 1 5(1) 5 6.

t 5 1.3 2 3t 5 0.

x 5 0,(185 , 0).

(185 ).x 5 3 2 3(21

5 ) 5

t 5 215 .1 1 5t 5 0,y 5 0,

y 5 0,

y 5 0,

(1, 22, 5).

1 5 5 6.z 5 (1)1

y 5 (1)2 2 2 5 0,x 5 (1)3 1 1 5 4,

x 2 1

35

y 1 2

25

z 2 5

1(51),

t 5 1

(22, 24, 4)z 5 (21)1 1 5 5 4.

y 5 (21)2 2 2 5 24,x 5 (21)3 1 1 5 22,

x 2 1

35

y 1 2

25

z 2 5

1(5 21).

t 5 21,

tPRx 2 1

35

y 1 2

25

z 2 5

1(5 t),

x 2 1

35

y 1 2

25

z 2 5

1

(0, 85),(21, 115 ),

y 5 55 5 1.3(1) 1 5y 2 8 5 0,x 5 1,y 5 8

5.

3(0) 1 5y 2 8 5 0,x 5 0,x 5 21.

y 5 115 ,5y 5 11.8 5 0,3(21) 1 5y 2

x 5 21,

(21, 5),

(1)(3, 22) 5 (5, 1).(2, 3) 1s 5 1,

(0)(3, 22) 5 (2, 3),(2, 3) 1s 5 0,

(21)(3, 22) 5 (21, 5).(2, 3) 1s 5 21,

(23, 4)

(25, 1),(27, 22),y 5 3(1) 1 1 5 4.

x 5 2(1) 2 5 5 23,t 5 1,y 5 3(0) 1 1 5 1,

x 5 2(0) 2 5 5 25,t 5 0,(21) 1 1 5 22.

y 5 3x 5 2(21) 2 5 5 27,t 5 21,

cos21( 8!5!14

) 8 17°.

0 (3, 2, 1) 0 5 "14.0 (2, 1, 0) 0 5 "55 8.

(2, 1, 0) ? (3, 2, 1) 5 2(3) 1 1(2) 1 0(1)(3, 2, 1).

(2, 1, 0)

P2(4, 23, 24)

P1(2, 3, 22)z 5 2 2 3(2) 5 24.

y 5 27 1 2(2) 5 23,x 5 22 1 3(2) 5 4,s 5 2,

z 5 23 2 (21) 5 22.y 5 4 1 (21) 5 3,

x 5 4 1 2(21) 5 2,t 5 21,

s 5 2.11s 5 22.228 1 11s 2 6(21) 5 0,

228 1 11s 2 6t 5 0,

t 5 21.1 1 t 5 0,5 1 1 t.(255 1 22s 2 11t)5 (22)(228 1 11s 2 6t) 1

(3, 2, 23)4(22)3P1P2

>

? (2, 1, 21)4 1 3P1P2

>

?

22(0) 1 0 5 0.

(22)3P1P2

>

? (2, 1, 21)4 1 3P1P2

>

? (3, 2, 23)4 5

255 1 22s 2 11t 5 0

2(211 1 2s 2 t) 1 (23)(5 2 3s 1 t) 5

P1P2

>

? (3, 2, 23) 5 3(26 1 3s 2 2t) 1

228 1 11s 2 6t 5 0.

1(211 1 2s 2 t) 1 (21)(5 2 3s 1 t) 5

P1P2

>

? (2, 1, 21) 5 2(26 1 3s 2 2t) 1

5 (26 1 3s 2 2t, 211 1 2s 2 t, 5 2 3s 1 t)2 (4 1 2t, 4 1 t, 23 2 t)

P1P2

>

5 (22 1 3s, 27 1 2s, 2 2 3s)

P1P2

>

? (3, 2, 23) 5 0.

P1P2

>

? (2, 1, 21) 5 0(3, 2, 23),

(2, 1, 21)


and Sothe angle the line makes with the y-axis is

The acute angle between

them is approximately5. Since the perpendicular line has as adirection vector, is a normal vector for thedesired line. So a Cartesian equation for this lineis for some constant C. C isfound by knowing that is a point on theline. So or Hence and the Cartesian equation is

6. Parallel lines have collinear direction vectors.Since the direction vector for the first line is

it may also be the direction vector forthe desired line. The symmetric equation for thisline having (0, 0, 2) as its origin point is

7. Sinceparallel lines have collinear direction vectors,

may be taken to be the direction vectorfor the parallel line. So the parametric equation with(1, 2, 5) as its origin point is

8. The direction vector for this line is The direction angles are found by finding the anglesthis vector makes with the coordinate axes. Thedirection vectors for the x-axis, y-axis, and z-axisare (1, 0, 0), (0, 1, 0), and (0, 0, 1), respectively.

and

so the angle the line makes with the x-axis is

so the angle the linemakes with the y-axis is

so the angle the linemakes with the z-axis is So the direction angles are approximately 79.3°,137.7°, and 49.7°.9. If a, b, c is a unit vector with direction vectors60°, 90°, and 30°, then

Yet

and So Similarlyso Also

so So

is a direction vector for the line, as well as

So the symmetric equation of the line with thisdirection vector and as an origin point is

10. The direction vectors for the x-axis, y-axis, and z-axis are (1, 0, 0), (0, 1, 0), and (0, 0, 1), respectively.The origin is a point on each of the axes, so it maybe taken as the origin point for each equation. So aparametric equation for the x-axis is

Similarly a parametric equation for they-axis is and a parametricequation for the z-axis is 11. a. The direction vector for the first line is

and the direction vector forthe second line is The lines areparallel if and only if the direction vectors arecollinear. The vectors are collinear only when oneis a multiple of the other, which happens only whenthe ratio between the coordinates is constant. Sothe direction vectors are parallel if and only if

If then

or

So If

then and since as

well, the ratios are a constant. So the lines are parallel if The lines are perpendicular if andonly if the dot product of the direction vectors is zero.

So the dot product is zero when or simply So if then the lines areperpendicular.12. The x-intercept occurs when so solve forthe x value when to find the point. At

so So

the x-intercept is at The y-intercept occurs

when So at the y-intercept,so So the y-intercept is at So the triangle with the origin has abase of 6 units and a height of 4 units. Hence the hypotenuse has a length of

So the perimeter is equal tounits. The area of the triangle is

13. a. Solving the Cartesian equation for y yieldsSo the direction of the line is (4, 23)y 5 23

4 x 1 6.

12 3 4 3 6 5 12.

17.24 1 6 1 "52 8

"52."42 1 62 5

(0, 24).

y 5 (22)(22) 2 8 5 24.

y 1 8

225 0 2 6

3 5 22,x 5 0.

(26, 0).

x 5 (24)3 1 6 5 26.x 2 6

35 0 1 8

22 5 24,

y 5 0,y 5 0,

y 5 0,

k 5 119,k 5 1

19.

238k 1 2 5 0,

5 238k 1 2.

(k 1 1)(23) 1 (3k 1 1)(210) 1 (k 2 3)(25)

(k 1 1, 3k 1 1, k 2 3) ? (23, 210, 25) 5

k 5 27.

27 2 325 5 2

3k 1 1

2105 2,

k 1 1

235

k 5 27,k 5 27.29k 2 3.210k 2 10 5

210(k 1 1) 5 23(3k 1 1)

k 1 1

235

3k 1 1

210,

k 1 1

235

3k 1 1

2105

k 2 3

25.

(23, 210, 25).

(k 1 1, 3k 1 1, k 2 3)

tPR.z 5 t,y 5 0,x 5 0,

tPR,z 5 0,y 5 t,x 5 0,

tPR.z 5 0,

y 5 0,x 5 t,

x 2 3

15

z 2 6

!3.y 5 24,

P(3, 24, 6)

(1, 0, "3).

(12, 0, !3

2 )

!32 5 c.(a, b, c) ? (0, 0, 1),cos (30°) 5

0 5 b.(a, b, c) ? (0, 1, 0),cos (90°) 5

a 5 12.(1, 0, 0) 5 a.(a, b, c) ?

cos (60°) 5 12(1, 0, 0).cos (60°) 5 (a, b, c) ?

)(

cos21( 71!117) 8 49.7°.

(2, 28, 7) ? (0, 0, 1) 5 7,

cos21( 281!117

) 8 137.7°.

(2, 28, 7) ? (0, 1, 0) 5 28,

cos21( 21!117) 8 79.3°.

(2, 28, 7) ? (1, 0, 0) 5 2,

0 (1, 0, 0) 0 5 0 (0, 1, 0) 0 5 0 (0, 0, 1) 0 5 "1 5 1.

0 (2, 28, 7) 0 5 "22 1 (28)2 1 72 5 "117,

(2, 28, 7).

tPR.z 5 5 1 t,y 5 2 2 9t,x 5 1 1 t,

(1, 29, 1)

KL>

5 (3, 25, 6) 2 (2, 4, 5) 5 (1, 29, 1).

x3

5y

245

z 2 2

4.

(3, 24, 4),

5x 2 7y 2 41 5 0.

C 5 241,

41 1 C 5 0.5(4) 2 7(23) 1 C 5 0

(4, 23)

5x 2 7y 1 C 5 0,

(5, 27)

(5, 27)

180° 2 141° 5 39°.

141°.cos21( 251!41) 8

0 (0, 1) 0 5 "1 5 1."41,0 (4, 25) 0 5


and the y-intercept, (0, 6), may be the originpoint of the line. So a vector equation is

b. The corresponding parametric equation for thevector equation in part a. is c. A direction vector for the x-axis is (1, 0).

andSo the angle between the line and the

x-axis is d. The normal vector for the line is (3, 4), which isa vector perpendicular to the line. So a line with theorigin as its origin point with a direction vector of(3, 4) is

14. is the slope of the line connecting and B(8, 4). Since is a point, the scalar equation can be found from

So

which gives as the scalar equation.is a direction

vector for the line and we may take to bethe origin point for the line. So a vector equation forthe line is Thecorresponding parametric equation is

15. The direction vector for the given line isSo a vector is normal to if


So if and then

So is normal to the line. Since

is a unit vector normal to thegiven line.16. a. Since the slope is a direction vector for the line is A parametric equation with anorigin point of is

b. The direction vector for the given line is A vector, perpendicular to

satisfies and and are notboth zero. So if and then So (1, 1) is a direction vector for a line perpendicularto the given line. A parametric equation for this perpendicular line with an origin point of isc. A direction vector for the line is

So we may use (0, 1) as a

direction vector. Also since the line connecting (0, 10) and (0, 7) has the origin as a point, the origin may be used as the origin point for the parametric equation. Hence a parametric equationfor the line is 17. a. The coordinate planes are and If the x coordinate of the line is zero,then so So the line intersects the yz-plane when Since

the lineintersects the yz-plane at (0, 8, 4). Similarly if the y coordinate of the line is zero, so

Since the line intersects the xz-plane at (6, 0, 0). If the zcoordinate of the line is zero, t is also 2. Hence theline intersects the xy-plane at (6, 0, 0).b. Since any line intersecting a coordinate axisintersects two coordinate planes at the same point,the only possible points for intersection with an axisare (0, 8, 4) and (6, 0, 0). (0, 8, 4) does not lie on acoordinate axis, but (6, 0, 0) lies on the x-axis. Sothe line intersects the x-axis at (6, 0, 0).

c.

18. a. Choose to be the origin point for theequations. So the vector equation is

Thecorresponding parametric equation is

and the symmetric

equation is b. Choose to be the origin point for the equations.So the vector equation is

The corresponding parametric equation isand the

symmetric equation is c. Choose to be the origin point for the equations.So the vector equation is

The corresponding parametric equation istPR.

r>

5 (0, 0, 6) 1 t(21, 5, 1),

P0

x 2 3

25

y 2 6

45

z 2 9

6.

tPR,z 5 9 1 6t,y 5 6 1 4t,x 5 3 1 2t,tPR.

(2, 4, 6),r>

5 (3, 6, 9) 1

P0

5z 2 8

1.

x 2 1

255

y 1 2

22

tPR,z 5 8 1 t,y 5 22 2 2t,x 5 1 2 5t,

tPR.t(25, 22, 1),r>

5 (1, 22, 8) 1

P0

y

(6, 0, 0)

5

5

5

10

(0, 8, 4)

z

x

(12, 28, 24) 1 2(23, 4, 2) 5 (6, 0, 0),t 5 2.

28 1 4t 5 0,

4(23, 4, 2) 5 (0, 8, 4),(12, 28, 24) 1

t 5 4.

t 5 4.12 2 3t 5 0,

z 5 0.

y 5 0,x 5 0,

tPR.y 5 t,x 5 0,

(0, 3).(0, 10) 2 (0, 7) 5

tPR.y 5 21 1 t,x 5 1 1 t,(1, 21)

(2, 24) ? (t1, t2) 5 0.t2 5 1,t1 5 1

(2, 22) ? (t1, t2) 5 2(t1) 2 2(t2)

t2t1(2, 22) ? (t1, t2) 5 0

(2, 22)(t1, t2),(2, 22).

tPR.y 5 10 2 2t,x 5 25 1 3t,(25, 10)

(3, 22).

223,

1!5

(2, 1) 5 ( 2!5

, 1!5)

0 (2, 1) 0 5 "5,(2, 1)

(2, 24) ? (t1, t2) 5 0.t2 5 1,t1 5 2

(2, 24) ? (t1, t2) 5 2(t1) 2 4(t2)

t2t1(2, 24) ? (t1, t2) 5 0

(2, 24)(t1, t2),(2, 24).

tPR.y 5 6 2 2t,x 5 24 1 12t,tPR.r

>

5 (24, 6) 1 t(12, 22),

A(24, 6)

AB>

5 (8, 4) 2 (24, 6) 5 (12, 22)

x 1 6y 2 32 5 0

6(y 2 6) 5 21(x 1 4),y 2 6

x 2 (24)5 21

6.

A(24, 6)A(24, 6)

4 2 68 2 (24) 5 22

12 5 216,

tPR.r>

5 t(3, 4),

cos21(45) 8 36.9°.

0 (1, 0) 0 5 1.

0 (4, 23) 0 5 "25 5 5,(1, 0) 5 4,(4, 23) ?

tPR.y 5 6 2 3t,x 5 4t,

tPR.t(4, 23),r>

5 (0, 6) 1


, and the symmetric

equation is

d. Choose to be the origin point for the equations.So the vector equation is

. The corresponding parametric equation is. Since the direction

vector has two zero coordinates, there is no symmetricequation for this line.19. A line parallel to the line connecting the points

and has a direction vector ofSince

collinear vectors of are also directionvectors for the line, is a direction vector.So the vector equation for a line with a directionvector of passing through the origin is

.20. The midpoint between (2, 6, 10) and is precisely The line connecting the midpoint and the given pointhas a direction vector of

So theparametric equations of the line through the desiredpoints is .21. The direction vector for the first line is and the direction vector for the second line is

So the directionvectors are collinear. The direction vectors arecollinear if and only if the lines are parallel, so theequations describe parallel lines.22. Since the point

lies on the line.

8.4 Vector and Parametric Equations of a Plane, pp. 459–460

1. a. plane; This is a vector equation of a plane in b. line; This is a vector equation of a line in c. line; This is a parametric equation for a line in d. plane; This is a parametric equation of a plane in

using (0, 0, 0) as 2. a. The first direction vector can be expressedwith integers as follows:

b. The second direction vector can be reduced asfollows:

c. The resulting equation of the plane using the twonew direction vectors is:

t,3. a. By inspection, if we choose weget the point b. Collecting the vector components of the n, andm, multiples we can rewrite the equation of theplane in vector form as:

Thus our direction vectors are:and

c.

Letting and we get:

d. Letting

We get the following parametric equations:

for and we get:

So our solution is and e. For the point the first two parametricequations are the same; yielding and however the third equation would then give:

which is not true. So there can be nosolution.4. a.

b.5 (3, 23, 21)

QR>

5 R 2 Q 5 (1 2 (22), 0 2 3, 1 2 2)

r>

5 (22, 3, 1) 1 t(0, 0, 1) 1 s(3, 23, 0)

5 (3, 23, 0)

PR>

5 R 2 P 5 (1 2 (22), 0 2 3, 1 2 1)

5 (0, 0, 1)

PQ>

5 Q 2 P 5 (22 2 (22), 3 2 3, 2 2 1)

R(1, 0, 1)Q(22, 3, 2),P(22, 3, 1),

28 5 27

28 5 21 2 3(0) 2 2(3)

28 5 21 2 3m 2 2n

n 5 3,m 5 0

B(0, 15, 28)

n 5 3.m 5 0

27 5 27

27 5 21 2 3(0) 2 2(3)

n 5 3m 5 0 27 5 21 2 3m 2 2n;

3 5 n.

15 5 5n 15 5 0 1 (23)m 1 5n

0 5 m.

0 5 0 1 2m 1 0n;

1 n(0, 5, 22)

A(0, 15, 27) 5 (0, 0, 21) 1 m(2, 23, 23)

r>

5 A(0, 15, 17)

5 (22, 217, 10)

5 (0, 0, 21) 1 (22, 3, 3) 1 (0, 220, 8)

(0, 5, 22)

r>

5 (0, 0, 21) 1 (21)(2, 23, 23) 1 (24)

n 5 24m 5 21

m, nPRr>

5 (0, 0, 21) 1 m(2, 23, 23) 1 n(0, 5, 22);

(0, 5, 22)(2, 23, 23)

m, nPRr>

5 (0, 0, 21) 1 m(2, 23, 23) 1 n(0, 5, 22);

(0, 0, 21).

n 5 m 5 0,

sPRr>

5 (2, 1, 3) 1 s(4, 224, 9) 1 t(1, 22, 5),

(6, 212, 30) 31

65 (1, 22, 5)

a1

3, 22,

3

4b 3 12 5 (4, 224, 9).

r>

0.R3

R3.

R3.

R3.

(7, 21, 8)

7 2 43 5 21 1 2

1 5 8 2 62 5 1,

(23, 29, 15) 5 23(1, 3, 25).

(1, 3, 25),

tPRz 5 1,y 5 28 2 13t,x 5 t,

(0, 28, 1) 2 (21, 5, 1) 5 (1, 213, 0).

(21, 5, 1).12 3(2, 6, 10) 1 (24, 4, 28)4 5

(24, 4, 28)

tPRr>

5 t(5, 25, 21),

(5, 25, 21)

(5, 25, 21)

(10, 210, 22)

(6, 25, 4) 2 (24, 5, 6) 5 (10, 210, 22).

(6, 25, 4)(24, 5, 6)

tPRz 5 22t,y 5 0,x 5 2,

tPRr>

5 (2, 0, 0) 1 t(0, 0, 22),

P0

x21

5y5

5z 2 6

1.

tPRz 5 6 1 t,y 5 5t,x 5 2t,


Using as the other direction vector:

Using as the other direction vector:,

5. a.does not represent a plane because the

direction vectors are the same. We can rewrite thesecond direction vector as:

And so we can rewrite the equation as:

This is an equation of a line in 6. a. The plane with direction vectors and that passes through the point

has a vector equation of:

The parametric equations are then:

b.

Using A(1, 0, 0) as our point with and asour direction vectors, our vector equation is:

And thus our parametric equations are:

c. using this andas our direction vectors and A(1, 1, 0)

as our point, the vector equation is:

The parametric equations are:

7. a.This gives the parametric equations:

Substituting for t gives:

which is true so and b.Gives the following parametric equations:

The third equation then says:

which is a false statement. So the pointis not on the plane.

8. a. Using the direction vectors and the point two

equations of intersecting lines on the plane in vectorform are:

b. When and it is easily seen that thesetwo lines both have the point in common.9. hasparametric equations:

The plane crosses the z-axis when both x and yequal 0.

And so the z-coordinate is:The plane crosses

the z-axis at the point (0, 0, 5)10. Using the point on the line and thepoint we get another direction vector:

The equation of the planehaving the given properties is then:

t, sPRr>

5 (2, 1, 3) 1 s(4, 1, 5) 1 t(3, 21, 2),

a>

5 Q 2 P 5 (3, 21, 2).

P(21, 2, 1),

Q(2, 1, 3)

z 5 6 1 3(21) 2 2(21) 5 5.

s 5 1 1 2(21) 5 21.

t 5 21.

0 5 15 1 15t 0 5 4 1 11(1 1 2t) 2 7t 0 5 4 1 11s 2 7t 0 5 1 2 s 1 2t 1 s 5 1 1 2t

z 5 6 1 3s 2 2ty 5 1 2 s 1 2tx 5 4 1 11s 2 7t

r>

5 (4, 1, 6) 1 s(11, 21, 3) 1 t(27, 2, 22)

(23, 5, 6)

t 5 0s 5 0

tPR p>

5 (23, 5, 6) 1 t(2, 1, 23);

sPR l>

5 (23, 5, 6) 1 s(21, 1, 2);

A(23, 5, 6),b>

5 (2, 1, 23)

a>

5 (21, 1, 2),

A(0, 5, 24)

24 5 172 ,

24 5 1 21

21 2(4)

24 5 1 2 s 1 2t

t 5 2 1 2 5 4.

t 5 2 1 4a1

2b

1

25 s.

3 5 6s 5 5 2s 1 (2 1 4s)

5 5 2s 1 t 0 5 2 1 4s 2 t 1 t 5 2 1 4s.

(0, 5, 24) 5 (2, 0, 1) 1 s(4, 2, 21) 1 t(21, 1, 2)

t 5 1.s 5 12 5 2;

2 5 1 2 1 1 2(1)

2 5 1 2 s 1 2tt 5 23 1 4(1) 5 1.

1 5 s.

6 5 6s3 5 2s 1 (23 1 4s)

3 5 2s 1 t.5 5 2 1 4s 2 t 1 t 5 23 1 4s.

(5, 3, 2) 5 (2, 0, 1) 1 s(4, 2, 21) 1 t(21, 1, 2)

t, sPRz 5 26t 1 2s,

y 5 1 1 4t 1 sx 5 1 1 3t 1 7s

t, sPRr>

5 (1, 1, 0) 1 t(3, 4, 26) 1 s(7, 1, 2),

a>

5 (7, 1, 2)

AB>

5 B 2 A 5 (3, 4, 26)

t, sPRz 5 s,

y 5 tx 5 1 2 t 2 s

t, sPRr>

5 (1, 0, 0) 1 t(21, 1, 0) 1 s(21, 0, 1),

AC>

AB>

AC>

5 (0, 0, 1) 2 (1, 0, 0) 5 (21, 0, 1)

AB>

5 (0, 1, 0) 2 (1, 0, 0) 5 (21, 1, 0)

t, sPRz 5 7 2 s;

y 5 2 1 t 1 4sx 5 21 1 4t 1 3s

t, sPRr>

5 (21, 2, 7) 1 t(4, 1, 0) 1 s(3, 4, 21),

A(21, 2, 7)

b>

5 (3, 4, 21),

a>

5 (4, 1, 0)

R3.

nPR 5 (1, 0, 21) 1 n(2, 3, 4),

5 (1, 0, 21) 1 (s 1 2t)(2, 3, 24)

r>

5 (1, 0, 21) 1 s(2, 3, 24) 1 2t(2, 3, 24)

(2)(2, 3, 24)

t, sPR,

r>

5 (1, 0, 21) 1 s(2, 3, 24) 1 t(4, 6, 28),

t, sPRr>

5 (1, 0, 1) 1 t(3, 23, 0) 1 s(3, 23, 21)

PR>

t, sPRr>

5 (22, 3, 22) 1 t(0, 0, 1) 1 s(3, 23, 21),

PQ>


11. Using the point and the point (0, 0, 0) on the line we get another direction vector of:

So the equation of the plane withthe given properties is:

12. a. The xy-plane in has no z-coordinate so twosets of direction vectors are: (1, 0, 0), (0, 1, 0) and(1, 1, 0),b. A vector equation for the xy-plane in is:


13. a. We can use the direction vectorsand and the

origin to write the vector equation of the plane:

b. Using andas direction vectors, the

vector equation of the plane is:

c. The two planes in parts a. and b. are parallelsince they have the same direction vectors.14. We simply need to show that the direction vectorscan be expressed as a linear combination of the othertwo:

15. The planehas

parametric equations:

Solving for the y-intercept:

Solving for the z-intercept:

The direction vector between the two points is then:

And the equation of the line between them is:

16. The fact that the planecontains both of the given lines

is easily seen when letting and respectively.

8.5 The Cartesian Equation of a Plane,pp. 468–469

1. a.b. In the Cartesian equation:

If the planepasses through the origin.c. Three coordinates:

2. a.b. In the Cartesian equation: So the planepasses through the origin.c. Three coordinates:3. a.b. In the Cartesian equation: So the planepasses through the origin.c. Three coordinates:4. a. which is equivalentto The Cartesian equation is:

Since the plane passesthrough the origin So the equation is:

b. is equivalent to so the Cartesian equation is:

and since the plane passesthrough the origin 5. Method 1: Let be a point on the plane.Then is a vector on theplane.

Method 2: so the Cartesian equationis:We know the point is on the plane andmust satisfy the equation, so:

This also gives the equation:x 1 7y 1 5z 2 43 5 0.

D 5 243.

43 1 D 5 0

(23) 1 7(3) 1 5(5) 1 D 5 0

(23, 3, 5)

x 1 7y 1 5z 1 D 5 0.

n>

5 (1, 7, 5)

x 1 7y 1 5z 2 43 5 0

(x 1 3) 1 7(y 2 3) 1 5(z 2 5) 5 0

n>

? PA>

5 0

PA>

5 (x 1 3, y 2 3, z 2 5)

A(x, y, z)

28x 1 12y 1 7z 5 0D 5 0.

28x 1 12y 1 7z 1 D 5 0,

n>

5 (28, 12, 7),

n>

5 (212,

34,

716)

x 1 5y 2 7z 5 0.

D 5 0.

x 1 5y 2 7z 1 D 5 0.

n>

5 (1, 5, 27).

n>

5 (15, 75, 2105)

(0, 0, 0), (0, 1, 0), (0, 0, 1)

D 5 0.

n>

5 (A, B, C) 5 (1, 0, 0)

(0, 0, 0), (5, 2, 0), (5, 2, 1)

D 5 0.

n>

5 (A, B, C) 5 (2, 25, 0)

(11, 21, 1),

(11, 21, 1),(0, 0, 0),

D 5 0Ax 1 By 1 Cz 1 D 5 0,

n>

5 (A, B, C) 5 (1, 27, 218)

t 5 0s 5 0

r>

5 OP>

0 1 sa>

1 tb>

tPRr>

5 (0, 3, 0) 1 t(0, 3, 2),

(0, 3, 0) 2 (0, 0, 22) 5 (0, 3, 2).

z 5 3 1 5(21) 1 3(0) 5 22.

n 5 021 5 m;

0 5 3 1 3m0 5 2 1 2m 2 (21 2 m)

n 5 21 2 m

y 5 2 1 2(0) 2 (21) 5 3

n 5 210 5 m;

0 5 4m0 5 3 1 5m 1 3(21 2 m)

0 5 3 1 5m 1 3n0 5 1 1 m 1 n 1 n 5 21 2 m

z 5 3 1 5m 1 3ny 5 2 1 2m 2 nx 5 1 1 m 1 n

r>

5 (1, 2, 3) 1 m(1, 2, 5) 1 n(1, 21, 3)

27

13(23, 2, 4) 2

17

13(24, 7, 1) 5 (21, 25, 7).

(24, 7, 1) 2 (23, 2, 4) 5 (21, 5, 23)

t, sPRr>

5 (22, 2, 3) 1 s(21, 2, 5) 1 t(3, 21, 7),

PR>

5 R 2 P 5 (3, 21, 7)

PQ>

5 Q 2 P 5 (21, 2, 5)

t, sPRr>

5 s(21, 2, 5) 1 t(3, 21, 7),

OC>

5 (3, 21, 7)OA>

5 (21, 2, 5)

t, sPRz 5 0,

y 5 tx 5 s

t, sPR.r>

5 s(1, 0, 0) 1 t(0, 1, 0),

R3

(21, 1, 0).

R3

m, nPR.r>

5 m(2, 21, 7) 1 n(22, 2, 3),

a>

5 (22, 2, 3).

A(22, 2, 3)


6. a.

Using the Cartesian equation is:

Using the point on the plane tosolve for D:

b.


Using the point on the plane to solvefor D:

c. There is only one simplified Cartesian equationthat satisfies the given information, so the equationsmust be the same.7.


Using the point on the plane to solve for D:

8. The point is on the line and thus alsoon the plane and we can get another direction vectorfrom:

Using as the otherdirection vector we can find the normal vector:

Our Cartesian equation is thus:

Using the point to determine D:

9. a.

So the unit normal vector is:

b.


c.


10. We know the point is on the plane,and can obtain another direction vector from:

Let be our otherdirection vector.

The Cartesian equation is then:21x 2 15y 2 z 1 D 5 0.

5 (21, 215, 21)

2 (1)(3), (1)(1) 2 (1)(2))

n>

5 AP>

3 a>

5 ((1)(3) 2 (26)(3), (26)(2)

a>

5 (2, 1, 3)AP>

5 (1, 1, 26).

P(1, 1, 5)

n>

@n> @

5 a3

13, 2

4

13,

12

13b

5 13

5 !169

@n> @ 5 !9 1 16 1 144

n>

5 (3, 24, 12)

3x 2 4y 1 12z 2 1 5 0

n>

@n> @

5 a4

!26, 2

3

!26,

1

!26b

5 !26

@n> @ 5 !16 1 9 1 1

n>

5 (4, 23, 1)

4x 2 3y 1 z 2 3 5 0

n>

@n> @

5 a2

3,

2

3, 2

1

3b

5 3

0 n>

0 5 !4 1 4 1 1

n>

5 (2, 2, 21)

2x 1 2y 2 z 2 1 5 0

20x 1 9y 1 7z 2 47 5 0 D 5 247.

47 1 D 5 0 20(1) 1 9(3) 1 7(0) 1 D 5 0.

(1, 3, 0)

20x 1 9y 1 7z 1 D 5 0.

5 (220, 29, 27) 5 21(20, 9, 7).

2 (1)(5), (1)(5) 2 (23)(24))

n>

5 PQ>

3 a>

5 ((23)(5) 2 (1)(5), (1)(24)

a>

5 (24, 5, 5)PQ>

5 (1, 23, 1).

Q(2, 0, 1)

7x 1 17y 2 13z 2 24 5 0

D 5 224.

24 1 D 5 0

7(1) 1 17(1) 2 13(0) 1 D 5 0

(1, 1, 0)

7x 1 17y 2 13z 1 D 5 0.

n>

5 (7, 17, 213)

5 (7, 17, 213)

2 (5)(21), (5)(22) 2 (1)(3))

AB>

3 AC>

5 ((1)(21) 2 (4)(22), (4)(3)

AC>

5 (3, 22, 21).

AB>

5 (5, 1, 4).

7x 1 19y 2 3z 2 28 5 0

D 5 228.

28 1 D 5 0

27 1 38 2 3 1 D 5 0

7(21) 1 19(2) 2 (3)(1) 1 D 5 0

P(21, 2, 1)

7x 1 19y 2 3z 1 D 5 0.

n>

5 (7, 19, 23)

5 (7, 19, 23).

2 (24)(4), (24)(1) 2 (1)(21))

QP>

3 PR>

5 ((1)(4) 2 (23)(1), (23)(21)

5 (21, 1, 4)

PR>

5 (22 2 (21), 3 2 2, 5 2 1)

5 (24, 1, 23)

QP>

5 (21 2 3, 2 2 1, 1 2 4)

7x 1 19y 2 3z 2 28 5 0

D 5 228.

28 1 D 5 0

214 1 57 2 15 1 D 5 0

7(22) 1 19(3) 2 3(5) 1 D 5 0

R(22, 3, 5)

7x 1 19y 2 3z 1 D 5 0.

n>

5 (7, 19, 23)

5 21(7, 19, 23).

5 (27, 219, 3)

2 (4)(1), (4)(2) 2 (21)(25))

PQ>

3 QR>

5 ((21)(1) 2 (3)(2), (3)(25)

QR>

5 (22 2 3, 3 2 1, 5 2 4) 5 (25, 2, 1)

5 (4, 21, 3)

PQ>

5 (3 2 (21), 1 2 2, 4 2 1)


Using the point to solve for D:

11. Since the normal vector is perpendicular to theplane, we can use the direction vector of the line asour normal vector:

The Cartesian equation is then:

We need the point to be on the plane so:

And the Cartesian equation of the planesatisfying the given conditions is:

12. a. To determine the angle between two planes,first determine their normal vectors. This is easilydone if the equations given are in Cartesian form.Once the normal vectors are known, and thenthe angle between the two planes can be determinedfrom the formula:

b.

13. a.

b. The parametric equations for the line are:

which give the following vector equation:Since the line and

normal vector are both perpendicular to the planewe may take:

The Cartesian equation for the plane is then:


And the Cartesian equation becomes:

14. a. and When is equivalent to:so the planes are parallel when b. When the planes are perpendicular

c. No the planes cannot ever be coincident. If theywere then they would also be parallel, so andwe would have the two equations:

Here all of the coefficients are equal except for theD values, which means that they don’t coincide.15. Since the plane passes through the points and it contains the line and the direction vector between them. The direction vector is:

The normal vector, must be perpendicular to thedirection vector and to the normal vector,

of the other plane, so:

Take and the Cartesian equation ofthe plane is:

Use the point (1, 4, 5) to determine D:

3x 1 5y 2 z 2 18 5 0

D 5 218.

18 1 D 5 0

3(1) 1 5(4) 2 5 1 D 5 0

3x 1 5y 2 z 1 D 5 0

n1

>

5 (3, 5, 21)

5 (26, 210, 2) 5 22(3, 5, 21)

2 (2)(1), (2)(21) 2 (22)(2))

n1

>

5 r>

3 n2

>

5 ((22)(1) 2 (24)(21), (24)(2)

n2

>

5 (2, 21, 1),

n1

>

,

r>

5 (2, 22, 24).

(3, 2, 1)

(1, 4, 5)

2x 1 4y 2 z 1 4 5 0 1 4x 1 8y 2 2z 1 8 5 0.

4x 1 8y 2 2z 1 1 5 0.

k 5 8,

k 5 210

45 2

5

2

10 1 4k 5 0

n1

>

? n2

>

5 8 1 4k 1 2 5 0

n1

>

? n2

>

5 0.

k 5 8.

n1

>

5 2(2, 4, 21),n1

>

k 5 8,

n2

>

5 (2, 4, 21).n1

>

5 (4, k, 22)

2x 2 3y 2 z 1 5 5 0.

D 5 5.

25 1 D 5 0

2(1) 2 (3)(2) 2 (1)(1) 1 D 5 0

P(1, 2, 1)

2x 2 3y 2 z 1 D 5 0.

n>

5 (2, 23, 21).

r>

5 (3, 21, 24) 1 t(22, 3, 1).

z 5 24 1 t, y 5 21 1 3t x 5 3 2 2t

u 5 cos21a5

!70b 5 53.3°

cos (u) 55

!70.

5 !70

@n1

> @ @n2

> @ 5 !14 ? !5

5 5

n1

>

? n2

>

5 1 1 4

n2

>

5 (1, 2, 0)n1

>

5 (1, 2, 23).

5 30°

u 5p

6

5!3

2

cos (u) 53

!12

5 !12.

n1

>

? n2

>

5 !2 ? !6

5 3

n1

>

? n2

>

5 2 1 0 1 1

n2

>

5 (2, 1, 21).n1

>

5 (1, 0, 21).

cos (u) 5n1

>

? n2

>

@n1

> @ @n2

> @.

n2

>

,n1

>

2x 2 4y 2 z 1 6 5 0.

D 5 6.

26 1 D 5 0

2(21) 2 4(1) 2 (0) 1 D 5 0.

(21, 1, 0)

2x 2 4y 2 z 1 D 5 0.

n>

5 (3, 22, 0) 2 (1, 2, 1) 5 (2, 24, 21).

21x 2 15y 2 z 2 1 5 0.

D 5 21.

1 1 D 5 0

21(1) 2 15(1) 2 (5) 1 D 5 0

(1, 1, 5)


16. Let be the normal vector of theunknown plane, and be the normalvector to the perpendicular plane. so we get:

We also know that the z-axis has the directionvector So:

The other constraint which we can choose is thelength of Since this is arbitrary (multiplicationby any scalar will give an equivalent normal vector)choose We have:

The equation of the plane is then:

17. The point equidistant from and

is the point

If every point in the plane is equidistant from these twopoint than the normal to the plane must point in thesame direction as the line connecting them:

The equation of the plane is thus:


We now have the equation of the plane:

Or equivalently:

8.6 Sketching Planes in pp. 476–477

1. a. A plane parallel to the yz-axis but two unitsaway, in the negative x direction.b. A plane parallel to the xz-axis but three unitsaway, in the positive y direction.c. A plane parallel to the xy-axis but 4 units away,in the positive z direction.2. The point of intersection of the three planes inproblem 1 must lie in every plane. Therefore thepoint of intersection is:3. The point must lie on the plane

since the point has an x-coordinate of 5, anddoesn’t have a y-coordinate of 6.4. In represents two lines,and In represents two planeswith the same equations.

5. a. i. x-intercept is when

Similarly the y-intercept is:

Since x and y cannot both be zero at the same time thereis no z-intercept. The plane is parallel to the z-axis.

y 5 6

3y 5 18

x 5 9

2x 5 18

y 5 z 5 0.

y

z

x

20

1

–1–2

2y

x

–2

x2 2 1 5 0R3,x 5 1.

x 5 21x2 2 1 5 0R2,

x 5 5,

p1:P(5, 23, 23)

(22, 3, 4)

R3,

8x 2 2y 2 16z 2 5 5 0.

4x 2 y 2 8z 25

25 0.

5

21 D 5 0 1 D 5 2

5

2

4(1) 23

22 0 1 D 5 0

(1, 32, 0)

4x 2 y 2 8z 1 D 5 0.

n>

5 (3, 1, 24) 2 (21, 2, 4) 5 (4, 21, 28).

(1, 32, 0).12((21, 2, 4) 1 (3, 1, 24)) 5

(3, 1, 24)

(21, 2, 4)

22

!5 x 1

1

!5 y 1 !3z 5 0.

A 5 22

!5.B 5

1

!5;

B2 51

5

4B2 1 B2 1 3 5 4

A2 1 B2 1 C2 5 4

!3

25

C2

1 C 5 !3.

@n1

> @ 5 2.

n1

>

.

5C

"A2 1 B2 1 C2.

5n1

>

? r>

@n1

> @ @r> @

cos (30°) 5!3

2

r>

5 (0, 0, 1).

A 5 22B A 1 2B 5 0.

n1

>

? n2

>

5 0

n2

>

5 (1, 2, 0)

n1

>

5 (A, B, C),


ii. x-intercept:

y-intercept:

z-intercept:

iii. There is no x-intercept since y and z cannot bothbe simultaneously zero.y-intercept:

z-intercept:

b. i. Since the plane is parallel to the z-axis onedirectional vector is: The other lies alongthe line so ii. We can find directional vectors by taking thedifference between two points, namely the interceptswe found in a. :or equivalently

orequivalently iii. Since the plane is parallel to the x-axis is one directional vector.

Or equivalently

6. a. i. Three points satisfyingthis equation are:ii. The line where this plane intersects the xy-planeis simply the line when

b.

7. has the solutions:So the three planes are the yz-plane, xz-plane, andthe xy-plane.

8. a.

b.

c.

d.

9. a.

b.

20

1

–1–2

2y

x

1–1–2

y(x 1 2) 5 0

xy 1 2y 5 0

y

z

x

y

z

x

y

z

x

y

z

x

z 5 0.y 5 0,x 5 0,xyz 5 0

y

z

x

2x 2 y 5 0.

z 5 0:

(0, 0, 0), (1, 2, 0), (0, 5, 1).

2x 2 y 1 5z 5 0.p:

(0, 1, 13).

(0, 3, 0) 2 (0, 0, 239) 5 (0, 3, 39).

(1, 0, 0)

(5, 0, 23).

(40, 0, 0) 2 (0, 0, 24) 5 (40, 0, 224)

(4, 3, 0).

(40, 0, 0) 2 (0, 230, 0) 5 (40, 30, 0)

(3, 22, 0).2x 1 3y 5 18,

(0, 0, 1).

z 5 239

2z 5 39

y 5 3

13y 5 39

z 5 24

5z 5 120

y 5 230

24y 5 120

x 5 40

3x 5 120


c.

10. a.

b.

c.

11. a. The plane with x-, y-, z- intercepts of

3, 4, and 6, respectively is

b. The plane with x- and z-intercepts of 5 and respectively, and which is parallel to the y-axis is

c. No x- or y-intercepts but with a z-intercept of

8 has the equation

Review Exercise, pp. 480–4831. Answers may vary. For example:

s,

2.

s,

t,

Both Cartesian equations are the same regardless ofwhich vectors are used.3. a. Answers may vary. For example:

tPR r>

5 (4, 3, 9) 1 t(7, 1, 1),

AB>

5 (7, 1, 1) 5 a>

B(4, 3, 9)A(23, 2, 8),

3x 1 y 2 z 2 6 5 0

D 5 26

3(1) 1 (2) 2 1(21) 1 D 5 0

(3)x 1 (1)y 1 (21)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

b>

3 c>

5 (1, 0, 3) 3 (2, 21, 5) 5 (3, 1, 21)

sPRr>

5 (1, 2, 21) 1 s(2, 21, 5) 1 t(1, 0, 3),

BC>

5 (1, 0, 3) 5 b>

AC>

5 (2, 21, 5) 5 c>

3x 1 y 2 z 2 6 5 0

D 5 26

3(1) 1 (2) 2 1(21) 1 D 5 0

(3)x 1 (1)y 1 (21)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

b>

3 a>

5 (1, 0, 3) 3 (1, 21, 2) 5 (3, 1, 21)

tPRr>

5 (1, 2, 21) 1 s(1, 21, 2) 1 t(1, 0, 3),

BC>

5 (1, 0, 3) 5 b>

AB>

5 (1, 21, 2) 5 a>

C(3, 1, 4)B(2, 1, 1),A(1, 2, 21),

z 5 21 1 2s 1 3t y 5 2 2 s x 5 1 1 s 1 t

tPR r>

5 (1, 2, 21) 1 s(1, 21, 2) 1 t(1, 0, 3),

r>

5 r0

>

1 sa>

1 tb>

BC>

5 (1, 0, 3) 5 b>

AB>

5 (1, 21, 2) 5 a>

C(3, 1, 4)B(2, 1, 1),A(1, 2, 21),

z8

5 1.

x5

2z7

5 1.

27,

x3

1y4

1z6

5 1.

y

z

x

y

z

x

y

z

x

y

z

x


b. Answers may vary. For example:

t,

t,c. There are no symmetric equations, because thereare two parameters.4. A line passing through andperpendicular to the plane with the equation

Since the line is perpendicularto the plane, the normal of the plane is the linesvector.

5. a.

b.

c.

6.

7. Since the plane is parallel to the yz-plane, itsdirection vectors are (0, 1, 0) and (0, 0, 1).

8.

9.

34x 1 32y 2 7z 2 229 5 0

D 5 2229

34(4) 1 32(4) 2 7(5) 1 D 5 0

(34)x 1 (32)y 1 (27)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

a>

3 b>

5 (34, 32, 27)

b>

5 (2, 23, 24)a>

5 (5, 24, 6),

sPR L2: r>

5 (4, 4, 5) 1 s(2, 23, 24),

sPRL1: r>

5 (4, 4, 5) 1 s(5, 24, 6),

3x 1 y 2 z 2 7 5 0

D 5 27

3(4) 1 1(23) 2 1(2) 1 D 5 0

(3)x 1 (1)y 1 (21)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

a>

3 b>

5 (24, 8, 28) 5 (3, 1, 21)

b>

5 (2, 26, 0) a>

5 (1, 1, 4),

b>

5 3(4 2 2), (23 2 3), (2 2 2)4 a>

5 (1, 1, 4),

tPR r>

5 (2, 3, 2) 1 t(1, 1, 4),

A 5 (4, 23, 2)

z 5 1 1 sy 5 2 1 t, x 5 21,

sPR r>

5 (21, 2, 1) 1 t(0, 1, 0) 1 s(0, 0, 1)t,

b>

5 (0, 0, 1) a>

5 (0, 1, 0),

r>

5 r0

>

1 ta>

1 sb>

A 5 (21, 2, 1)

19x 2 7y 2 8z 5 0

D 5 0

19(0) 2 7(0) 2 8(0) 1 D 5 0

19x 2 7y 2 8z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

n>

5 (19, 27, 28)

c 5 28b 5 27,a 5 19,

2a 1 2b 1 3c 5 0

(2, 2, 3) ? (a, b, c) 5 n>

tPRr>

5 (3, 7, 1) 1 t(2, 2, 3),

3y 1 z 2 7 5 0

D 5 27

(0)(1) 1 (3)(2) 1 (1)(1) 1 D 5 0

(0)x 1 (3)y 1 (1)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

AB>

3 n>

5 (0, 3, 1)

n>

5 (1, 0, 0)

AB>

5 (1, 21, 3)

B(2, 1, 4)A(1, 2, 1),

3x 1 5y 2 2z 2 7 5 0

D 5 27

(3)(3) 1 (5)(0) 1 (22)(1) 1 D 5 0

(3)x 1 (5)y 1 (22)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

n>

3 AB>

5 (3, 5, 22)

n>

5 (1, 21, 21)

AB>

5 (23, 1, 22)

B(0, 1, 21)A(3, 0, 1),

x 2 3y 2 3z 2 3 5 0

2x 1 3y 2 3 1 3z 1 6 5 0

1 (3)(z 1 2) 5 0

(21)(x 2 0) 1 (3)(y 2 1)

Ax 1 By 1 Cz 1 D 5 0

n>

5 (21, 3, 3)

P(0, 1, 22)

x 2 7

25

y 2 1

235

z 1 2

1

x 2 x0

a5

y 2 y0

b5

z 2 z0

c

z 5 22 1 ty 5 1 2 3t, x 5 7 1 2t,tPR r

>

5 (7, 1, 22) 1 t(2, 23, 1),

r>

5 r0

>

1 tm>

m>

5 (2, 23, 1)

1 5 0.2x 2 3y 1 z 2

A(7, 1, 22)

sPRz 5 9 1 t 1 3s,

y 5 3 1 t 1 2s, x 5 4 1 7t 1 3s,

z 5 z0 1 ta3 1 tb3

y 5 y0 1 ta2 1 tb2, x 5 x0 1 ta1 1 tb1,

sPR r>

5 (4, 3, 9) 1 t(7, 1, 1) 1 s(3, 2, 3),

CB>

5 (6, 4, 6) 5 (3, 2, 3) 5 b>

AB>

5 (7, 1, 1) 5 a>

C(22, 21, 3)B(4, 3, 9),A(23, 2, 8),

x 2 4

75

y 2 3

15

z 2 9

1

x 2 x0

a5

y 2 y0

b5

z 2 z0

c

tPRz 5 9 1 t, x 5 4 1 7t, y 5 3 1 t,

z 5 z0 1 tc x 5 x0 1 ta, y 5 y0 1 tb,


10. Answers may vary. For example: Since the line isperpendicular to the plane. The normal of the plane isthe directional vector of the line.

11. Answers may vary. For example: Use the dotproduct and cross product to find two points that areorthogonal to the normal of the plane. Then use anypoint from the plane.

s,

12. Answers may vary. For example: The x-interceptis and z-intercept is (0, 0, 7). Find thedirectional vector from these points and use a pointone of the intercepts.

13. The two direction vectors for these lines are

So the lines and are parallel (they aren’t thesame line, as a point on is not apoint on ). Take one of the direction vectors forthe plane to be the vector andfind another by computing the vector with tail at

(a point on ) and head at (a point on ). This is the vector

The point is on the plane, so the vectorequation of the plane is

s, .The parametric form for the plane is

s,Finally, to find the Cartesian equation of the plane,compute the cross product of the direction vectors.

So the Cartesian equation is of the form.

To find the value of D, substitute in the point on theplane

So the Cartesian equation is

14. a.

b.

y

z

x

y

x

z

18x 2 19y 1 15z 2 145 5 0

D 5 2145

18(3) 2 19(24) 1 15(1) 1 D 5 0

(3, 24, 1).

18x 2 19y 1 15z 1 D 5 0

5 (18, 219, 15)

2 1(21), 1(3) 2 4(23))

5 (23(21) 2 (3)(25), 4(25)

a>

3 v>

5 (1, 23, 25) 3 (4, 3, 21)

tPRz 5 1 2 5s 2 t,y 5 24 2 3s 1 3tx 5 3 1 s 1 4t,

tPRr>

5 (3, 24, 1) 1 s(1, 23, 25) 1 t(4, 3, 21),

(3, 24, 1)

5 (4, 3, 21)

v>

5 (7, 21, 0) 2 (3, 24, 1)

L2

(7, 21, 0)L1(3, 24, 1)

a>

5 (1, 23, 25),

L2

L1,(3, 24, 1),

L2L1

b>

5 (2, 26, 210) 5 2a>

a>

5 (1, 23, 25)

z 5 7 1 2ty 5 0,x 5 t,tPR r

>

5 (0, 0, 7) 1 t(1, 0, 2),

tPR r>

5 r0 1 ta,

v>

5 (3.5, 0, 7) 5 (1, 0, 2)

v>

5 3(0 2 3.5), (0 2 0), (7 2 0)4

B 5 (0, 0, 7) A 5 (23.5, 0, 0),

(23.5, 0, 0)

z 5 6 1 3s 2 ty 5 25t,x 5 s 1 3t,tPR1 t(3, 25, 21),r

>

5 (0, 0, 6) 1 s(1, 0, 3)

5 (3, 25, 21)

(3, 2, 21) 3 (1, 0, 3) 5 (6, 210, 22)

a>

5 (1, 0, 3)

3a 1 2b 2 c 5 0

(a, b, c) ? (3, 2, 21) 5 0

a>

? (3, 2, 21) 5 0

3x 1 2y 2 z 1 6 5 0

x 2 2

35

y 2 3

225

z 1 3

1

z 5 23 1 sy 5 3 2 2s,x 5 2 1 3s,

sPRr>

5 (2, 3, 23) 1 s(3, 22, 1),

n>

5 (3, 22, 1)

3x 2 2y 1 z 5 0

A(2, 3, 23)


c.

d.

e.

15. a. Answers may vary. For example:

t,

b. Answers may vary. For example: The normal ofthe plane is the direction vector of the line, since itis perpendicular to the plane. Then find using theCartesian form of a plane.

c. Answers may vary. For example: Since the planeis parallel to the z-axis, one of its direction vectorsis (0, 0, 1).

t,

d. Answers may vary. For example:

t,

16. They are in the same plane because both planeshave the same normal vectors and Cartesian equations.

s,

u,

2x 2 3y 1 z 1 1 5 0

D 5 1

2(1) 2 3(21) 1 (26) 1 D 5 0

D 5 1

2(1) 2 3(2) 1 (3) 1 D 5 0

2x 2 3y 1 z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

(1, 1, 1) 3 (2, 5, 11) 5 (6, 29, 3) 5 (2, 23, 1)

(23, 5, 21) 3 (0, 1, 3) 5 (26, 9, 23) 5 (2, 23, 1)

vPRL2: r

>

5 (1, 21, 26) 1 u(1, 1, 1) 1 v(2, 5, 11),

tPRL1: r>

5 (1, 2, 3) 1 s(23, 5, 21) 1 t(0, 1, 3),

78x 1 10y 2 12z 2 168 5 0

D 5 2168

78(1) 1 10(3) 2 12(25) 1 D 5 0

(78)x 1 (10)y 1 (212)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

AB>

3 BC>

5 (78, 10, 212)

z 5 25 1 9t 2 sy 5 3 1 3t 2 9s,x 5 1 1 t 1 s,

sPRr>

5 (1, 3, 25) 1 t(1, 3, 9) 1 s(1, 29, 21),

r>

5 r0 1 ta>

1 sb>

BC>

5 (1, 29, 21)

AB>

5 (1, 3, 9)

C(3, 23, 3)B(2, 6, 4),A(1, 3, 25),

3x 1 y 2 13 5 0

D 5 213

3(4) 1 1(1) 1 D 5 0

(3)x 1 (1)y 1 (0)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

(1, 23, 5) ? (0, 0, 1) 5 (23, 21, 0) 5 (3, 1, 0)

z 5 21 1 5t 1 sy 5 1 2 3t,x 5 4 1 t,sPRr

>

5 (4, 1, 21) 1 t(1, 23, 5) 1 s(0, 01),

AB>

5 (1, 23, 5)

B(5, 22, 4)A(4, 1, 21),

24x 1 11z 2 18 5 0

D 5 218

24(1) 1 11(2) 1 D 5 0

(24)x 1 (0)y 1 (11)z 1 D 5 0

Ax 1 Bx 1 Cx 1 D 5 0

BC>

5 (24, 0, 11)

C 5 (22, 1, 5)B 5 (2, 1, 26),A(1, 1, 2)

15x 2 8y 1 2z 2 41 5 0

D 5 241

15(3) 2 8(1) 1 2(2) 1 D 5 0

(15)x 1 (28)y 1 (2)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

a>

3 p>

5 (2, 3, 23) 3 (2, 4, 1) 5 (15, 28, 2)

z 5 2 1 t 2 3sy 5 1 1 4t 1 3s,x 5 3 1 2t 1 2s,

sPRr>

5 (3, 1, 2) 1 t(2, 4, 1) 1 s(2, 3, 23),

a>

5 (2, 4, 1)

5 (0, 3, 1) 1 t(2, 4, 1)

L: r>

5 2ti 1 (4t 1 3)j 1 (t 1 1)k

PQ>

5 p>

5 (2, 3, 23)

Q(3, 1, 2)

P(1, 22, 5)

y

z

x

y

z

x

y

z

x


17. A point B on the line will have coordinatesThen

For this vector to be perpendicular to it wouldhave zero dot product with the direction vector for

So

So and the point B is

18. a. The plane is parallel to the z-axis through thepoints (3, 0, 0) and b. The plane is parallel to the y-axis through thepoints (6, 0, 0) and c. The plane is parallel to the x-axis through thepoints (0, 3, 0) and 19. a. To determine which points lie on the line, justsee if there is a t-value such that the coordinateworks.

There is no value of t that satisfies the equations.


There is no value of t that satisfies the equations.Only A lies on the line.b.

20. a.

b.

c.

d.

21. a.

b.

22. a. i. The given line is not parallel to the planebecause (3, 0, 2) is a point on the line and the plane.

u 5 90°

cos u 50

("3)("14)

n2 5 (2, 3, 21) n1 5 (1, 21, 21),

cos u 50 n1 ? n2 0

0 n1 0 0 n2 0

L2: 2x 1 3y 2 z 1 4 5 0

L1: x 2 y 2 z 2 1 5 0

u 5 44.2°

cos u 56

("14)("5)

n2 5 (1, 2, 0)

n1 5 (2, 3, 21),

cos u 50 n1 ? n2 0

0 n1 0 0 n2 0

L2:x 1 2y 1 4 5 0

L1: 2x 1 3y 2 z 1 9 5 0

u 5 90°

cos u 50

("96)("14)

n2 5 (21, 2, 3) n1 5 (4, 8, 24),

cos u 50 n1 ? n2 0

0 n1 0 0 n2 0

L2: (x, y, z) 5 (1, 5, 24) 1 t(21, 2, 3)

L1: (x, y, z) 5 (4, 7, 21) 1 t(4, 8, 24)

u 5 37.4°

cos u 516

("29)("14)

n2 5 (2, 3, 1) n1 5 (3, 4, 22),

cos u 50 n1 ? n2 0

0 n1 0 0 n2 0

z 5 27 1 sy 5 3s, L2: x 5 21 1 2s,

z 5 22ty 5 1 1 4t,L1: x 5 21 1 3t, u 5 59.0°

cos u 53

("17)("2)

cos u 50 n1 ? n2 0

0 n1 0 0 n2 0

n2 5 (1, 21)

n1 5 (1, 4)

y 5 2x 1 3y 5 4x 1 2,

u 5 45.0°

cos u 513

("26)("13)

n2 5 (2, 23)n1 5 (1, 5),

cos u 50 n1 ? n2 0

0 n1 0 0 n2 0

L2: x 2 2

25

1 2 y3

L1: x 2 1

15

y 2 3

5

b 5 3 1 t 5 21

a 5 2t 5 28

t 5 24

23 5 1 1 tb 5 3 1 t,a 5 2t,z 5 1 1 ty 5 3 1 t,x 5 2t,

2 5 1 1 t6 5 3 1 t,6 5 2t,D(6, 6, 2)

2 5 1 1 t5 5 3 1 t,4 5 2t,C(4, 5, 2)

1 5 1 1 t2 5 3 1 t,22 5 2t,B(22, 2, 1)

t 5 1

2 5 1 1 t4 5 3 1 t,2 5 2t,A(2, 4, 2)

z 5 1 1 ty 5 3 1 t,x 5 2t,

(0, 0, 26).

(0, 0, 22).

(0, 22, 0).

Ba2 1 2a7

3b , 1 1

7

3, 2 2

7

3b 5 Ba

20

3,

10

3, 2

1

3b.

t 5 146 5 7

3,

5 214 1 6t 5 26 1 4t 2 3 1 t 2 5 1 t 5 (2, 1, 21) ? (23 1 2t, 23 1 t, 5 2 t)

0 5 v>

? AB>

v>

5 (2, 1, 21).L2,

L2,

5 (23 1 2t, 23 1 t, 5 2 t) AB

>

5 (2 1 2t, 1 1 t, 2 2 t) 2 (5, 4, 23)

tPRB(2 1 2t, 1 1 t, 2 2 t),

L2


ii. Substitute the expressions for the components ofthe parametric equation of the line into the equationof the plane.

This last statement is never true. So the line and theplane have no points in common. Therefore, the lineis parallel to the plane.iii. Use the symmetric equation to rewrite x and z interms of y.

Substitute into the equation of the plane.

This equation has a solution. Therefore, the line andplane have a point in common and are not parallel.b. i. Substitute the expressions for the componentsof the parametric equation of the line into theequation of the plane.

This last statement is always true. So every point onthe line is also in the plane. Therefore, the line liesin the plane.ii. The line is parallel to the plane, and so does notlie in it.iii. is a point that lies on the line thatdoes not lie in the plane. Therefore, the line doesnot lie in the plane.23.

24. One direction vector for the plane is (2, 4, 1) and (1, 4, 4) are on the plane, so anotherdirection vector is So the parametric equations are

25. A plane has two parameters, because a planegoes in two different directions unlike a line thatonly goes in one direction.26. This equation will always pass through the origin, because you can always set and to obtain (0, 0, 0).

27. a. They do not form a plane, because these threepoints are collinear.

b. They do not form a plane, because the point lieson the line.

28. If a is the x-intercept, b is the y-intercept, and cis the z-intercept, this means that (a, 0, 0), (0, b, 0),and (0, 0, c) are points on the plane. So

are direction vectors for the plane. So a normal forthis plane is

So the Cartesian equation of the plane is of the form

Substitute the x-intercept, (a, 0, 0), into thisequation to determine the value of D.

So the Cartesian equation of this plane isor

29. If the normal vector is then theCartesian equation of the plane will be of the form

To determine the value of D, substitute the point(which is on the plane) into this equation.

So the Cartesian equation of the plane is

30. a., b.

t,

t,

z 5 2 2 4t 1 3sy 5 23 1 7t 2 2s,x 5 1 2 3t 1 5s,

sPRr>

5 (1, 23, 2) 1 t(23, 7, 24) 1 s(5, 22, 3),

sPRr>

5 r>

0 1 ta>

1 sb>

,

BC>

5 (5, 22, 3)

AB>

5 (23, 7, 24)

C(3, 2, 1)B(22, 4, 22),A(1, 23, 2),

6x 2 5y 1 12z 1 46 5 0.

D 5 46

6(5) 2 5(8) 1 12(23) 1 D 5 0

(5, 8, 23)

6x 2 5y 1 12z 1 D 5 0

(6, 25, 12),

bcx 1 acy 1 abz 5 abcbcx 1 acy 1 abz 2 abc 5 0

D 5 2abc bc(a) 1 ac(0) 1 ab(0) 1 D 5 0

bcx 1 acy 1 abz 1 D 5 0

5 (bc, ac, ab)

a(b) 2 0(0))

5 (0(2c) 2 b(2c), 0(2c) 2 a(2c),

u>

3 v>

5 (a, 0, 2c) 3 (0, b, 2c)

5 (0, b, 2c)

v>

5 (0, b, 0) 2 (0, 0, c)

5 (a, 0, 2c)

u>

5 (a, 0, 0) 2 (0, 0, c)

5 (8, 27, 5)

r>

5 (4, 9, 23) 1 4(1, 24, 2)

r>

5 (4, 9, 23) 1 t(1, 24, 2)

r>

5 (21, 2, 1) 1 t(3, 1, 22)

(x, y, z) 5 (a 2 a, b 2 b, c 2 c) 5 (0, 0, 0)

(x, y, z) 5 (a, b, c) 1 0(d, e, f ) 2 1(a, b, c)

t 5 21 s 5 0,

(x, y, z) 5 (a, b, c) 1 s(d, e, f ) 1 t(a, b, c)

t 5 21s 5 0

tPR.z 5 4 2 3s 1 t, s,y 5 4 2 t,x 5 1 1 s 1 3t,

(2, 4, 1) 2 (1, 4, 4) 5 (1, 0, 23).

(3, 21, 1).

(x, y, z) 5 (4, 5, 8) 2 (4, 5, 6)

1 2(26, 6, 21)

(x, y, z) 5 (4, 1, 6) 1 4(3, 22, 1)

(x, y, z) 5 (4, 1, 6) 1 p(3, 22, 1) 1 q(26, 6, 21)

(5, 27, 1)

0 5 0

12 1 4t 2 2t 2 2 2 2t 2 10 5 0

4(3 1 t) 1 (22t) 2 (2 1 2t) 2 10 5 0

214y 2 96 5 0

216y 2 92 1 y 1 y 1 6 2 10 5 0

4(24y 2 23) 1 y 2 (2y 2 6) 2 10 5 0

z 5 2y 2 6

x 5 24y 2 23

215 5 0

212t 2 5 1 2t 1 10t 2 10 5 0

4(23t) 1 (25 1 2t) 2 (210t) 2 10 5 0


c. To find the Cartesian equation of the plane,a normal vector is needed. This can be found bycomputing the cross product of the direction vectorsfound in parts a. and b.

So the Cartesian equation has the form

Since is a point on this plane, we cansubstitute it in to determine the value of D.

So the Cartesian equation of this plane is

d. Substituting into the Cartesian equationfound in part c., we get

This means that is not on the plane.31. a. The normal vector to the given plane is

so any plane parallel to this one musthave this same normal vector. So if a parallel planecontains the point (0, 0, 0), it will have the form

Substitute in the point (0, 0, 0) to determine thevalue of D.


b. Reasoning as in part a., if we want the pointto be in our parallel plane we find D in

the following way:

So the Cartesian equation of the plane in this case is

c. Reasoning as in parts a. and b., if we want thepoint to be in our parallel plane we findD in the following way:


32. a. The direction vector for is (2, 1) and foris This means that

and are parallel, and since they have the point(11, 0) in common (take in and in ),

these lines are coincident. So the angle betweenthem is b. The parametric equations of these lines are

So a point of intersection satisfies

or

or

So the point of intersection is

The point of intersection is at (for )and (for ).The direction vector for is (3, 4), and for is

So the angle between these lines satisfies

It would also have been correct to report the supplement of this angle, or roughly 93.18°, asthe answer in this case.33. a.

tPR r>

5 (1, 3, 5) 1 t(22, 24, 210),

r>

5 r>

0 1 ta>

P(1, 3, 5)

8 86.82°

5 cos21 a

1

5"3b

u 5 cos21 a

(3, 4) ? (3, 22)

0 (3, 4) 0 0 (3, 22) 0b

cos u 5(3, 4) ? (3, 22)

0 (3, 4) 0 0 (3, 22) 0

u(3, 22).

L2L1

L1t 5 32

L2s 5 232(3

2, 5) 5 5

y 5 21 1 4a3

2b

53

2

x 5 23 1 3a3

2b

53

2

5 23

21 3

t 5 s 1 3

s 5 23

2

6s 5 29

4(s 1 3) 1 2s 5 3

4t 1 2s 5 3

t 5 s 1 3

4t 2 2s 5 3

3t 2 3s 5 9

21 1 4t 5 2 2 2s23 1 3t 5 6 1 3s

sPRy 5 2 2 2s,L2: x 5 6 1 3s,

tPRy 5 21 1 4t,L1: x 5 23 1 3t,

u 5 0°.

L2s 5 6L1t 5 3

L2

L1(22, 21) 5 21(2, 1).L2

L1

4x 2 2y 1 5z 2 22 5 0.

D 5 222

4(2) 2 2(22) 1 5(2) 1 D 5 0

(2, 22, 2)

4x 2 2y 1 5z 1 19 5 0.

D 5 19

4(21) 2 2(5) 1 5(21) 1 D 5 0

(21, 5, 21)

4x 2 2y 1 5z 5 0.

D 5 0

4(0) 2 2(0) 1 5(0) 1 D 5 0

4x 2 2y 1 5z 1 D 5 0.

(4, 22, 5),

(3, 5, 24)

13(3) 2 11(5) 2 29(24) 1 12 5 100 2 0

(3, 5, 24)

13x 2 11y 2 29z 1 12 5 0.

D 5 12

13(1) 2 11(23) 2 29(2) 1 D 5 0

(1, 23, 2)

13x 2 11y 2 29z 1 D 5 0.

5 (13, 211, 229)

2 (23)(3), (23)(22) 2 5(7))

5 (7(3) 2 (22)(24), 5(24)

AB>

3 BC>

5 (23, 7, 24) 3 (5, 22, 3)


b.

c.

d. Since its parallel to the x-axis, its direction vectoris (1, 0, 0).

e. Find a perpendicular vector use the dot product.

f. Since the line is perpendicular to the plane, theline’s directional vector is the normal of the plane.Use the cross product to find the vector.

34. a. This plane will be of the form

To find D, substitute in


b. The direction vector for this line is(we can use this as one of the direction

vectors for the plane), and a point on this line isSo a second direction vector for the

plane will be

So a normal vector for this plane is

The Cartesian equation of this plane has the form

Substitute in to determine D.

The Cartesian equation of this plane is

c. This plane, being parallel to the xy-plane, iscompletely determined by a fixed z-coordinate (thex- and y- coordinates are allowed to be anything atall). Since it passes through the point theequation of this plane is Written in Cartesianform, this is d. Since this plane is to be parallel to

it will have the same normal vector, So this plane will be of theform Since is on this plane, we can substitutethis in to determine the value of D.


e. Since this plane is perpendicular to the yz-plane,it is completely determined by its intersection withthe yz-plane, which will be a line with y-intercept 4 and z-intercept This means that y and z arerelated by because of the y-intercept of 4. We can find the value of m by using the z-intercept of

So y and z are related via and theCartesian equation of the plane is (x is allowed to be anything here.)f. A normal vector, for this plane will beperpendicular to the normal vector for the plane

(A, B, C),

y 2 2z 2 4 5 0.

y 5 2z 1 4,

m 5 2

0 5 m(22) 1 4

22.

y 5 mz 1 4

22.

3x 1 y 2 4z 1 26 5 0.

D 5 26

3(24) 1 2 2 4(4) 1 D 5 0

P(24, 2, 4)

3x 1 y 2 4z 1 D 5 0.

(3, 1, 24).

3x 1 y 2 4z 1 8 5 0,

z 2 3 5 0.

z 5 3.

P(3, 3, 3),

29x 1 27y 1 24z 2 86 5 0.

D 5 286

29(22) 1 27(0) 1 24(6) 1 D 5 0

P(22, 0, 6)

29x 1 27y 1 24z 1 D 5 0.

5 (29, 27, 24)

2 6(25))

2 3(25), 3(22)

2 (22)(2), 6(2)

(3, 25, 2) 3 (6, 22, 25) 5 ((25)(25)

5 (6, 22, 25)

v>

5 (4, 22, 1) 2 P(22, 0, 6)

(4, 22, 1).

(3, 25, 2)

2x 2 4y 1 5z 1 23 5 0.

D 5 23

2(22) 2 4(6) 1 5(1) 1 D 5 0

P(22, 6, 1).

2x 2 4y 1 5z 1 D 5 0.

z 5 5 1 6ty 5 3 1 t, x 5 1,

r>

5 (1, 3, 5) 1 t(0, 1, 6)

AB>

3 BC>

5 (0, 27, 242) 5 (0, 1, 6) 5 n>

BC>

5 (26, 6, 21)

AB>

5 (21, 26, 1)

C(23, 2, 1)B(3, 24, 2),A(4, 2, 1),

tPR r>

5 (1, 3, 5) 1 t(0, 6, 4),

n>

5 (0, 6, 4)

c 5 4b 5 6, a 5 0,

23a 1 4b 2 6c 1 0

(23, 4, 26) ? (a, b, c) 5 0

z 5 5y 5 3, x 5 1 1 t,tPR r

>

5 (1, 3, 5) 1 t(1, 0, 0),

r>

5 r>

0 1 ta>

n>

5 (1, 0, 0)

P(1, 3, 5),

x 2 1

265

x 2 3

2135

x 2 5

14

z 5 5 1 14ty 5 3 2 13t, x 5 1 2 6t,tPR r

>

5 (1, 3, 5) 1 t(26, 213, 14),

r>

5 r>

0 1 ta>

RS>

5 (26, 213, 14)

P(1, 3, 5)

x 2 1

285

x 2 3

65

x 2 5

22

z 5 5 2 2ty 5 3 1 6t, x 5 1 2 8t,tPR r

>

5 (1, 3, 5) 1 t(28, 6, 22),

r>

5 r>

0 1 ta>

PQ>

5 (28, 6, 22)

Q(27, 9, 3)P(1, 3, 5),

x 2 1

225

y 2 3

245

z 2 5

210

z 5 5 2 10ty 5 3 2 4t, x 5 1 2 2t,


which is Also,will be perpendicular to the direction

vector for the line contained in the plane we seek.This direction vector is (3, 1, 2), and so this meanswe can take

So the Cartesian equation will have the form

Since is on this plane (take in the line this plane isto contain), we can substitute this in to determinethe value of D.


Chapter 8 Test, p. 4841. a. i and can be the direction vectors forthis plane and A(1, 2, 4) can be the origin point.

This gives a vector equation ofs, .

The corresponding parametric equation for thisplane is

s, .ii. The corresponding Cartesian equation is found bytaking the cross product of the two direction vectors.

So is a normal vector for the plane, sothe plane has the form forsome constant D. To find D, we know that A(1, 2, 4)is a point on the plane, so

So orSo the Cartesian equation for the plane is

b. A point (x, y, z) is on the plane if and only if Since

the point is not on the plane.

2. a. Since for all (x, y, z) on the

plane, it holds true for the given points. So

or Similarly

and implies that and

So the equation of the plane is b. If both sides are multiplied by the least commonmultiple of the denominators, then an equivalentequation for the plane is Hence(6, 4, 3) is a normal vector for this plane.3. a. Since the origin is a point on the plane and

is a point on theplane, (2, 1, 3) is a direction vector for the plane.

is a point on theplane and (2, 1, 3) is another point on the plane,so is a directionalvector for the plane as well. (2, 1, 3) and (1, 2, 5)are not collinear, because the ratios between thecoordinates are not equal. Since the origin is a pointon the plane, a vector equation for the plane is

s, .b. To find the Cartesian equation for the plane, thenormal vector is determined by the cross product ofthe two direction vectors from part a.

So the Cartesian equation for the plane has the form for some constant D.Since the origin is a point on the plane,

so Thus theequation is 4. a. and are each direction vectors for the planes. The vectors are notcollinear since the ratios of the coordinates arenot equal. is a point on the plane,so a vector equation for the plane is


So the Cartesian equation for the plane has the form for some constant D.3x 2 13y 1 2z 1 D 5 0,

5 (3, 213, 2)

2 (2)(21), (2)1 2 (0)5)

2 (23)1, (23)5

(2, 0, 23) 3 (5, 1, 21) 5 ((0)(21)

tPRr>

5 (4, 23, 5) 1 s(2, 0, 23) 1 t(5, 1, 21),

(4, 23, 5)

(5, 1, 21)(2, 0, 23)

2x 2 7y 1 3z 5 0.

D 5 0.3(0) 1 D 5 0,2 (0) 2 7(0) 1

2x 2 7y 1 3z 1 D 5 0,

5 (21, 27, 3)

2 (2)5, (2)2 2 (1)1)

(2, 1, 3) 3 (1, 2, 5) 5 ((1)5 2 (3)2, (3)1

tPRr>

5 s(2, 1, 3) 1 t(1, 2, 5),

(3, 3, 8) 2 (2, 1, 3) 5 (1, 2, 5)

(2, 1, 3) 1 1(1, 2, 5) 5 (3, 3, 8)

(2, 1, 3) 1 0(1, 2, 5) 5 (2, 1, 3)

6x 1 4y 1 3z 5 12.

x2

1y3

1z4

5 1.

c 5 4.b 5 30

a1

0

b1

4

c5 1

0

a1

3

b1

0

c5 1a 5 2.

2

a1

0

b1

0

c5 1

xa

1yb

1zc

5 1

(1, 21, 212)

8(212) 2 28 5 227 2 0,2(1) 2 3(21) 1

2x 2 3y 1 8z 2 28 5 0.

2x 2 3y 1 8z 2 28 5 0.

D 5 228.

28 1 D 5 0,2(1) 2 3(2) 1 8(4) 1 D 5 0.

2x 2 3y 1 8z 1 D 5 0,

(2, 23, 8)

5 (2, 23, 8)

2 (1)0, (1)2 2 (22)3)

AB>

3 AC>

5 ((22)0 2 (21)2, (21)3

tPRz 5 4 2 s,

y 5 2 2 2s 1 2t,x 5 1 1 s 1 3t,

tPRt(3, 2, 0),r>

5 (1, 2, 4) 1 s(1, 22, 21) 1

5 (3, 2, 0)

AC>

5 (4, 4, 4) 2 (1, 2, 4)

5 (1, 22, 21)

AB>

5 (2, 0, 3) 2 (1, 2, 4)

AC>

AB>

25x 1 y 1 7z 1 18 5 0.

D 5 18

25(2) 1 (21) 1 7(21) 1 D 5 0

25x 1 y 1 7z 1 D 5 0.

5 (25, 1, 7)

2 (1)(1), 3(1) 2 1(2), 1(1) 2 3(22))

(A, B, C) 5 (1, 22, 1) 3 (3, 1, 2) 5 ((22)(2)

(A, B, C)

(1, 22, 1).x 2 2y 1 z 5 6,


Since the is a point on the plane,so

So Thus the equation is

5. a. The line intersects the yz-plane when

If then so

and Thus the point is

b. The direction vector is the same, sothe equivalent symmetric equation for the line is

6. a. The angle between two planes is determinedby the dot product of their normal vectors. The normal vector of the first plane is and the normal vector of the second plane is

and

So the angle between the planes

is The acute angle is b. i. The planes are parallel if and only if thecorresponding normal vectors are parallel. The normalvector of the first plane is and the normalvector of the second plane is The vectorsare parallel if and only if the ratios between the

coordinates are equal. Suppose so then So the vectors can be parallel onlywhen Since as well, the vectors areparallel at ii. The planes are perpendicular when their normalvectors are perpendicular. The vectors are perpendicular when their dot product is equal to zero.

So if then the dot product of the two normal vectors is equal to zero. Hence the planesare perpendicular at

c. The first plane in b. intersects the y-axis at the point(0, d, 0), where d satisfies So The second plane in b. intersects the y-axis at the point where e satisfies

So Since theplanes intersect the y-axis only once and the pointsare different, the equations can never represent thesame plane.7. a.

b. The equation for the plane can be written asSo for any real number t,

so the point (0, 0, t) is onthe graph. So the z-axis is on the plane. Also theplane cuts across the xy-plane along the line

So the origin is a point, as well as

c. The equation for the plane can be written asFor any real number t,

so (0, 0, t) is on theplane. Since this is true for all real numbers, the z-axis is on the plane.

B(0) 1 0(t) 5 0,A(0) 1

Ax 1 By 1 0z 5 0.

y

z

x

(22, 1, 0).

2x 1 y 5 0.

(0) 1 0(t) 5 0,2(0) 1

2x 1 y 1 0z 5 0.

4 60

642

–2–4–6

2–4–6 –2

x

y

e 5 24.5.2e 1 8(0) 5 9.k(0) 2

(0, e, 0),

d 5 25.

2(0) 2 d 1 k(0) 5 5.

k 5 215.

k 5 215,

5 10k 1 2

(2, 21, k) ? (k, 22, 8) 5 2k 2 1(22) 1 8k

k 5 4.

84 5 2k 5 4.

k 5 4.

k2

5 2221 5 2,

(k, 22, 8).

(2, 21, k)

70.5°.cos21( 21!3!3) 8 109.5°

0 (1, 1, 21) 0 5 "3.

(1, 1, 21) ? (1, 21, 1) 5 21

(1, 21, 1).

(1, 1, 21)

x4

5y 2 5

225 z 1 1

2.

(4, 22, 1)

(0, 5, 212).

z 5 212.(22) 1 4 5 5

y 5 (212)

y 2 4

225 z 5 0 2 2

4 5 212,x 5 0,

x 5 0.

3x 2 13y 1 2z 2 61 5 0.

D 5 261.61 1 D 5 0.

2(5) 1 D 5 0,3(4) 2 13(23) 1

(4, 23, 5)


Review of Prerequisite Skills,pp. 424–425

1. a.

b.

2. a. The points A, B, and C are collinear if andonly if the vectors and are collinear.

So and so and are collinear.b. The points J, K, and L are collinear if and only if the vectors and are collinear.

So and so and are collinear.c. The points A, B, and C are collinear if and only if the vectors and are collinear.

So and so and are collinear.d. The points R, S, and T are collinear if and only if the vectors and are collinear.

Since the ratios of the components are not equal,and are not collinear. So R, S, and T do not

lie on the same line.3. ABC is a right triangle if its sides,

and satisfy the Pythagorean theorem.

So

So

So

.Since ABC is a right triangle.4. The vectors and are perpendicular if

So if then 5. a. A vector, is perpendicular to if



b. A vector, is perpendicular to ifand and are not both zero.


c. A vector, is perpendicular to ifand are not all zero.

5 27(t1) 2 4(t2) 1 0(t3)

c>

? (t1, t2, t3) 5 (27, 24, 0) ? (t1, t2, t3)

t3t2,t1,c>

? (t1, t2, t3) 5 0

c>

(t1, t2, t3),

b>

.(5, 6)

a>

? (t1, t2) 5 0.t2 5 6,t1 5 5

5 6(t1) 2 5(t2)

b>

? (t1, t2) 5 (6, 25) ? (t1, t2)

t2t1b>

? (t1, t2) 5 0

b>

(t1, t2),

a>

.(3, 1)

a>

? (t1, t2) 5 0.t2 5 1,t1 5 3

5 1(t1) 2 3(t2)

a>

? (t1, t2) 5 (1, 23) ? (t1, t2)

t2t1a>

? (t1, t2) 5 0

a>

(t1, t2),

u>

? v>

5 0.t 5 18,

5 t 2 18

5 2t 1 (21)t 1 3(26)

u>

? v>

5 (t, 21, 3) ? (2, t, 26)

u>

? v>

5 0.

v>

u>

@AB>

@ 2 1 @BC>

@ 2 5 @AC>

@ 2,5 "14

@BC>

@ 5 "32 1 (22)2 1 (21)2

5 (3, 22, 21)

BC>

5 (5, 3, 2) 2 (2, 5, 3)

5 "41.

@AC>

@ 5 "42 1 (23)2 1 42

5 (4, 23, 4)

AC>

5 (5, 3, 2) 2 (1, 6, 22)

5 "27.

@AB>

@ 5 "12 1 (21)2 1 52

5 (1, 21, 5)

AB>

5 (2, 5, 3) 2 (1, 6, 22)

@BC>

@ ,@AC

>

@ ,@AB>

@ ,

RT>

RS>

5 (1, 2, 3)

RT>

5 (2, 4, 0) 2 (1, 2, 23)

5 (3, 21, 6)

RS>

5 (4, 1, 3) 2 (1, 2, 23)

RT>

RS>

AC>

AB>

AC>

5 2AB>

,

5 2(3, 5, 21)

5 (6, 10, 22)

AC>

5 (7, 12, 21) 2 (1, 2, 1)

5 (3, 5, 21)

AB>

5 (4, 7, 0) 2 (1, 2, 1)

AC>

AB>

JL>

JK>

JK>

5 12 JL

>

,

51

2(8, 2)

5 (4, 1)

JL>

5 (0, 4) 2 (24, 3)

5 (8, 2)

JK>

5 (4, 5) 2 (24, 3)

JL>

JK>

AC>

AB>

AC>

5 23AB>

,

5 23(3, 5)

5 (29, 215)

AC>

5 (28, 218) 2 (1, 23)

5 (3, 5)

AB>

5 (4, 2) 2 (1, 23)

AC>

AB>

5 (13, 212, 241)

5 (10, 215, 220) 1 (3, 3, 221)

1 (3 3 1, 3 3 1, 3 3 (27))

5 (5 3 2, 5 3 (23), 5 3 (24))

5(2, 23, 24) 1 3(1, 1, 27)

5 (2, 29, 6)

1 2 (25))5 (3 2 1, 22 2 7,

(3, 22, 1) 2 (1, 7, 25)

CHAPTER 8Equations of Lines and Planes


So if and thenSo is perpendicular

to 6. The area of a parallelogram formed by two vectors is determined by the magnitude of the crossproduct of the vectors.

7. a.

So is a vector perpendicular toboth vectors.b.

So is a vector perpendicular to bothvectors.8.

9. a.

b.

c.

d.

10. a.

b.

c.

d.

11. a. The y-intercept occurs when

So the y-intercept is The slope is equal to b.

So the y-intercept is The slope is equal to c.

So So the y-intercept is The slope is equal to d.

So So the y–intercept is 3. The slope is equal to a. –d.

12. Any positive scalar multiple of a vector is acollinear vector in the same direction. Answers mayvary. For example:a.b. 3(25, 4, 3) 5 (215, 12, 9)

2(4, 7) 5 (8, 14)

20

64y = –2x –5

4x – 8y = 8

5x = 5y – 15

3x – 5y + 1 = 0

2

–2–4–6

–4–8 –6–10 –2 x

y

55 5 1.

5y 5 15

5(0) 5 5y 2 15

5x 5 5y 2 15

35.

2125 5 1

5.

25y 5 21

3(0) 2 5y 1 1 5 0

3x 2 5y 1 1 5 0

48 5 1

2.21.

5 21

y 58

28

4(0) 2 8y 5 8

4x 2 8y 5 8

22.25.

5 25

5 22(0) 2 5

y 5 22x 2 5

x 5 0.

5 (4, 25, 24)

p>

5 (4, 0, 24) 2 (0, 5, 0)

5 (22, 8, 25)

p>

5 (1, 2, 4) 2 (3, 26, 9)

5 (210, 214)

p>

5 (27, 26) 2 (3, 8)

5 (7, 3)

p>

5 (4, 8) 2 (23, 5)

5 (24, 5, 4)

p>

5 (0, 5, 0) 2 (4, 0, 24)

5 (2, 28, 5)

p>

5 (3, 26, 9) 2 (1, 2, 4)

5 (10, 14)

p>

5 (3, 8) 2 (27, 26)

5 (27, 23)

p>

5 (23, 5) 2 (4, 8)

y

z

x

A

B

C D

(0, 0, 23)

5 0

1 (0)(23)5 (22)(0) 1 (21)(0)

b>

? (222, 28, 213)

5 0

a>

? (0, 0, 23) 5 (21)(0) 1 (22)(0) 1 (0)(23)

5 (0, 0, 23)

2 (21)(0), (21)(21) 2 (22)(22))

a>

3 b>

5 ((22)(0) 2 (0)(21), (0)(22)

(222, 28, 213)

5 0

5 266 1 40 1 26

1 (22)(213)

b>

? (222, 28, 213) 5 (3)(222) 1 (25)(28)

5 0

5 244 2 8 1 52

1 (24)(213)

a>

? (222, 28, 213) 5 (2)(222) 1 (1)(28)

5 (222, 28, 213)

2 (2)(22), (2)(25) 2 (1)(3))

a>

3 b>

5 ((1)(2) 2 (24)(25), (24)(3)

5 "2802

5 "(229)2 1 352 1 (226)2

5 0 (229, 35, 226) 0 A 5 area of parallelogram

5 (229, 35, 226)

(4)(1) 2 (10)(3))

(9)(3) 2 (4)(2),5 ((10)(22) 2 (9)(1),

(4, 10, 9) 3 (3, 1, 22)

c>

.

(24, 7, 0)c>

? (t1, t2, t3) 5 0.

t3 5 0,t2 5 7,t1 5 24,


c.

d.13. To simplify can be written in algebraic notation. So a.

b.

So

c.

d.

e. is merely the negative of So

f.

14. The dot product of two vectors yields a realnumber, while the cross product of two vectorsgives another vector.

8.1 Vector and ParametricEquations of a Line in pp. 433–434

1. Direction vectors for a line are unique only up toscalar multiplication. So since each of the givenvectors is just a scalar multiple of each is anacceptable direction vectors for the line.

2. a. Simply find x and y coordinates for three values of t. Three possible values are and At and

At and At and So (1, 5), and

are three points on the line.b. Find the t value when the y-coordinate is 15. Sosolve for t.

If the Sois a point on the line.

3. Answers may vary. For example:a. is a point on the line and is adirection vector for the line.b. is a point on the line and is adirection vector for the line.c. is a point on the line and (0, 2) is a directionvector for the line.d. is a point on the line and is adirection vector for the line.4. Answers may vary. For example: One possibleline has A(2, 1) as its origin point and as itsdirection vector, while another has as itsorigin point and as its direction vector.

So the first case is .

The second case is 5. a. Find the t value when the y-coordinate is 18.So solve for t.

If the So is apoint on the line.b. Answers may vary. For example: A directionalvector for the line is Since is apoint on the line, a possible vector equation is

.c. Answers may vary. For example: We may take

to find another point on the line. Soand Hence

is a point on the line. So another vectorequation is .6. Answers may vary. For example:a. Three different s values will yield three different points on the line. If thens 5 21,

tPRr>

5 (22, 4) 1 t(21, 2),

(22, 4)

y 5 4 1 2(0) 5 4.x 5 22 2 0 5 22

t 5 0

tPRr>

5 (29, 18) 1 t(21, 2),

R(29, 18)(21, 2).

R(29, 18)x 5 22 2 7 5 29.t 5 7,

t 5 7

2t 5 14

18 5 4 1 2t

sPR.q>

5 (23, 5) 1 s(5, 24),

BA>

5 (2, 1) 2 (23, 5) 5 (5, 24)

tPRr>

5 (2, 1) 1 t(25, 4),

AB>

5 (23, 5) 2 (2, 1) 5 (25, 4)

BA>

B(23, 5)

AB>

(25, 0)(0, 6)

(4, 1)

(2, 27)(1, 3)

(2, 1)(3, 4)

P(214, 15)

x 5 1 1 3(25) 5 214.t 5 25,

t 5 25

22t 5 10

15 5 5 2 2t

(4, 3)

(22, 7),y 5 5 2 2(1) 5 3.

x 5 1 1 3(1) 5 4t 5 1,y 5 5 2 2(0) 5 5.

x 5 1 1 3(0) 5 1t 5 0,y 5 5 2 2(21) 5 7.

x 5 1 1 3(21) 5 22t 5 21,t 5 1.

t 5 0,t 5 21,

(13,

16)

R2,

5 (55, 40, 2140)

2 (220)(24)) (12)(25)

(21)(24) 2 (12)(23),

2 (21)(25), (2u>

1 v>

) 3 (u>

2 2v>

) 5 ((220)(23)

5 (24, 25, 23)

5 (4, 29, 21) 2 (8, 24, 2)

u>

2 2v>

5 (4, 29, 21) 2 2(4, 22, 1)

5 (12, 220, 21)

5 (8, 218, 22) 1 (4, 22, 1)

2u>

1 v>

5 2(4, 29, 21) 1 (4, 22, 1)

5 (11, 8, 228)

v>

3 u>

5 2 (211, 28, 28)

u>

3 v>

.v>

3 u>

5 (211, 28, 28)

2 (4)(1), (4)(22) 2 (29)(4))

u>

3 v>

5 ((29)(1) 2 (21)(22), (21)(4)

5 77

1 (0)(22)

(u>

1 v>

) ? (u>

2 v>

) 5 (8)(0) 1 (211)(27)

5 (0, 27, 22)

u>

2 v>

5 (4, 29, 21) 2 (4, 22, 1)

5 (8, 211, 0)

u>

1 v>

5 (4, 29, 21) 1 (4, 22, 1)

5 233

5 216 2 18 1 1

2v>

? u>

5 (24)(4) 1 (2)(29) 1 (21)(21)

5 (24, 2, 21)

2v>

5 21(4, 22, 1)

5 33

5 16 1 18 2 1

u>

? v>

5 (4)(4) 1 (29)(22) 1 (21)(1)

v>

5 (4, 22, 1)

v>

4(25i>

1 8j>

1 2k>

) 5 220i>

1 32j>

1 8k>

1

2(2i

>

1 6j>

2 4k>

) 5 i>

1 3j>

2 2k>


If then and if then Hence

and are three points on the line.

b. is a line that passes through theorigin different from the line in part a.c. If then So

, is a line that passesthrough the origin with a direction vector of Hence this describes the same line as part a.7. One can multiply a direction vector by a constantto keep the same line, but multiplying the pointyields a different line.8. a.

b. is a possible direction vector for this lineand Q is a point on the line.


parametric equation is .9. a.

b. is a possible direction vector for this lineand M is a point on the line.


parametric equation is .10. a. A line perpendicular to L would have adirection vector that is perpendicular to the direction vector of L. is perpendicular to if


So if and then So an equation for a line with as a directionvector and P(2, 0) as a line is

.b. The line intersects the y-axis when the x coordinateis zero. The x coordinate is zero, when or

The y coordinate at this point is orSo the line intersects the y-axis at the point

11. The line crosses the x-axis, when soor So the x coordinate at this

point is The line crossesthe y-axis, when so or

So the y coordinate at this point isSo the triangle formed by the

origin, A(6, 0), and B(0, 3) is a right triangle with abase of six units and a height of three units. So thearea is 12. First all the relevant vectors are found.

a.

b.

c.

13. a. Find the t values such that x and y coordinatessatisfy or similarly

So when or LetA be the point where So x coordinate of Ais and the y coordinate is Let B be the point where So x coordinateof B is and the y coordinate is

So A is and B is

b. and

hence the length of

or about 24.045 "578,

5 "(217)2 1 (217)2

AB 5 @AB>

@AB

>

5 (212, 25) 2 (5, 12) 5 (217, 217)

(212, 25)(5, 12)9 2 14 5 25.

2 2 14 5 212,

t 5 214.

9 1 3 5 12.2 1 3 5 5,

t 5 3.

t 5 214.t 5 3x2 1 y2 2 169 5 0,

5 (t 2 3)(t 1 14)

5 t2 1 11t 2 42

1 t2 2 169

5 4 1 4t 1 t2 1 81 1 18t x2 1 y2 2 169 5 (2 1 t)2 1 (9 1 t)2 2 169

x2 1 y2 2 169 5 0.x2 1 y2 5 169

AC>

5 (24, 6) 52

3(26, 9) 5

2

3 AD

>

AD>

5 (26, 9) 5 3(22, 3) 5 3AB>

AC>

5 (24, 6) 5 2(22, 3) 5 2AB>

5 (26, 9)

AD>

5 ((1, 2) 1 3(22, 3)) 2 ((1, 2) 1 0(22, 3))

5 (24, 6)

AC>

5 ((1, 2) 1 2(22, 3)) 2 ((1, 2) 1 0(22, 3))

5 (22, 3)

AB>

5 ((1, 2) 1 1(22, 3)) 2 ((1, 2) 1 0(22, 3))

12(3)(6) 5 9.

y 5 8 1 (25) 5 3.

s 5 25.

210 2 2s 5 0,x 5 0,

x 5 210 2 2(28) 5 6.

s 5 28.8 1 s 5 0,

y 5 0,

(0, 21.2).

y 5 21.2.

0 1 3tt 5 20.4.

2 1 5t 5 0,

tPRr>

5 (2, 0) 1 t(5, 23),

(5, 23)

(3, 5) ? (t1, t2) 5 0.t2 5 23,t1 5 5

(3, 5) ? (t1, t2) 5 3(t1) 1 5(t2)

t2t1(3, 5) ? (t1, t2) 5 0

(3, 5)(t1, t2)

tPRy 5 5,x 5 4 1 5t,tPRr

>

5 (4, 5) 1 t(5, 0),

MN>

5 (9, 5) 2 (4, 5) 5 (5, 0)

(4, 5)

MN>

–6 6 80

42

–2

N(9, 5)

M(4, 5)

68

10y

x

2 4–4 –2

tPRy 5 7 1 2t,x 5 0,

tPRr>

5 (0, 7) 1 t(0, 2),

QR>

5 (0, 9) 2 (0, 7) 5 (0, 2)

(0, 7)

QR>

–6 60

42

–2

Q(0, 7)R(0, 9)

68

10y

x

2 4–4 –2

(3, 4).

tPRr>

5 (9, 12) 1 t(3, 4),

(9, 12) 1 t(3, 4) 5 (0, 0).t 5 23,

tPRr>

5 t(1, 1),

(3, 4)(0, 0),

(23, 24),s(3, 4) 5 (3, 4).s 5 1,

s(3, 4) 5 (0, 0)s 5 0,s(3, 4) 5 (23, 24).


14. In the parametric form, the second equationbecomes . If t is solved

for in this equation, we obtain and

Setting these two expressions equal to each other,

the line is described by or by

simplifying, So the second

equation describes a line with a slope of . If y is

solved for in the first expression, we see thatis on the second line but not the

first. Hence both equations are lines with slope of with no point in common and must be parallel.

8.2 Cartesian Equation of a Line,pp. 443–444

1. a. is a direction vector parallel tothe line.b. For a vector perpendicular to the line, a suitable

has to be found, such that is a such a vector.c. If then so (0, 9) is a point on thegiven line.d. A direction vector was found in part a., so a vector equation for a parallel line passing through A is . The corresponding parametric equation is

.e. A direction vector was found in part b., so a vector equation for a perpendicular line passingthrough is .The corresponding parametric equation is

.2. a.-b.

c. Switching the components of the direction vector with the coordinates of the point on the lineproduces a different line.

3. a. A direction vector parallel to the line is and if then So is a point onthe line. So a vector equation for the line is

. The correspondingparametric equation is .b. A direction vector parallel to the line is (2, 3),and if then So (0, 5) is a point on theline. So a vector equation for the line is

. The correspondingparametric equation is .c. The equation describes a horizontal linein the xy-plane, so a direction vector parallel to this line is Also is a point on this line, so a vector equation for the line is

, which gives a parametric equation of .d. The equation describes a vertical line inthe xy-plane, so a direction vector parallel to thisline is Also is a point on this line, so avector equation for the line is

, which gives a parametric equation of .

4. If the two lines have direction vectors that arecollinear and share a point in common, then the twolines are coincident. In this example, both have

as a parallel direction vector and both haveas a point on the line. Hence the two lines

are coincident.5. a. The normal vectors for the lines are and which are collinear. Since in twodimensions, any two direction vector perpendicularto are collinear, the lines have collineardirection vectors. Hence the lines are parallel.b. The lines will be coincident if they share a common point. is a point in the first line. So the lines are coincident if and only if

or equivalently So only if are the lines coincident.6. Since the normal vector is the Cartesianequation of the line is for someconstant k. Since is a point on the graph,

So So the equation of the line is 7. So the slope of this line is equal to Hence the equation for the line satisfies

or by multiplying both sides by y 2 5

x 2 (23)5 21,

4 2 522 2 (23) 5 21.

4x 1 5y 2 21 5 0.

k 5 4 2 25 5 221.

4(21) 1 5(5) 1 k 5 0.

A(21, 5)

4x 1 5y 1 k 5 0,

(4, 5),

k 5 12,

k 5 12.4(0) 2 6(2) 1 k 5 0,

(0, 2)

(2, 23)

(4, 26),

(2, 23)

(24, 0)

(3, 2)

tPRy 5 t,x 5 4,tPR

r>

5 (4, 0) 1 t(0, 1),

(4, 0)(0, 1).

x 5 4

tPRy 5 21,x 5 t,tPRr

>

5 (0, 21) 1 t(1, 0),

(0, 21)(1, 0).

y 5 21


>

5 (0, 5) 1 t(2, 3),

y 5 5.x 5 0,


>

5 (0, 26) 1 t(8, 7),

(0, 26)y 5 26.x 5 0,

(8, 7),

–6 60

42

–2

68

10y

x

2 4–4 –2

rq

tPRy 5 1 1 6t,x 5 22 1 5t,

tPRr>

5 (22, 1) 1 t(5, 6),B(22, 1)

tPRy 5 9 2 5t,x 5 7 1 6t,

tPRr>

5 (7, 9) 1 t(6, 25),(7, 9)

y 5 9,x 5 0,

n>

5 (5, 6)m>

? n>

5 0.n>

m>

5 (6, 25)

23

(1, 6)y 5 23x 1 5.

23

y 2 6 5 23x 2 2

3.

x 2 1

65 y 2 6

4,

t 5y 2 6

4.t 5

x 2 1

6

tPRy 5 6 1 4t,x 5 1 1 6t,


Moving everythingthe left hand side yields or

which is the equation in Cartesianform.8. So the directional vector of the line is collinearwith the normal vector and so has slopeequal to Furthermore is a point on theline. Hence the equation for the line satisfies

or by multiplying both sides by

Moving everything to the left side yields or

which is the equation inCartesian form.9. a.

b. First solve for t in both coordinates. So

and Then set these two sides equal to

each other to obtain or simply

So or

10. The acute angle of the intersection between twovectors and is found by taking the inverse

cosine of the absolute value of

a.and So the acute angle is

b. and


c. The direction vector for the first line is anda direction vector for the second is

and


d. A direction vector for the second line is and


e.

and So the acute angle is

f. has a direction vector of and thedirection vector for the second line is

and


11. a.

b. The normal vectors are and (1, 2).

and


and the obtuse angle is

12. a. Let the coordinates of C be (x, y). They mustsatisfy the equation Rewrite this equation in Cartesian form. The slopeis . The equation is of the form

. Substitute into the equation to solve for b.

The equation of the line is .If C is the vertex of the right triangle, and must be perpendicular, meaning that their dot product must be 0.

So Substitute for y.24

3 x 2 2

224 2 5x 1 x2 1 8 2 6y 1 y2 5 0.

(23 2 x)(8 2 x) 1 (2 2 y)(4 2 y) 5 0

CA>

? CB>

5 (23 2 x)(8 2 x) 1 (2 2 y)(4 2 y)

CB>

5 (8 2 x, 4 2 y)

CA>

5 (23 2 x, 2 2 y)

CB>

CA>

y 5 243 x 2 2

22 5 b 6 5 8 1 b

6 5 24

3(26) 1 b

(26, 6)y 5 243 x 1 b

m 5 243

(x, y) 5 (26, 6) 1 t(3, 24).

180°245° 5 135°.

cos21( 5!10!5) 5 45°

0 (1, 2) 0 5 "5.

0 (1, 23) 0 5 "10,(1, 23) ? (1, 2) 5 25,

(1, 23)

4 60

642

–2 2–4–6 –2

x

y

cos21( 1!1!5) 8 63°.

0 (2, 1) 0 5 "5.

0 (0, 1) 0 5 "1,(0, 1) ? (2, 1) 5 1,

(2, 1).

(0, 1)x 5 3

cos21( 13!29!17

) 8 54°.

0 (24, 1) 0 5 "17.

0 (2, 25) 0 5 "29,(2, 25) ? (24, 1) 5 213,

cos21( 8!20!5) 8 37°.

0 (2, 1) 0 5 "5.

0 (2, 4) 0 5 "20,(2, 4) ? (2, 1) 5 8,

(2, 1).

cos21( 5!25!5) 8 63°.

0 (4, 23) 0 5 "25.

0 (2, 1) 0 5 "5,(2, 1) ? (4, 23) 5 5,

(4, 23).

(2, 1)

cos21( 29!41!37) 8 42°.

0 (1, 26) 0 5 "37.

0 (25, 4) 0 5 "41,(25, 4) ? (1, 26) 5 229,

cos21( 3!29!17) 8 82°.

0 (24, 21) 0 5 "17.

0 (2, 25) 0 5 "29,(2, 25) ? (24, 21) 5 23,

a>

? b>

@a>

@ @b>

@.

b>

a>

4x 2 y 2 14 5 0.

212 1 4x 5 y 1 224(3 2 x) 5 y 1 2.

3 2 x 5y 1 2

24,

t 5y 1 2

24.

t 5 3 2 x

4 60

642

–2–4–6

2–4–6 –2

x

y

2x 1 y 2 16 5 0,

y 2 2 1 2x 2 14 5 0,

y 2 2 5 22(x 2 7).

x 2 7,y 2 2

x 2 75 22,

P(7, 2)22.

(2, 24),

x 1 y 2 2 5 0,

y 2 5 1 x 1 3 5 0,

y 2 5 5 21(x 1 3).x 2 (23),


So When When But then C would have thesame coordinates as A. This would not produce aright triangle. So the coordinates of C are .b.

c.

Since the dot product of the vectors is 0, the vectorsare perpendicular, and 13. The sum of the interior angles of a quadrilateralis 360°. The normals make 90° angles with theirrespective lines at A and C. The angle of thequadrilateral at B is 180° . Let x represent themeasure of the interior angle of the quadrilateralat O.

Therefore, the angle between the normals is thesame as the angle between the lines.

14. The normal vector for the first line is and (1, k) for the second.

and

So and We

obtain after squaring both sides that

So or simplySolving by the quadratic equation

gives

8.3 Vector, Parametric, and SymmetricEquations of a Line in pp. 449–450

1. a. A point on this line is b. A point on this line is c. A point on this line is d. A point on this line is e. A point on this line is

f. A point on this line is 2. a. A direction vector is b. A direction vector is c. A direction vector is d. A direction vector is e. A direction vector is (0, 0, 2).

f. A direction vector is which if multipliedby the least common denominator, 4, yields a vectorof 3. a.is a direction vector, as well as

So, is one possible

vector equation is another.

b. The parametric equation corresponding with thefirst vector equation is

. The second parametric equation is

4. a.So is a direction vector for the equation,and so may be used as the direction vector.Hence , is a vectorequation for a line containing the points

and b. The corresponding parametric equation is

.c. Since two of the coordinates in the directionvector are zero, a symmetric equation cannot exist.

tPRz 5 24,y 5 5,x 5 21 1 t,

B(2, 5, 24).A(A21, 5, 24)

tPRr>

5 (21, 5, 4) 1 t(1, 0, 0),

(1, 0, 0)

(3, 0, 0)

AB>

5 (2, 5, 24) 2 (21, 5, 24) 5 (3, 0, 0).

sPR.z 5 5 2 s,y 5 23 1 5s,x 5 3 2 4s,

tPRz 5 4 1 t,y 5 2 2 5t,x 5 21 1 4t,

sPRq>

5 (3, 23, 5) 1 s(24, 5, 21),

tPRt(4, 25, 1),r>

5 (21, 2, 4) 1

(3, 23, 5) 5 (24, 5, 21).BA>

5 (21, 2, 4) 2

AB>

5 (3, 23, 5) 2 (21, 2, 4) 5 (4, 25, 1)

(2, 21, 2).

(12, 2

14,

12),

(21, 0, 2).

(3, 24, 21).

(2, 1, 21).

(21, 1, 9).

(13, 2

34,

25).

(3, 22, 21).

(22, 23, 1).

(22, 1, 3).

(1, 21, 3).

(23, 1, 8).

R3,

k 5 2 6 "3.

k2 2 4k 1 1 5 0.

2 2 4k 1 2k2 5 1 1 k2

1 2 2k 1 k2

2(1 1 k2)5 1

4.

cos (60°) 5 0.5.cos (60°) 51 2 k

!2!1 1 k2

0 (1, k) 0 5 "1 1 k2.0 (1, 21) 0 5 "2,

(1, 21) ? (1, k) 5 1 2 k,

(1, 21)

x 5 u

360° 2 u 1 x 5 360°

90° 1 90° 1 180° 2 u 1 x 5 360°

2 u

/ACB 5 90°.

CA>

? CB>

5 (23)(8) 1 (4)(6)

5 224 1 24

5 0

5 (8, 6)

CB>

5 (8 2 0, 4 2 (22))

5 (23, 4)

CA>

5 (23 2 0, 2 2 (22))

60

42

–2–4

C(0, –2)

A(–3, 2)

68

10y

x

2 4–2

B(8, 4)

(0, 22)

x 5 23, y 5 2.

x 5 0, y 5 22.

x 5 0 or x 5 23.

224 2 5x 1 x2 1 8 2 6y 1 y2 5 0

216 2 5x 1 x2 2 6a24

3x 2 2b 1 a2

4

3x 2 2b

2

5 0

216 2 5x 1 x2 1 8x 1 12 116

9x2 1

16

3x 1 4 5 0

25

9x2 1 3x 1

16

3x 5 0

25x2 1 75x 5 0

25x(x 1 3) 5 0


5. a. So is avector equation for the line and the correspondingparametric equation is


b.is a direction vector for the line. So

is a vector equation for the line and the corresponding parametricequation is . So the


c.is a direction vector for the line. Since

is also a direction vector for this line. So

is a vector equationfor the line and the corresponding parametric equationis . So the


d.is a direction vector for the line. So

is a vector equationfor the line and the corresponding parametric equationis . Since two of thecoordinates in the direction vector are zero, there isno symmetric equation for this line.e. is adirection vector for the line. So

is a vector equation for the line and thecorresponding parametric equation is


f. The direction vector for the z-axis is soa line parallel to the z-axis has (0, 0, 1) as a directionvector. So is a vectorequation for the line and the corresponding parametricequation is . Since twoof the coordinates in the direction vector are zero,there is no symmetric equation for this line.

6. a. So the first line is given by

. If x, y, and z are

solved for in terms of t, the corresponding parametricequations is

. So the first line has a direction vector ofThe second line is given by

If x and y are solved

for in terms of s, and areobtained. So the parametric equation for the secondline is andso has a direction vector of b.

and So the angle between the

two lines is

7. The directional vector of the first line isSo is a

directional vector for the first line as well. Sinceis also the directional vector of the

second line, the lines are the same if the lines sharea point. is a point on the second line. Since

is a point on thefirst line as well. Hence the lines are the same.8. a. The line that passes through with adirectional vector of is given by theparametric equation is

. So the y coordinate is equal to only whenAt and

So is a pointon the line. So the y coordinate is equal to 5 onlywhen At and

So is apoint on the line.b. Since the point A occurs when and pointB occurs when the line segment connectingthe two points is precisely all the t values between

and 5. So the equation is

9. The direction vector for the first line isand for the second line is

The lines are perpendicular precisely whenSo

So ifthen and

the lines are perpendicular.10. a. Three different points occur at three differentvalues of t. At the corresponding point onthe line is At the corresponding point on the line ist 5 1,

(4, 22, 5) 2 (24, 26, 8) 5 (8, 4, 23).

t 5 21,

(k, 2, k 2 1) ? (22, 0, 1) 5 0,k 5 21,

1(k 2 1) 5 2k 2 1.5 22(k) 1 0(2) 1

(k, 2, k 2 1) ? (22, 0, 1)

(k, 2, k 2 1) ? (22, 0, 1) 5 0.

(22, 0, 1).(k, 2, k 2 1)

22 # t # 5.z 5 3 2 6t,y 5 t,x 5 23t,22

t 5 5,

t 5 22,

B(215, 5, 227)z 5 3 2 6(5) 5 227.

x 5 23(5) 5 215t 5 5,t 5 5.

A(6, 22, 15)z 5 3 2 6(22) 5 15.

x 5 23(22) 5 6t 5 22,t 5 22.

22tPRz 5 3 2 6t,y 5 t,x 5 23t,

(23, 1, 26)

(0, 0, 3)

(1, 1, 3)1 5 1 1 78 5 1 1 1

2 5 3 2 522 ,

(1, 1, 3)

(24, 21, 1)

(24, 21, 1)(8, 2, 22) 5 22(24, 21, 1).

cos21( 2!2!3

) 8 35.3°.

0 (1, 21, 1) 0 5 "3.

0 (1, 21, 0) 0 5 "20(1) 5 2.5 1(1) 2 1(21) 1

(1, 21, 1) ? (1, 21, 0)

(1, 21, 0).

sPR,z 5 5,y 5 11 2 s,x 5 27 1 s,

y 5 11 2 sx 5 27 1 s,

x 1 7

15

y 2 11

21(5s), z 5 5.

(1, 21, 1).

tPR

z 5 7 1 t,y 5 10 2 t,x 5 26 1 t,

x 1 6

15 y 2 10

215 z 2 7

1(5t)

tPRz 5 4 1 t,y 5 2,x 5 1,

tPRr>

5 (1, 2, 4) 1 t(0, 0, 1),

(0, 0, 1),

x24

5y3, z 5 0.

tPRz 5 0,y 5 3t,x 5 24t,

tPRr>

5 t(24, 3, 0),

XO>

5 (24, 3, 0) 2 (0, 0, 0) 5 (24, 3, 0)

tPRz 5 0,y 5 t,x 5 21,

tPRt(0, 1, 0),r>

5 (21, 0, 0) 1

DE>

5 (21, 1, 0) 2 (21, 0, 0) 5 (0, 1, 0)

x 5 22.y 2 3

15

z1,

tPRz 5 t,y 5 3 1 t,x 5 22,

tPRt(0, 1, 1),r>

5 (22, 3, 0) 1

(0, 1, 1)(0, 26, 26) 5 26(0, 1, 1),

MN>

5 (22, 4, 7) 2 (22, 22, 1) 5 (0, 26, 26)

x 5 21.y 2 1

15

z1,

tPRz 5 t,y 5 1 1 t,x 5 21,

tPRt(0, 1, 1),r>

5 (21, 1, 0) 1

AB>

5 (21, 2, 1) 2 (21, 1, 0) 5 (0, 1, 1)

x 1 1

35

y 2 2

225

z 2 1

1.

tPRz 5 1 1 t,y 5 2 2 2t,x 5 21 1 3t,

tPR,r>

5 (21, 2, 1) 1 t(3, 22, 1),



The pointat the origin is b. Three different points occur at three different values of s. At the corresponding point onthe line is when

and Atthe corresponding point on the line is when

andSo and (1, 1, 3) are

two points on the line. The point at the origin is

c. is actually equal to

for any . So we can

pick different t values to obtain different points onthe lines. At the corresponding point on theline is found by solving for x, y, and z, in the

equation So

and So is a point onthe line. At and solving for x, y, and z, in the

equation yields

andSo (2, 1, 4) is a point on the line.

Also the point at the origin is


for any . So we

can pick different t values to obtain different pointson the lines. At the corresponding point onthe line is found by solving for x, y, and z, in the

equation So

andSo is a point

on the line. At and solving for x, y, and z, in

the equation yields

and So is a point on the

line. Also the point at the origin is 11. For part a. the corresponding parametricequation is

. The corresponding symmetric equation isx 2 4

245

y 1 2

265

z 2 5

8.

tPRz 5 5 1 8t,y 5 22 2 6t,x 5 4 2 4t,

(24, 2, 3).

(24, 5, 8)z 5 (1)5 1 3 5 8.

y 5 (1)(3) 1 2 5 5,x 5 24,

y 2 2

35

z 2 3

5(51),x 5 24,

t 5 1

(24, 21, 22)z 5 (21)5 1 3 5 22.

y 5 (21)(3) 1 2 5 21,x 5 24,

y 2 2

35 z 2 3

5(5 21).x 5 24,

t 5 21,

tPRy 2 2

35

z 2 3

5(5t),x 5 24,

y 2 2

35 z 2 3

5x 5 24,

(21, 2, 0).

z 5 (1)4 5 4.

y 5 (1)(21) 1 2 5 1,x 5 (1)3 2 1 5 2,

x 1 1

35

y 2 2

215

z4

(51),

t 5 1

(24, 3, 24)z 5 (21)4 5 24.

y 5 (21)(21) 1 2 5 3,x 5 (21)3 2 1 5 24,

x 1 1

35

y 2 2

215

z4

(5 21).

t 5 21,

tPRx 1 1

35 y 2 2

215 z

4( 5 t),

x 1 1

35 y 2 2

215 z

4

(24, 2, 9).

(29, 3, 15)z 5 9 2 6(1) 5 3.

y 5 2 2 (1) 5 1,x 5 24 1 5(1) 5 1,

s 5 1,

z 5 9 2 6(21) 5 15.y 5 2 2 (21) 5 3,

x 5 24 1 5(21) 5 29,

s 5 21,

(4, 22, 5).

(4, 22, 5) 1 (24, 26, 8) 5 (0, 28, 13). For part b. the corresponding vector equation isThe

corresponding symmetric equation is

For part c. the point at the origin is and thedirection vector is So the correspondingvector equation is ,and parametric equation

For part d. the point at the origin is and adirection vector is So the correspondingvector equation is ,and parametric equation

.12. The direction vector of the first line is

and the direction vector of the secondline is The cross product of these two vectors gives a vector that is perpendicular to bothdirection vectors.

So a line with a direction vector of isperpendicular to the two initial lines. A parametricequation of such a line passing through the point

is .

13. Since and ifthen

or equivalently

So if then or Also if or then So the only two points occur at and

At and or

At and or

14. Let andbe two such points

for some real numbers s and t. So isperpendicular to the lines and and so sinceL2,L1

P1P2

>

P2(22 1 3s, 27 1 2s, 2 2 3s)

P1(4 1 2t, 4 1 t, 23 2 t)(2, 1, 2).z 5 2,y 5 5 1 (24) 5 1,

x 5 10 1 2(24) 5 2,s 5 24,

(22, 21, 2).z 5 2,y 5 5 1 (26) 5 21,

x 5 10 1 2(26) 5 22,s 5 26,s 5 24.

s 5 26

x2 1 y2 1 z2 5 9.s 5 24,s 5 26

s 5 24.s 5 26x2 1 y2 1 z2 5 9,

5 5(s 1 6)(s 1 4).

5 5s2 1 50s 1 120

(10 1 2s)2 1 (5 1 s)2 1 (2)2 2 9

(2)2 2 9 5 0.

(10 1 2s)2 1 (5 1 s)2 1(2)2 5 9

(10 1 2s)2 1 (5 1 s)2 1x2 1 y2 1 z2 5 9,

z 5 2,y 5 5 1 s,x 5 10 1 2s,

tPRz 5 13t,y 5 25 1 25t,x 5 2 2 34t,(2, 25, 0)

(234, 25, 13)

5 (234, 25, 13)

5 (228 2 6, 9 1 16, 28 1 21)

5 ((27)4 2 (3)2, (3)3 2 (24)4, (24)2 2 (27)3)

(24, 27, 3) 3 (3, 2, 4)

(3, 2, 4).

(24, 27, 3)

tPRz 5 3 1 5t,y 5 2 1 3t,x 5 24,

tPRr>

5 (24, 2, 3) 1 t(0, 3, 5),

(0, 3, 5).

(24, 2, 3)

tPR.z 5 4t,y 5 2 2 t,x 5 21 1 3t,

tPRr>

5 (21, 2, 0) 1 t(3, 21, 4)

(3, 21, 4).

(21, 2, 0)

x 1 4

55

y 2 2

215

z 2 9

26.

sPR.r>

5 (24, 2, 9) 1 s(5, 21, 26),

the direction vectors for the lines are andrespectively, and

So

So

So

Yet

So or Since

or So At

and At

and So andare the points that work.

15. The direction vector for the first line is and the direction vector for the second line is

and Sothe angle between the two lines is

Chapter 8 Mid-Chapter Review,pp. 451–452

1. a. Any three different t values yield three differentpoints. At

At and at So and

are three points on the line.b. Pick any three s values. At

Atand at So

(2, 3), and (5, 1) are three points on the line.c. Pick three different x values and solve for y toobtain the three points. At

or So when

Similarly at orAt or

So three points on the line are and (1, 1).


for any . So we can pick different t values to obtain different pointson the lines. At the corresponding point on the line is found by solving for x, y, and z, in the

equation So and

So is a point onthe line. At and solving for x, y, and z, in the

equation yields and

So (4, 0, 6) is another point onthe line. Also the point at the origin is 2. a. The x-intercept occurs when so solvefor the t values when to find the point. At

so So

So the x-intercept is at

The y-intercept occurs when or

So at the y-intercept, SoSo the y-intercept is at (0, 6).

b. The x-intercept occurs when so solve forthe s values when to find the point. At

so So

So the x-intercept is at The y-intercept occurs when or So at the y-intercept, So So the y-intercept is at 3. The direction vector for the first line is and the direction vector for the second is (2, 1).

and

So the angle between the lines is

The acute angle between the lines is approximately 4. The direction vector for the x-axis is (1, 0) and thedirection vector for the y-axis is (0, 1). The directionvector of the line is

and So the angle the line makes with the x-axis is

(0, 1) 5 25,(4, 25) ?( 41!41) 8 51°.cos21

0 (1, 0) 0 5 "1 5 1.0 (4, 25) 0 5 "41,

(4, 25) ? (1, 0) 5 4,(4, 25).

180° 2 93.2° 5 86.8°.

cos21( 21!5!65) 8 93.2°.

0 (2, 1) 0 5 "5.

0 (24, 7) 0 5 "65,(24, 7) ? (2, 1) 5 21,

(24, 7)

(0, 23).

y 5 3 2 2(3) 5 23.t 5 3.

26 1 2s 5 0.x 5 0,

(2143 , 0).

x 5 26 1 2(23) 5 214

3 .s 5 23.3 2 2s 5 0,

y 5 0,y 5 0,

y 5 0,

y 5 1 1 5(1) 5 6.

t 5 1.3 2 3t 5 0.

x 5 0,(185 , 0).

(185 ).x 5 3 2 3(21

5 ) 5

t 5 215 .1 1 5t 5 0,y 5 0,

y 5 0,

y 5 0,

(1, 22, 5).

1 5 5 6.z 5 (1)1

y 5 (1)2 2 2 5 0,x 5 (1)3 1 1 5 4,

x 2 1

35

y 1 2

25

z 2 5

1(51),

t 5 1

(22, 24, 4)z 5 (21)1 1 5 5 4.

y 5 (21)2 2 2 5 24,x 5 (21)3 1 1 5 22,

x 2 1

35

y 1 2

25

z 2 5

1(5 21).

t 5 21,

tPRx 2 1

35

y 1 2

25

z 2 5

1(5 t),

x 2 1

35

y 1 2

25

z 2 5

1

(0, 85),(21, 115 ),

y 5 55 5 1.3(1) 1 5y 2 8 5 0,x 5 1,y 5 8

5.

3(0) 1 5y 2 8 5 0,x 5 0,x 5 21.

y 5 115 ,5y 5 11.8 5 0,3(21) 1 5y 2

x 5 21,

(21, 5),

(1)(3, 22) 5 (5, 1).(2, 3) 1s 5 1,

(0)(3, 22) 5 (2, 3),(2, 3) 1s 5 0,

(21)(3, 22) 5 (21, 5).(2, 3) 1s 5 21,

(23, 4)

(25, 1),(27, 22),y 5 3(1) 1 1 5 4.

x 5 2(1) 2 5 5 23,t 5 1,y 5 3(0) 1 1 5 1,

x 5 2(0) 2 5 5 25,t 5 0,(21) 1 1 5 22.

y 5 3x 5 2(21) 2 5 5 27,t 5 21,

cos21( 8!5!14

) 8 17°.

0 (3, 2, 1) 0 5 "14.0 (2, 1, 0) 0 5 "55 8.

(2, 1, 0) ? (3, 2, 1) 5 2(3) 1 1(2) 1 0(1)(3, 2, 1).

(2, 1, 0)

P2(4, 23, 24)

P1(2, 3, 22)z 5 2 2 3(2) 5 24.

y 5 27 1 2(2) 5 23,x 5 22 1 3(2) 5 4,s 5 2,

z 5 23 2 (21) 5 22.y 5 4 1 (21) 5 3,

x 5 4 1 2(21) 5 2,t 5 21,

s 5 2.11s 5 22.228 1 11s 2 6(21) 5 0,

228 1 11s 2 6t 5 0,

t 5 21.1 1 t 5 0,5 1 1 t.(255 1 22s 2 11t)5 (22)(228 1 11s 2 6t) 1

(3, 2, 23)4(22)3P1P2

>

? (2, 1, 21)4 1 3P1P2

>

?

22(0) 1 0 5 0.

(22)3P1P2

>

? (2, 1, 21)4 1 3P1P2

>

? (3, 2, 23)4 5

255 1 22s 2 11t 5 0

2(211 1 2s 2 t) 1 (23)(5 2 3s 1 t) 5

P1P2

>

? (3, 2, 23) 5 3(26 1 3s 2 2t) 1

228 1 11s 2 6t 5 0.

1(211 1 2s 2 t) 1 (21)(5 2 3s 1 t) 5

P1P2

>

? (2, 1, 21) 5 2(26 1 3s 2 2t) 1

5 (26 1 3s 2 2t, 211 1 2s 2 t, 5 2 3s 1 t)2 (4 1 2t, 4 1 t, 23 2 t)

P1P2

>

5 (22 1 3s, 27 1 2s, 2 2 3s)

P1P2

>

? (3, 2, 23) 5 0.

P1P2

>

? (2, 1, 21) 5 0(3, 2, 23),

(2, 1, 21)


, and the symmetric

equation is

d. Choose to be the origin point for the equations.So the vector equation is

. The corresponding parametric equation is. Since the direction

vector has two zero coordinates, there is no symmetricequation for this line.19. A line parallel to the line connecting the points

and has a direction vector ofSince

collinear vectors of are also directionvectors for the line, is a direction vector.So the vector equation for a line with a directionvector of passing through the origin is

.20. The midpoint between (2, 6, 10) and is precisely The line connecting the midpoint and the given pointhas a direction vector of

So theparametric equations of the line through the desiredpoints is .21. The direction vector for the first line is and the direction vector for the second line is

So the directionvectors are collinear. The direction vectors arecollinear if and only if the lines are parallel, so theequations describe parallel lines.22. Since the point

lies on the line.

8.4 Vector and Parametric Equations of a Plane, pp. 459–460

1. a. plane; This is a vector equation of a plane in b. line; This is a vector equation of a line in c. line; This is a parametric equation for a line in d. plane; This is a parametric equation of a plane in

using (0, 0, 0) as 2. a. The first direction vector can be expressedwith integers as follows:

b. The second direction vector can be reduced asfollows:

c. The resulting equation of the plane using the twonew direction vectors is:

t,3. a. By inspection, if we choose weget the point b. Collecting the vector components of the n, andm, multiples we can rewrite the equation of theplane in vector form as:

Thus our direction vectors are:and

c.

Letting and we get:

d. Letting

We get the following parametric equations:

for and we get:

So our solution is and e. For the point the first two parametricequations are the same; yielding and however the third equation would then give:

which is not true. So there can be nosolution.4. a.

b.5 (3, 23, 21)

QR>

5 R 2 Q 5 (1 2 (22), 0 2 3, 1 2 2)

r>

5 (22, 3, 1) 1 t(0, 0, 1) 1 s(3, 23, 0)

5 (3, 23, 0)

PR>

5 R 2 P 5 (1 2 (22), 0 2 3, 1 2 1)

5 (0, 0, 1)

PQ>

5 Q 2 P 5 (22 2 (22), 3 2 3, 2 2 1)

R(1, 0, 1)Q(22, 3, 2),P(22, 3, 1),

28 5 27

28 5 21 2 3(0) 2 2(3)

28 5 21 2 3m 2 2n

n 5 3,m 5 0

B(0, 15, 28)

n 5 3.m 5 0

27 5 27

27 5 21 2 3(0) 2 2(3)

n 5 3m 5 0 27 5 21 2 3m 2 2n;

3 5 n.

15 5 5n 15 5 0 1 (23)m 1 5n

0 5 m.

0 5 0 1 2m 1 0n;

1 n(0, 5, 22)

A(0, 15, 27) 5 (0, 0, 21) 1 m(2, 23, 23)

r>

5 A(0, 15, 17)

5 (22, 217, 10)

5 (0, 0, 21) 1 (22, 3, 3) 1 (0, 220, 8)

(0, 5, 22)

r>

5 (0, 0, 21) 1 (21)(2, 23, 23) 1 (24)

n 5 24m 5 21

m, nPRr>

5 (0, 0, 21) 1 m(2, 23, 23) 1 n(0, 5, 22);

(0, 5, 22)(2, 23, 23)

m, nPRr>

5 (0, 0, 21) 1 m(2, 23, 23) 1 n(0, 5, 22);

(0, 0, 21).

n 5 m 5 0,

sPRr>

5 (2, 1, 3) 1 s(4, 224, 9) 1 t(1, 22, 5),

(6, 212, 30) 31

65 (1, 22, 5)

a1

3, 22,

3

4b 3 12 5 (4, 224, 9).

r>

0.R3

R3.

R3.

R3.

(7, 21, 8)

7 2 43 5 21 1 2

1 5 8 2 62 5 1,

(23, 29, 15) 5 23(1, 3, 25).

(1, 3, 25),

tPRz 5 1,y 5 28 2 13t,x 5 t,

(0, 28, 1) 2 (21, 5, 1) 5 (1, 213, 0).

(21, 5, 1).12 3(2, 6, 10) 1 (24, 4, 28)4 5

(24, 4, 28)

tPRr>

5 t(5, 25, 21),

(5, 25, 21)

(5, 25, 21)

(10, 210, 22)

(6, 25, 4) 2 (24, 5, 6) 5 (10, 210, 22).

(6, 25, 4)(24, 5, 6)

tPRz 5 22t,y 5 0,x 5 2,

tPRr>

5 (2, 0, 0) 1 t(0, 0, 22),

P0

x21

5y5

5z 2 6

1.

tPRz 5 6 1 t,y 5 5t,x 5 2t,


Using as the other direction vector:

Using as the other direction vector:,

5. a.does not represent a plane because the

direction vectors are the same. We can rewrite thesecond direction vector as:

And so we can rewrite the equation as:

This is an equation of a line in 6. a. The plane with direction vectors and that passes through the point

has a vector equation of:

The parametric equations are then:

b.

Using A(1, 0, 0) as our point with and asour direction vectors, our vector equation is:

And thus our parametric equations are:

c. using this andas our direction vectors and A(1, 1, 0)

as our point, the vector equation is:


7. a.This gives the parametric equations:

Substituting for t gives:

which is true so and b.Gives the following parametric equations:

The third equation then says:

which is a false statement. So the pointis not on the plane.

8. a. Using the direction vectors and the point two

equations of intersecting lines on the plane in vectorform are:

b. When and it is easily seen that thesetwo lines both have the point in common.9. hasparametric equations:

The plane crosses the z-axis when both x and yequal 0.

And so the z-coordinate is:The plane crosses

the z-axis at the point (0, 0, 5)10. Using the point on the line and thepoint we get another direction vector:

The equation of the planehaving the given properties is then:

t, sPRr>

5 (2, 1, 3) 1 s(4, 1, 5) 1 t(3, 21, 2),

a>

5 Q 2 P 5 (3, 21, 2).

P(21, 2, 1),

Q(2, 1, 3)

z 5 6 1 3(21) 2 2(21) 5 5.

s 5 1 1 2(21) 5 21.

t 5 21.

0 5 15 1 15t 0 5 4 1 11(1 1 2t) 2 7t 0 5 4 1 11s 2 7t 0 5 1 2 s 1 2t 1 s 5 1 1 2t

z 5 6 1 3s 2 2ty 5 1 2 s 1 2tx 5 4 1 11s 2 7t

r>

5 (4, 1, 6) 1 s(11, 21, 3) 1 t(27, 2, 22)

(23, 5, 6)

t 5 0s 5 0

tPR p>

5 (23, 5, 6) 1 t(2, 1, 23);

sPR l>

5 (23, 5, 6) 1 s(21, 1, 2);

A(23, 5, 6),b>

5 (2, 1, 23)

a>

5 (21, 1, 2),

A(0, 5, 24)

24 5 172 ,

24 5 1 21

21 2(4)

24 5 1 2 s 1 2t

t 5 2 1 2 5 4.

t 5 2 1 4a1

2b

1

25 s.

3 5 6s 5 5 2s 1 (2 1 4s)

5 5 2s 1 t 0 5 2 1 4s 2 t 1 t 5 2 1 4s.

(0, 5, 24) 5 (2, 0, 1) 1 s(4, 2, 21) 1 t(21, 1, 2)

t 5 1.s 5 12 5 2;

2 5 1 2 1 1 2(1)

2 5 1 2 s 1 2tt 5 23 1 4(1) 5 1.

1 5 s.

6 5 6s3 5 2s 1 (23 1 4s)

3 5 2s 1 t.5 5 2 1 4s 2 t 1 t 5 23 1 4s.

(5, 3, 2) 5 (2, 0, 1) 1 s(4, 2, 21) 1 t(21, 1, 2)

t, sPRz 5 26t 1 2s,

y 5 1 1 4t 1 sx 5 1 1 3t 1 7s

t, sPRr>

5 (1, 1, 0) 1 t(3, 4, 26) 1 s(7, 1, 2),

a>

5 (7, 1, 2)

AB>

5 B 2 A 5 (3, 4, 26)

t, sPRz 5 s,

y 5 tx 5 1 2 t 2 s

t, sPRr>

5 (1, 0, 0) 1 t(21, 1, 0) 1 s(21, 0, 1),

AC>

AB>

AC>

5 (0, 0, 1) 2 (1, 0, 0) 5 (21, 0, 1)

AB>

5 (0, 1, 0) 2 (1, 0, 0) 5 (21, 1, 0)

t, sPRz 5 7 2 s;

y 5 2 1 t 1 4sx 5 21 1 4t 1 3s

t, sPRr>

5 (21, 2, 7) 1 t(4, 1, 0) 1 s(3, 4, 21),

A(21, 2, 7)

b>

5 (3, 4, 21),

a>

5 (4, 1, 0)

R3.

nPR 5 (1, 0, 21) 1 n(2, 3, 4),

5 (1, 0, 21) 1 (s 1 2t)(2, 3, 24)

r>

5 (1, 0, 21) 1 s(2, 3, 24) 1 2t(2, 3, 24)

(2)(2, 3, 24)

t, sPR,

r>

5 (1, 0, 21) 1 s(2, 3, 24) 1 t(4, 6, 28),

t, sPRr>

5 (1, 0, 1) 1 t(3, 23, 0) 1 s(3, 23, 21)

PR>

t, sPRr>

5 (22, 3, 22) 1 t(0, 0, 1) 1 s(3, 23, 21),

PQ>


11. Using the point and the point (0, 0, 0) on the line we get another direction vector of:

So the equation of the plane withthe given properties is:

12. a. The xy-plane in has no z-coordinate so twosets of direction vectors are: (1, 0, 0), (0, 1, 0) and(1, 1, 0),b. A vector equation for the xy-plane in is:


13. a. We can use the direction vectorsand and the

origin to write the vector equation of the plane:

b. Using andas direction vectors, the

vector equation of the plane is:

c. The two planes in parts a. and b. are parallelsince they have the same direction vectors.14. We simply need to show that the direction vectorscan be expressed as a linear combination of the othertwo:

15. The planehas

parametric equations:

Solving for the y-intercept:

Solving for the z-intercept:

The direction vector between the two points is then:

And the equation of the line between them is:

16. The fact that the planecontains both of the given lines

is easily seen when letting and respectively.

8.5 The Cartesian Equation of a Plane,pp. 468–469

1. a.b. In the Cartesian equation:

If the planepasses through the origin.c. Three coordinates:

2. a.b. In the Cartesian equation: So the planepasses through the origin.c. Three coordinates:3. a.b. In the Cartesian equation: So the planepasses through the origin.c. Three coordinates:4. a. which is equivalentto The Cartesian equation is:

Since the plane passesthrough the origin So the equation is:

b. is equivalent to so the Cartesian equation is:

and since the plane passesthrough the origin 5. Method 1: Let be a point on the plane.Then is a vector on theplane.

Method 2: so the Cartesian equationis:We know the point is on the plane andmust satisfy the equation, so:

This also gives the equation:x 1 7y 1 5z 2 43 5 0.

D 5 243.

43 1 D 5 0

(23) 1 7(3) 1 5(5) 1 D 5 0

(23, 3, 5)

x 1 7y 1 5z 1 D 5 0.

n>

5 (1, 7, 5)

x 1 7y 1 5z 2 43 5 0

(x 1 3) 1 7(y 2 3) 1 5(z 2 5) 5 0

n>

? PA>

5 0

PA>

5 (x 1 3, y 2 3, z 2 5)

A(x, y, z)

28x 1 12y 1 7z 5 0D 5 0.

28x 1 12y 1 7z 1 D 5 0,

n>

5 (28, 12, 7),

n>

5 (212,

34,

716)

x 1 5y 2 7z 5 0.

D 5 0.

x 1 5y 2 7z 1 D 5 0.

n>

5 (1, 5, 27).

n>

5 (15, 75, 2105)

(0, 0, 0), (0, 1, 0), (0, 0, 1)

D 5 0.

n>

5 (A, B, C) 5 (1, 0, 0)

(0, 0, 0), (5, 2, 0), (5, 2, 1)

D 5 0.

n>

5 (A, B, C) 5 (2, 25, 0)

(11, 21, 1),

(11, 21, 1),(0, 0, 0),

D 5 0Ax 1 By 1 Cz 1 D 5 0,

n>

5 (A, B, C) 5 (1, 27, 218)

t 5 0s 5 0

r>

5 OP>

0 1 sa>

1 tb>

tPRr>

5 (0, 3, 0) 1 t(0, 3, 2),

(0, 3, 0) 2 (0, 0, 22) 5 (0, 3, 2).

z 5 3 1 5(21) 1 3(0) 5 22.

n 5 021 5 m;

0 5 3 1 3m0 5 2 1 2m 2 (21 2 m)

n 5 21 2 m

y 5 2 1 2(0) 2 (21) 5 3

n 5 210 5 m;

0 5 4m0 5 3 1 5m 1 3(21 2 m)

0 5 3 1 5m 1 3n0 5 1 1 m 1 n 1 n 5 21 2 m

z 5 3 1 5m 1 3ny 5 2 1 2m 2 nx 5 1 1 m 1 n

r>

5 (1, 2, 3) 1 m(1, 2, 5) 1 n(1, 21, 3)

27

13(23, 2, 4) 2

17

13(24, 7, 1) 5 (21, 25, 7).

(24, 7, 1) 2 (23, 2, 4) 5 (21, 5, 23)

t, sPRr>

5 (22, 2, 3) 1 s(21, 2, 5) 1 t(3, 21, 7),

PR>

5 R 2 P 5 (3, 21, 7)

PQ>

5 Q 2 P 5 (21, 2, 5)

t, sPRr>

5 s(21, 2, 5) 1 t(3, 21, 7),

OC>

5 (3, 21, 7)OA>

5 (21, 2, 5)

t, sPRz 5 0,

y 5 tx 5 s

t, sPR.r>

5 s(1, 0, 0) 1 t(0, 1, 0),

R3

(21, 1, 0).

R3

m, nPR.r>

5 m(2, 21, 7) 1 n(22, 2, 3),

a>

5 (22, 2, 3).

A(22, 2, 3)


6. a.


Using the point on the plane tosolve for D:

b.


Using the point on the plane to solvefor D:

c. There is only one simplified Cartesian equationthat satisfies the given information, so the equationsmust be the same.7.


Using the point on the plane to solve for D:

8. The point is on the line and thus alsoon the plane and we can get another direction vectorfrom:

Using as the otherdirection vector we can find the normal vector:

Our Cartesian equation is thus:

Using the point to determine D:

9. a.


b.


c.


10. We know the point is on the plane,and can obtain another direction vector from:

Let be our otherdirection vector.

The Cartesian equation is then:21x 2 15y 2 z 1 D 5 0.

5 (21, 215, 21)

2 (1)(3), (1)(1) 2 (1)(2))

n>

5 AP>

3 a>

5 ((1)(3) 2 (26)(3), (26)(2)

a>

5 (2, 1, 3)AP>

5 (1, 1, 26).

P(1, 1, 5)

n>

@n> @

5 a3

13, 2

4

13,

12

13b

5 13

5 !169

@n> @ 5 !9 1 16 1 144

n>

5 (3, 24, 12)

3x 2 4y 1 12z 2 1 5 0

n>

@n> @

5 a4

!26, 2

3

!26,

1

!26b

5 !26

@n> @ 5 !16 1 9 1 1

n>

5 (4, 23, 1)

4x 2 3y 1 z 2 3 5 0

n>

@n> @

5 a2

3,

2

3, 2

1

3b

5 3

0 n>

0 5 !4 1 4 1 1

n>

5 (2, 2, 21)

2x 1 2y 2 z 2 1 5 0

20x 1 9y 1 7z 2 47 5 0 D 5 247.

47 1 D 5 0 20(1) 1 9(3) 1 7(0) 1 D 5 0.

(1, 3, 0)

20x 1 9y 1 7z 1 D 5 0.

5 (220, 29, 27) 5 21(20, 9, 7).

2 (1)(5), (1)(5) 2 (23)(24))

n>

5 PQ>

3 a>

5 ((23)(5) 2 (1)(5), (1)(24)

a>

5 (24, 5, 5)PQ>

5 (1, 23, 1).

Q(2, 0, 1)

7x 1 17y 2 13z 2 24 5 0

D 5 224.

24 1 D 5 0

7(1) 1 17(1) 2 13(0) 1 D 5 0

(1, 1, 0)

7x 1 17y 2 13z 1 D 5 0.

n>

5 (7, 17, 213)

5 (7, 17, 213)

2 (5)(21), (5)(22) 2 (1)(3))

AB>

3 AC>

5 ((1)(21) 2 (4)(22), (4)(3)

AC>

5 (3, 22, 21).

AB>

5 (5, 1, 4).

7x 1 19y 2 3z 2 28 5 0

D 5 228.

28 1 D 5 0

27 1 38 2 3 1 D 5 0

7(21) 1 19(2) 2 (3)(1) 1 D 5 0

P(21, 2, 1)

7x 1 19y 2 3z 1 D 5 0.

n>

5 (7, 19, 23)

5 (7, 19, 23).

2 (24)(4), (24)(1) 2 (1)(21))

QP>

3 PR>

5 ((1)(4) 2 (23)(1), (23)(21)

5 (21, 1, 4)

PR>

5 (22 2 (21), 3 2 2, 5 2 1)

5 (24, 1, 23)

QP>

5 (21 2 3, 2 2 1, 1 2 4)

7x 1 19y 2 3z 2 28 5 0

D 5 228.

28 1 D 5 0

214 1 57 2 15 1 D 5 0

7(22) 1 19(3) 2 3(5) 1 D 5 0

R(22, 3, 5)

7x 1 19y 2 3z 1 D 5 0.

n>

5 (7, 19, 23)

5 21(7, 19, 23).

5 (27, 219, 3)

2 (4)(1), (4)(2) 2 (21)(25))

PQ>

3 QR>

5 ((21)(1) 2 (3)(2), (3)(25)

QR>

5 (22 2 3, 3 2 1, 5 2 4) 5 (25, 2, 1)

5 (4, 21, 3)

PQ>

5 (3 2 (21), 1 2 2, 4 2 1)



11. Since the normal vector is perpendicular to theplane, we can use the direction vector of the line asour normal vector:

The Cartesian equation is then:

We need the point to be on the plane so:

And the Cartesian equation of the planesatisfying the given conditions is:

12. a. To determine the angle between two planes,first determine their normal vectors. This is easilydone if the equations given are in Cartesian form.Once the normal vectors are known, and thenthe angle between the two planes can be determinedfrom the formula:

b.

13. a.

b. The parametric equations for the line are:

which give the following vector equation:Since the line and

normal vector are both perpendicular to the planewe may take:

The Cartesian equation for the plane is then:


And the Cartesian equation becomes:

14. a. and When is equivalent to:so the planes are parallel when b. When the planes are perpendicular

c. No the planes cannot ever be coincident. If theywere then they would also be parallel, so andwe would have the two equations:

Here all of the coefficients are equal except for theD values, which means that they don’t coincide.15. Since the plane passes through the points and it contains the line and the direction vector between them. The direction vector is:

The normal vector, must be perpendicular to thedirection vector and to the normal vector,

of the other plane, so:

Take and the Cartesian equation ofthe plane is:

Use the point (1, 4, 5) to determine D:

3x 1 5y 2 z 2 18 5 0

D 5 218.

18 1 D 5 0

3(1) 1 5(4) 2 5 1 D 5 0

3x 1 5y 2 z 1 D 5 0

n1

>

5 (3, 5, 21)

5 (26, 210, 2) 5 22(3, 5, 21)

2 (2)(1), (2)(21) 2 (22)(2))

n1

>

5 r>

3 n2

>

5 ((22)(1) 2 (24)(21), (24)(2)

n2

>

5 (2, 21, 1),

n1

>

,

r>

5 (2, 22, 24).

(3, 2, 1)

(1, 4, 5)

2x 1 4y 2 z 1 4 5 0 1 4x 1 8y 2 2z 1 8 5 0.

4x 1 8y 2 2z 1 1 5 0.

k 5 8,

k 5 210

45 2

5

2

10 1 4k 5 0

n1

>

? n2

>

5 8 1 4k 1 2 5 0

n1

>

? n2

>

5 0.

k 5 8.

n1

>

5 2(2, 4, 21),n1

>

k 5 8,

n2

>

5 (2, 4, 21).n1

>

5 (4, k, 22)

2x 2 3y 2 z 1 5 5 0.

D 5 5.

25 1 D 5 0

2(1) 2 (3)(2) 2 (1)(1) 1 D 5 0

P(1, 2, 1)

2x 2 3y 2 z 1 D 5 0.

n>

5 (2, 23, 21).

r>

5 (3, 21, 24) 1 t(22, 3, 1).

z 5 24 1 t, y 5 21 1 3t x 5 3 2 2t

u 5 cos21a5

!70b 5 53.3°

cos (u) 55

!70.

5 !70

@n1

> @ @n2

> @ 5 !14 ? !5

5 5

n1

>

? n2

>

5 1 1 4

n2

>

5 (1, 2, 0)n1

>

5 (1, 2, 23).

5 30°

u 5p

6

5!3

2

cos (u) 53

!12

5 !12.

n1

>

? n2

>

5 !2 ? !6

5 3

n1

>

? n2

>

5 2 1 0 1 1

n2

>

5 (2, 1, 21).n1

>

5 (1, 0, 21).

cos (u) 5n1

>

? n2

>

@n1

> @ @n2

> @.

n2

>

,n1

>

2x 2 4y 2 z 1 6 5 0.

D 5 6.

26 1 D 5 0

2(21) 2 4(1) 2 (0) 1 D 5 0.

(21, 1, 0)

2x 2 4y 2 z 1 D 5 0.

n>

5 (3, 22, 0) 2 (1, 2, 1) 5 (2, 24, 21).

21x 2 15y 2 z 2 1 5 0.

D 5 21.

1 1 D 5 0

21(1) 2 15(1) 2 (5) 1 D 5 0

(1, 1, 5)


16. Let be the normal vector of theunknown plane, and be the normalvector to the perpendicular plane. so we get:

We also know that the z-axis has the directionvector So:

The other constraint which we can choose is thelength of Since this is arbitrary (multiplicationby any scalar will give an equivalent normal vector)choose We have:

The equation of the plane is then:

17. The point equidistant from and

is the point

If every point in the plane is equidistant from these twopoint than the normal to the plane must point in thesame direction as the line connecting them:

The equation of the plane is thus:


We now have the equation of the plane:

Or equivalently:

8.6 Sketching Planes in pp. 476–477

1. a. A plane parallel to the yz-axis but two unitsaway, in the negative x direction.b. A plane parallel to the xz-axis but three unitsaway, in the positive y direction.c. A plane parallel to the xy-axis but 4 units away,in the positive z direction.2. The point of intersection of the three planes inproblem 1 must lie in every plane. Therefore thepoint of intersection is:3. The point must lie on the plane

since the point has an x-coordinate of 5, anddoesn’t have a y-coordinate of 6.4. In represents two lines,and In represents two planeswith the same equations.

5. a. i. x-intercept is when

Similarly the y-intercept is:

Since x and y cannot both be zero at the same time thereis no z-intercept. The plane is parallel to the z-axis.

y 5 6

3y 5 18

x 5 9

2x 5 18

y 5 z 5 0.

y

z

x

20

1

–1–2

2y

x

–2

x2 2 1 5 0R3,x 5 1.

x 5 21x2 2 1 5 0R2,

x 5 5,

p1:P(5, 23, 23)

(22, 3, 4)

R3,

8x 2 2y 2 16z 2 5 5 0.

4x 2 y 2 8z 25

25 0.

5

21 D 5 0 1 D 5 2

5

2

4(1) 23

22 0 1 D 5 0

(1, 32, 0)

4x 2 y 2 8z 1 D 5 0.

n>

5 (3, 1, 24) 2 (21, 2, 4) 5 (4, 21, 28).

(1, 32, 0).12((21, 2, 4) 1 (3, 1, 24)) 5

(3, 1, 24)

(21, 2, 4)

22

!5 x 1

1

!5 y 1 !3z 5 0.

A 5 22

!5.B 5

1

!5;

B2 51

5

4B2 1 B2 1 3 5 4

A2 1 B2 1 C2 5 4

!3

25

C2

1 C 5 !3.

@n1

> @ 5 2.

n1

>

.

5C

"A2 1 B2 1 C2.

5n1

>

? r>

@n1

> @ @r> @

cos (30°) 5!3

2

r>

5 (0, 0, 1).

A 5 22B A 1 2B 5 0.

n1

>

? n2

>

5 0

n2

>

5 (1, 2, 0)

n1

>

5 (A, B, C),


ii. x-intercept:

y-intercept:

z-intercept:

iii. There is no x-intercept since y and z cannot bothbe simultaneously zero.y-intercept:

z-intercept:

b. i. Since the plane is parallel to the z-axis onedirectional vector is: The other lies alongthe line so ii. We can find directional vectors by taking thedifference between two points, namely the interceptswe found in a. :or equivalently

orequivalently iii. Since the plane is parallel to the x-axis is one directional vector.

Or equivalently

6. a. i. Three points satisfyingthis equation are:ii. The line where this plane intersects the xy-planeis simply the line when

b.

7. has the solutions:So the three planes are the yz-plane, xz-plane, andthe xy-plane.

8. a.

b.

c.

d.

9. a.

b.

20

1

–1–2

2y

x

1–1–2

y(x 1 2) 5 0

xy 1 2y 5 0

y

z

x

y

z

x

y

z

x

y

z

x

z 5 0.y 5 0,x 5 0,xyz 5 0

y

z

x

2x 2 y 5 0.

z 5 0:

(0, 0, 0), (1, 2, 0), (0, 5, 1).

2x 2 y 1 5z 5 0.p:

(0, 1, 13).

(0, 3, 0) 2 (0, 0, 239) 5 (0, 3, 39).

(1, 0, 0)

(5, 0, 23).

(40, 0, 0) 2 (0, 0, 24) 5 (40, 0, 224)

(4, 3, 0).

(40, 0, 0) 2 (0, 230, 0) 5 (40, 30, 0)

(3, 22, 0).2x 1 3y 5 18,

(0, 0, 1).

z 5 239

2z 5 39

y 5 3

13y 5 39

z 5 24

5z 5 120

y 5 230

24y 5 120

x 5 40

3x 5 120


c.

10. a.

b.

c.





8 has the equation


s,

2.

s,

t,


tPR r>

5 (4, 3, 9) 1 t(7, 1, 1),

AB>

5 (7, 1, 1) 5 a>

B(4, 3, 9)A(23, 2, 8),

3x 1 y 2 z 2 6 5 0

D 5 26

3(1) 1 (2) 2 1(21) 1 D 5 0

(3)x 1 (1)y 1 (21)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

b>

3 c>

5 (1, 0, 3) 3 (2, 21, 5) 5 (3, 1, 21)

sPRr>

5 (1, 2, 21) 1 s(2, 21, 5) 1 t(1, 0, 3),

BC>

5 (1, 0, 3) 5 b>

AC>

5 (2, 21, 5) 5 c>

3x 1 y 2 z 2 6 5 0

D 5 26

3(1) 1 (2) 2 1(21) 1 D 5 0

(3)x 1 (1)y 1 (21)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

b>

3 a>

5 (1, 0, 3) 3 (1, 21, 2) 5 (3, 1, 21)

tPRr>

5 (1, 2, 21) 1 s(1, 21, 2) 1 t(1, 0, 3),

BC>

5 (1, 0, 3) 5 b>

AB>

5 (1, 21, 2) 5 a>

C(3, 1, 4)B(2, 1, 1),A(1, 2, 21),

z 5 21 1 2s 1 3t y 5 2 2 s x 5 1 1 s 1 t

tPR r>

5 (1, 2, 21) 1 s(1, 21, 2) 1 t(1, 0, 3),

r>

5 r0

>

1 sa>

1 tb>

BC>

5 (1, 0, 3) 5 b>

AB>

5 (1, 21, 2) 5 a>

C(3, 1, 4)B(2, 1, 1),A(1, 2, 21),

z8

5 1.

x5

2z7

5 1.

27,

x3

1y4

1z6

5 1.

y

z

x

y

z

x

y

z

x

y

z

x

















or Similarly










5 (3, 213, 2)

2 (2)(21), (2)1 2 (0)5)

2 (23)1, (23)5

(2, 0, 23) 3 (5, 1, 21) 5 ((0)(21)

tPRr>

5 (4, 23, 5) 1 s(2, 0, 23) 1 t(5, 1, 21),

(4, 23, 5)

(5, 1, 21)(2, 0, 23)

2x 2 7y 1 3z 5 0.

D 5 0.3(0) 1 D 5 0,2 (0) 2 7(0) 1

2x 2 7y 1 3z 1 D 5 0,

5 (21, 27, 3)

2 (2)5, (2)2 2 (1)1)

(2, 1, 3) 3 (1, 2, 5) 5 ((1)5 2 (3)2, (3)1

tPRr>

5 s(2, 1, 3) 1 t(1, 2, 5),

(3, 3, 8) 2 (2, 1, 3) 5 (1, 2, 5)

(2, 1, 3) 1 1(1, 2, 5) 5 (3, 3, 8)

(2, 1, 3) 1 0(1, 2, 5) 5 (2, 1, 3)

6x 1 4y 1 3z 5 12.

x2

1y3

1z4

5 1.

c 5 4.b 5 30

a1

0

b1

4

c5 1

0

a1

3

b1

0

c5 1a 5 2.

2

a1

0

b1

0

c5 1

xa

1yb

1zc

5 1

(1, 21, 212)

8(212) 2 28 5 227 2 0,2(1) 2 3(21) 1

2x 2 3y 1 8z 2 28 5 0.

2x 2 3y 1 8z 2 28 5 0.

D 5 228.

28 1 D 5 0,2(1) 2 3(2) 1 8(4) 1 D 5 0.

2x 2 3y 1 8z 1 D 5 0,

(2, 23, 8)

5 (2, 23, 8)

2 (1)0, (1)2 2 (22)3)

AB>

3 AC>

5 ((22)0 2 (21)2, (21)3

tPRz 5 4 2 s,

y 5 2 2 2s 1 2t,x 5 1 1 s 1 3t,

tPRt(3, 2, 0),r>

5 (1, 2, 4) 1 s(1, 22, 21) 1

5 (3, 2, 0)

AC>

5 (4, 4, 4) 2 (1, 2, 4)

5 (1, 22, 21)

AB>

5 (2, 0, 3) 2 (1, 2, 4)

AC>

AB>

25x 1 y 1 7z 1 18 5 0.

D 5 18

25(2) 1 (21) 1 7(21) 1 D 5 0

25x 1 y 1 7z 1 D 5 0.

5 (25, 1, 7)

2 (1)(1), 3(1) 2 1(2), 1(1) 2 3(22))

(A, B, C) 5 (1, 22, 1) 3 (3, 1, 2) 5 ((22)(2)

(A, B, C)

(1, 22, 1).x 2 2y 1 z 5 6,


Since the is a point on the plane,so

So Thus the equation is

5. a. The line intersects the yz-plane when

If then so

and Thus the point is

b. The direction vector is the same, sothe equivalent symmetric equation for the line is

6. a. The angle between two planes is determinedby the dot product of their normal vectors. The normal vector of the first plane is and the normal vector of the second plane is

and

So the angle between the planes

is The acute angle is b. i. The planes are parallel if and only if thecorresponding normal vectors are parallel. The normalvector of the first plane is and the normalvector of the second plane is The vectorsare parallel if and only if the ratios between the

coordinates are equal. Suppose so then So the vectors can be parallel onlywhen Since as well, the vectors areparallel at ii. The planes are perpendicular when their normalvectors are perpendicular. The vectors are perpendicular when their dot product is equal to zero.

So if then the dot product of the two normal vectors is equal to zero. Hence the planesare perpendicular at

c. The first plane in b. intersects the y-axis at the point(0, d, 0), where d satisfies So The second plane in b. intersects the y-axis at the point where e satisfies

So Since theplanes intersect the y-axis only once and the pointsare different, the equations can never represent thesame plane.7. a.

b. The equation for the plane can be written asSo for any real number t,

so the point (0, 0, t) is onthe graph. So the z-axis is on the plane. Also theplane cuts across the xy-plane along the line

So the origin is a point, as well as

c. The equation for the plane can be written asFor any real number t,

so (0, 0, t) is on theplane. Since this is true for all real numbers, the z-axis is on the plane.

B(0) 1 0(t) 5 0,A(0) 1

Ax 1 By 1 0z 5 0.

y

z

x

(22, 1, 0).

2x 1 y 5 0.

(0) 1 0(t) 5 0,2(0) 1

2x 1 y 1 0z 5 0.

4 60

642

–2–4–6

2–4–6 –2

x

y

e 5 24.5.2e 1 8(0) 5 9.k(0) 2

(0, e, 0),

d 5 25.

2(0) 2 d 1 k(0) 5 5.

k 5 215.

k 5 215,

5 10k 1 2

(2, 21, k) ? (k, 22, 8) 5 2k 2 1(22) 1 8k

k 5 4.

84 5 2k 5 4.

k 5 4.

k2

5 2221 5 2,

(k, 22, 8).

(2, 21, k)

70.5°.cos21( 21!3!3) 8 109.5°

0 (1, 1, 21) 0 5 "3.

(1, 1, 21) ? (1, 21, 1) 5 21

(1, 21, 1).

(1, 1, 21)

x4

5y 2 5

225 z 1 1

2.

(4, 22, 1)

(0, 5, 212).

z 5 212.(22) 1 4 5 5

y 5 (212)

y 2 4

225 z 5 0 2 2

4 5 212,x 5 0,

x 5 0.

3x 2 13y 1 2z 2 61 5 0.

D 5 261.61 1 D 5 0.

2(5) 1 D 5 0,3(4) 2 13(23) 1

(4, 23, 5)


c.

10. a.

b.

c.





8 has the equation


s,

2.

s,

t,


tPR r>

5 (4, 3, 9) 1 t(7, 1, 1),

AB>

5 (7, 1, 1) 5 a>

B(4, 3, 9)A(23, 2, 8),

3x 1 y 2 z 2 6 5 0

D 5 26

3(1) 1 (2) 2 1(21) 1 D 5 0

(3)x 1 (1)y 1 (21)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

b>

3 c>

5 (1, 0, 3) 3 (2, 21, 5) 5 (3, 1, 21)

sPRr>

5 (1, 2, 21) 1 s(2, 21, 5) 1 t(1, 0, 3),

BC>

5 (1, 0, 3) 5 b>

AC>

5 (2, 21, 5) 5 c>

3x 1 y 2 z 2 6 5 0

D 5 26

3(1) 1 (2) 2 1(21) 1 D 5 0

(3)x 1 (1)y 1 (21)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

b>

3 a>

5 (1, 0, 3) 3 (1, 21, 2) 5 (3, 1, 21)

tPRr>

5 (1, 2, 21) 1 s(1, 21, 2) 1 t(1, 0, 3),

BC>

5 (1, 0, 3) 5 b>

AB>

5 (1, 21, 2) 5 a>

C(3, 1, 4)B(2, 1, 1),A(1, 2, 21),

z 5 21 1 2s 1 3t y 5 2 2 s x 5 1 1 s 1 t

tPR r>

5 (1, 2, 21) 1 s(1, 21, 2) 1 t(1, 0, 3),

r>

5 r0

>

1 sa>

1 tb>

BC>

5 (1, 0, 3) 5 b>

AB>

5 (1, 21, 2) 5 a>

C(3, 1, 4)B(2, 1, 1),A(1, 2, 21),

z8

5 1.

x5

2z7

5 1.

27,

x3

1y4

1z6

5 1.

y

z

x

y

z

x

y

z

x

y

z

x


b. Answers may vary. For example:

t,

t,c. There are no symmetric equations, because thereare two parameters.4. A line passing through andperpendicular to the plane with the equation

Since the line is perpendicularto the plane, the normal of the plane is the linesvector.

5. a.

b.

c.

6.

7. Since the plane is parallel to the yz-plane, itsdirection vectors are (0, 1, 0) and (0, 0, 1).

8.

9.

34x 1 32y 2 7z 2 229 5 0

D 5 2229

34(4) 1 32(4) 2 7(5) 1 D 5 0

(34)x 1 (32)y 1 (27)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

a>

3 b>

5 (34, 32, 27)

b>

5 (2, 23, 24)a>

5 (5, 24, 6),

sPR L2: r>

5 (4, 4, 5) 1 s(2, 23, 24),

sPRL1: r>

5 (4, 4, 5) 1 s(5, 24, 6),

3x 1 y 2 z 2 7 5 0

D 5 27

3(4) 1 1(23) 2 1(2) 1 D 5 0

(3)x 1 (1)y 1 (21)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

a>

3 b>

5 (24, 8, 28) 5 (3, 1, 21)

b>

5 (2, 26, 0) a>

5 (1, 1, 4),

b>

5 3(4 2 2), (23 2 3), (2 2 2)4 a>

5 (1, 1, 4),

tPR r>

5 (2, 3, 2) 1 t(1, 1, 4),

A 5 (4, 23, 2)

z 5 1 1 sy 5 2 1 t, x 5 21,

sPR r>

5 (21, 2, 1) 1 t(0, 1, 0) 1 s(0, 0, 1)t,

b>

5 (0, 0, 1) a>

5 (0, 1, 0),

r>

5 r0

>

1 ta>

1 sb>

A 5 (21, 2, 1)

19x 2 7y 2 8z 5 0

D 5 0

19(0) 2 7(0) 2 8(0) 1 D 5 0

19x 2 7y 2 8z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

n>

5 (19, 27, 28)

c 5 28b 5 27,a 5 19,

2a 1 2b 1 3c 5 0

(2, 2, 3) ? (a, b, c) 5 n>

tPRr>

5 (3, 7, 1) 1 t(2, 2, 3),

3y 1 z 2 7 5 0

D 5 27

(0)(1) 1 (3)(2) 1 (1)(1) 1 D 5 0

(0)x 1 (3)y 1 (1)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

AB>

3 n>

5 (0, 3, 1)

n>

5 (1, 0, 0)

AB>

5 (1, 21, 3)

B(2, 1, 4)A(1, 2, 1),

3x 1 5y 2 2z 2 7 5 0

D 5 27

(3)(3) 1 (5)(0) 1 (22)(1) 1 D 5 0

(3)x 1 (5)y 1 (22)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

n>

3 AB>

5 (3, 5, 22)

n>

5 (1, 21, 21)

AB>

5 (23, 1, 22)

B(0, 1, 21)A(3, 0, 1),

x 2 3y 2 3z 2 3 5 0

2x 1 3y 2 3 1 3z 1 6 5 0

1 (3)(z 1 2) 5 0

(21)(x 2 0) 1 (3)(y 2 1)

Ax 1 By 1 Cz 1 D 5 0

n>

5 (21, 3, 3)

P(0, 1, 22)

x 2 7

25

y 2 1

235

z 1 2

1

x 2 x0

a5

y 2 y0

b5

z 2 z0

c

z 5 22 1 ty 5 1 2 3t, x 5 7 1 2t,tPR r

>

5 (7, 1, 22) 1 t(2, 23, 1),

r>

5 r0

>

1 tm>

m>

5 (2, 23, 1)

1 5 0.2x 2 3y 1 z 2

A(7, 1, 22)

sPRz 5 9 1 t 1 3s,

y 5 3 1 t 1 2s, x 5 4 1 7t 1 3s,

z 5 z0 1 ta3 1 tb3

y 5 y0 1 ta2 1 tb2, x 5 x0 1 ta1 1 tb1,

sPR r>

5 (4, 3, 9) 1 t(7, 1, 1) 1 s(3, 2, 3),

CB>

5 (6, 4, 6) 5 (3, 2, 3) 5 b>

AB>

5 (7, 1, 1) 5 a>

C(22, 21, 3)B(4, 3, 9),A(23, 2, 8),

x 2 4

75

y 2 3

15

z 2 9

1

x 2 x0

a5

y 2 y0

b5

z 2 z0

c

tPRz 5 9 1 t, x 5 4 1 7t, y 5 3 1 t,

z 5 z0 1 tc x 5 x0 1 ta, y 5 y0 1 tb,


10. Answers may vary. For example: Since the line isperpendicular to the plane. The normal of the plane isthe directional vector of the line.

11. Answers may vary. For example: Use the dotproduct and cross product to find two points that areorthogonal to the normal of the plane. Then use anypoint from the plane.

s,

12. Answers may vary. For example: The x-interceptis and z-intercept is (0, 0, 7). Find thedirectional vector from these points and use a pointone of the intercepts.

13. The two direction vectors for these lines are

So the lines and are parallel (they aren’t thesame line, as a point on is not apoint on ). Take one of the direction vectors forthe plane to be the vector andfind another by computing the vector with tail at

(a point on ) and head at (a point on ). This is the vector

The point is on the plane, so the vectorequation of the plane is

s, .The parametric form for the plane is

s,Finally, to find the Cartesian equation of the plane,compute the cross product of the direction vectors.

So the Cartesian equation is of the form.

To find the value of D, substitute in the point on theplane

So the Cartesian equation is

14. a.

b.

y

z

x

y

x

z

18x 2 19y 1 15z 2 145 5 0

D 5 2145

18(3) 2 19(24) 1 15(1) 1 D 5 0

(3, 24, 1).

18x 2 19y 1 15z 1 D 5 0

5 (18, 219, 15)

2 1(21), 1(3) 2 4(23))

5 (23(21) 2 (3)(25), 4(25)

a>

3 v>

5 (1, 23, 25) 3 (4, 3, 21)

tPRz 5 1 2 5s 2 t,y 5 24 2 3s 1 3tx 5 3 1 s 1 4t,

tPRr>

5 (3, 24, 1) 1 s(1, 23, 25) 1 t(4, 3, 21),

(3, 24, 1)

5 (4, 3, 21)

v>

5 (7, 21, 0) 2 (3, 24, 1)

L2

(7, 21, 0)L1(3, 24, 1)

a>

5 (1, 23, 25),

L2

L1,(3, 24, 1),

L2L1

b>

5 (2, 26, 210) 5 2a>

a>

5 (1, 23, 25)

z 5 7 1 2ty 5 0,x 5 t,tPR r

>

5 (0, 0, 7) 1 t(1, 0, 2),

tPR r>

5 r0 1 ta,

v>

5 (3.5, 0, 7) 5 (1, 0, 2)

v>

5 3(0 2 3.5), (0 2 0), (7 2 0)4

B 5 (0, 0, 7) A 5 (23.5, 0, 0),

(23.5, 0, 0)

z 5 6 1 3s 2 ty 5 25t,x 5 s 1 3t,tPR1 t(3, 25, 21),r

>

5 (0, 0, 6) 1 s(1, 0, 3)

5 (3, 25, 21)

(3, 2, 21) 3 (1, 0, 3) 5 (6, 210, 22)

a>

5 (1, 0, 3)

3a 1 2b 2 c 5 0

(a, b, c) ? (3, 2, 21) 5 0

a>

? (3, 2, 21) 5 0

3x 1 2y 2 z 1 6 5 0

x 2 2

35

y 2 3

225

z 1 3

1

z 5 23 1 sy 5 3 2 2s,x 5 2 1 3s,

sPRr>

5 (2, 3, 23) 1 s(3, 22, 1),

n>

5 (3, 22, 1)

3x 2 2y 1 z 5 0

A(2, 3, 23)


c.

d.

e.

15. a. Answers may vary. For example:

t,

b. Answers may vary. For example: The normal ofthe plane is the direction vector of the line, since itis perpendicular to the plane. Then find using theCartesian form of a plane.

c. Answers may vary. For example: Since the planeis parallel to the z-axis, one of its direction vectorsis (0, 0, 1).

t,

d. Answers may vary. For example:

t,

16. They are in the same plane because both planeshave the same normal vectors and Cartesian equations.

s,

u,

2x 2 3y 1 z 1 1 5 0

D 5 1

2(1) 2 3(21) 1 (26) 1 D 5 0

D 5 1

2(1) 2 3(2) 1 (3) 1 D 5 0

2x 2 3y 1 z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

(1, 1, 1) 3 (2, 5, 11) 5 (6, 29, 3) 5 (2, 23, 1)

(23, 5, 21) 3 (0, 1, 3) 5 (26, 9, 23) 5 (2, 23, 1)

vPRL2: r

>

5 (1, 21, 26) 1 u(1, 1, 1) 1 v(2, 5, 11),

tPRL1: r>

5 (1, 2, 3) 1 s(23, 5, 21) 1 t(0, 1, 3),

78x 1 10y 2 12z 2 168 5 0

D 5 2168

78(1) 1 10(3) 2 12(25) 1 D 5 0

(78)x 1 (10)y 1 (212)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

AB>

3 BC>

5 (78, 10, 212)

z 5 25 1 9t 2 sy 5 3 1 3t 2 9s,x 5 1 1 t 1 s,

sPRr>

5 (1, 3, 25) 1 t(1, 3, 9) 1 s(1, 29, 21),

r>

5 r0 1 ta>

1 sb>

BC>

5 (1, 29, 21)

AB>

5 (1, 3, 9)

C(3, 23, 3)B(2, 6, 4),A(1, 3, 25),

3x 1 y 2 13 5 0

D 5 213

3(4) 1 1(1) 1 D 5 0

(3)x 1 (1)y 1 (0)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

(1, 23, 5) ? (0, 0, 1) 5 (23, 21, 0) 5 (3, 1, 0)

z 5 21 1 5t 1 sy 5 1 2 3t,x 5 4 1 t,sPRr

>

5 (4, 1, 21) 1 t(1, 23, 5) 1 s(0, 01),

AB>

5 (1, 23, 5)

B(5, 22, 4)A(4, 1, 21),

24x 1 11z 2 18 5 0

D 5 218

24(1) 1 11(2) 1 D 5 0

(24)x 1 (0)y 1 (11)z 1 D 5 0

Ax 1 Bx 1 Cx 1 D 5 0

BC>

5 (24, 0, 11)

C 5 (22, 1, 5)B 5 (2, 1, 26),A(1, 1, 2)

15x 2 8y 1 2z 2 41 5 0

D 5 241

15(3) 2 8(1) 1 2(2) 1 D 5 0

(15)x 1 (28)y 1 (2)z 1 D 5 0

Ax 1 By 1 Cz 1 D 5 0

a>

3 p>

5 (2, 3, 23) 3 (2, 4, 1) 5 (15, 28, 2)

z 5 2 1 t 2 3sy 5 1 1 4t 1 3s,x 5 3 1 2t 1 2s,

sPRr>

5 (3, 1, 2) 1 t(2, 4, 1) 1 s(2, 3, 23),

a>

5 (2, 4, 1)

5 (0, 3, 1) 1 t(2, 4, 1)

L: r>

5 2ti 1 (4t 1 3)j 1 (t 1 1)k

PQ>

5 p>

5 (2, 3, 23)

Q(3, 1, 2)

P(1, 22, 5)

y

z

x

y

z

x

y

z

x


17. A point B on the line will have coordinatesThen

For this vector to be perpendicular to it wouldhave zero dot product with the direction vector for

So

So and the point B is

18. a. The plane is parallel to the z-axis through thepoints (3, 0, 0) and b. The plane is parallel to the y-axis through thepoints (6, 0, 0) and c. The plane is parallel to the x-axis through thepoints (0, 3, 0) and 19. a. To determine which points lie on the line, justsee if there is a t-value such that the coordinateworks.



There is no value of t that satisfies the equations.Only A lies on the line.b.

20. a.

b.

c.

d.

21. a.

b.

22. a. i. The given line is not parallel to the planebecause (3, 0, 2) is a point on the line and the plane.

u 5 90°

cos u 50

("3)("14)

n2 5 (2, 3, 21) n1 5 (1, 21, 21),

cos u 50 n1 ? n2 0

0 n1 0 0 n2 0

L2: 2x 1 3y 2 z 1 4 5 0

L1: x 2 y 2 z 2 1 5 0

u 5 44.2°

cos u 56

("14)("5)

n2 5 (1, 2, 0)

n1 5 (2, 3, 21),

cos u 50 n1 ? n2 0

0 n1 0 0 n2 0

L2:x 1 2y 1 4 5 0

L1: 2x 1 3y 2 z 1 9 5 0

u 5 90°

cos u 50

("96)("14)

n2 5 (21, 2, 3) n1 5 (4, 8, 24),

cos u 50 n1 ? n2 0

0 n1 0 0 n2 0

L2: (x, y, z) 5 (1, 5, 24) 1 t(21, 2, 3)

L1: (x, y, z) 5 (4, 7, 21) 1 t(4, 8, 24)

u 5 37.4°

cos u 516

("29)("14)

n2 5 (2, 3, 1) n1 5 (3, 4, 22),

cos u 50 n1 ? n2 0

0 n1 0 0 n2 0

z 5 27 1 sy 5 3s, L2: x 5 21 1 2s,

z 5 22ty 5 1 1 4t,L1: x 5 21 1 3t, u 5 59.0°

cos u 53

("17)("2)

cos u 50 n1 ? n2 0

0 n1 0 0 n2 0

n2 5 (1, 21)

n1 5 (1, 4)

y 5 2x 1 3y 5 4x 1 2,

u 5 45.0°

cos u 513

("26)("13)

n2 5 (2, 23)n1 5 (1, 5),

cos u 50 n1 ? n2 0

0 n1 0 0 n2 0

L2: x 2 2

25

1 2 y3

L1: x 2 1

15

y 2 3

5

b 5 3 1 t 5 21

a 5 2t 5 28

t 5 24

23 5 1 1 tb 5 3 1 t,a 5 2t,z 5 1 1 ty 5 3 1 t,x 5 2t,

2 5 1 1 t6 5 3 1 t,6 5 2t,D(6, 6, 2)

2 5 1 1 t5 5 3 1 t,4 5 2t,C(4, 5, 2)

1 5 1 1 t2 5 3 1 t,22 5 2t,B(22, 2, 1)

t 5 1

2 5 1 1 t4 5 3 1 t,2 5 2t,A(2, 4, 2)

z 5 1 1 ty 5 3 1 t,x 5 2t,

(0, 0, 26).

(0, 0, 22).

(0, 22, 0).

Ba2 1 2a7

3b, 1 1

7

3, 2 2

7

3b 5 Ba

20

3,

10

3, 2

1

3b.

t 5 146 5 7

3,

5 214 1 6t 5 26 1 4t 2 3 1 t 2 5 1 t 5 (2, 1, 21) ? (23 1 2t, 23 1 t, 5 2 t)

0 5 v>

? AB>

v>

5 (2, 1, 21).L2,

L2,

5 (23 1 2t, 23 1 t, 5 2 t) AB

>

5 (2 1 2t, 1 1 t, 2 2 t) 2 (5, 4, 23)

tPRB(2 1 2t, 1 1 t, 2 2 t),

L2


ii. Substitute the expressions for the components ofthe parametric equation of the line into the equationof the plane.

This last statement is never true. So the line and theplane have no points in common. Therefore, the lineis parallel to the plane.iii. Use the symmetric equation to rewrite x and z interms of y.

Substitute into the equation of the plane.

This equation has a solution. Therefore, the line andplane have a point in common and are not parallel.b. i. Substitute the expressions for the componentsof the parametric equation of the line into theequation of the plane.

This last statement is always true. So every point onthe line is also in the plane. Therefore, the line liesin the plane.ii. The line is parallel to the plane, and so does notlie in it.iii. is a point that lies on the line thatdoes not lie in the plane. Therefore, the line doesnot lie in the plane.23.

24. One direction vector for the plane is (2, 4, 1) and (1, 4, 4) are on the plane, so anotherdirection vector is So the parametric equations are

25. A plane has two parameters, because a planegoes in two different directions unlike a line thatonly goes in one direction.26. This equation will always pass through the origin, because you can always set and to obtain (0, 0, 0).

27. a. They do not form a plane, because these threepoints are collinear.

b. They do not form a plane, because the point lieson the line.

28. If a is the x-intercept, b is the y-intercept, and cis the z-intercept, this means that (a, 0, 0), (0, b, 0),and (0, 0, c) are points on the plane. So

are direction vectors for the plane. So a normal forthis plane is

So the Cartesian equation of the plane is of the form

Substitute the x-intercept, (a, 0, 0), into thisequation to determine the value of D.

So the Cartesian equation of this plane isor

29. If the normal vector is then theCartesian equation of the plane will be of the form

To determine the value of D, substitute the point(which is on the plane) into this equation.


30. a., b.

t,

t,

z 5 2 2 4t 1 3sy 5 23 1 7t 2 2s,x 5 1 2 3t 1 5s,

sPRr>

5 (1, 23, 2) 1 t(23, 7, 24) 1 s(5, 22, 3),

sPRr>

5 r>

0 1 ta>

1 sb>

,

BC>

5 (5, 22, 3)

AB>

5 (23, 7, 24)

C(3, 2, 1)B(22, 4, 22),A(1, 23, 2),

6x 2 5y 1 12z 1 46 5 0.

D 5 46

6(5) 2 5(8) 1 12(23) 1 D 5 0

(5, 8, 23)

6x 2 5y 1 12z 1 D 5 0

(6, 25, 12),

bcx 1 acy 1 abz 5 abcbcx 1 acy 1 abz 2 abc 5 0

D 5 2abc bc(a) 1 ac(0) 1 ab(0) 1 D 5 0

bcx 1 acy 1 abz 1 D 5 0

5 (bc, ac, ab)

a(b) 2 0(0))

5 (0(2c) 2 b(2c), 0(2c) 2 a(2c),

u>

3 v>

5 (a, 0, 2c) 3 (0, b, 2c)

5 (0, b, 2c)

v>

5 (0, b, 0) 2 (0, 0, c)

5 (a, 0, 2c)

u>

5 (a, 0, 0) 2 (0, 0, c)

5 (8, 27, 5)

r>

5 (4, 9, 23) 1 4(1, 24, 2)

r>

5 (4, 9, 23) 1 t(1, 24, 2)

r>

5 (21, 2, 1) 1 t(3, 1, 22)

(x, y, z) 5 (a 2 a, b 2 b, c 2 c) 5 (0, 0, 0)

(x, y, z) 5 (a, b, c) 1 0(d, e, f ) 2 1(a, b, c)

t 5 21 s 5 0,

(x, y, z) 5 (a, b, c) 1 s(d, e, f ) 1 t(a, b, c)

t 5 21s 5 0

tPR.z 5 4 2 3s 1 t, s,y 5 4 2 t,x 5 1 1 s 1 3t,

(2, 4, 1) 2 (1, 4, 4) 5 (1, 0, 23).

(3, 21, 1).

(x, y, z) 5 (4, 5, 8) 2 (4, 5, 6)

1 2(26, 6, 21)

(x, y, z) 5 (4, 1, 6) 1 4(3, 22, 1)

(x, y, z) 5 (4, 1, 6) 1 p(3, 22, 1) 1 q(26, 6, 21)

(5, 27, 1)

0 5 0

12 1 4t 2 2t 2 2 2 2t 2 10 5 0

4(3 1 t) 1 (22t) 2 (2 1 2t) 2 10 5 0

214y 2 96 5 0

216y 2 92 1 y 1 y 1 6 2 10 5 0

4(24y 2 23) 1 y 2 (2y 2 6) 2 10 5 0

z 5 2y 2 6

x 5 24y 2 23

215 5 0

212t 2 5 1 2t 1 10t 2 10 5 0

4(23t) 1 (25 1 2t) 2 (210t) 2 10 5 0


c. To find the Cartesian equation of the plane,a normal vector is needed. This can be found bycomputing the cross product of the direction vectorsfound in parts a. and b.

So the Cartesian equation has the form

Since is a point on this plane, we cansubstitute it in to determine the value of D.


d. Substituting into the Cartesian equationfound in part c., we get

This means that is not on the plane.31. a. The normal vector to the given plane is

so any plane parallel to this one musthave this same normal vector. So if a parallel planecontains the point (0, 0, 0), it will have the form

Substitute in the point (0, 0, 0) to determine thevalue of D.


b. Reasoning as in part a., if we want the pointto be in our parallel plane we find D in

the following way:


c. Reasoning as in parts a. and b., if we want thepoint to be in our parallel plane we findD in the following way:


32. a. The direction vector for is (2, 1) and foris This means that

and are parallel, and since they have the point(11, 0) in common (take in and in ),

these lines are coincident. So the angle betweenthem is b. The parametric equations of these lines are

So a point of intersection satisfies

or

or

So the point of intersection is

The point of intersection is at (for )and (for ).The direction vector for is (3, 4), and for is

So the angle between these lines satisfies

It would also have been correct to report the supplement of this angle, or roughly 93.18°, asthe answer in this case.33. a.

tPR r>

5 (1, 3, 5) 1 t(22, 24, 210),

r>

5 r>

0 1 ta>

P(1, 3, 5)

8 86.82°

5 cos21 a

1

5"3b

u 5 cos21 a

(3, 4) ? (3, 22)

0 (3, 4) 0 0 (3, 22) 0b

cos u 5(3, 4) ? (3, 22)

0 (3, 4) 0 0 (3, 22) 0

u(3, 22).

L2L1

L1t 5 32

L2s 5 232(3

2, 5) 5 5

y 5 21 1 4a3

2b

53

2

x 5 23 1 3a3

2b

53

2

5 23

21 3

t 5 s 1 3

s 5 23

2

6s 5 29

4(s 1 3) 1 2s 5 3

4t 1 2s 5 3

t 5 s 1 3

4t 2 2s 5 3

3t 2 3s 5 9

21 1 4t 5 2 2 2s23 1 3t 5 6 1 3s

sPRy 5 2 2 2s,L2: x 5 6 1 3s,

tPRy 5 21 1 4t,L1: x 5 23 1 3t,

u 5 0°.

L2s 5 6L1t 5 3

L2

L1(22, 21) 5 21(2, 1).L2

L1

4x 2 2y 1 5z 2 22 5 0.

D 5 222

4(2) 2 2(22) 1 5(2) 1 D 5 0

(2, 22, 2)

4x 2 2y 1 5z 1 19 5 0.

D 5 19

4(21) 2 2(5) 1 5(21) 1 D 5 0

(21, 5, 21)

4x 2 2y 1 5z 5 0.

D 5 0

4(0) 2 2(0) 1 5(0) 1 D 5 0

4x 2 2y 1 5z 1 D 5 0.

(4, 22, 5),

(3, 5, 24)

13(3) 2 11(5) 2 29(24) 1 12 5 100 2 0

(3, 5, 24)

13x 2 11y 2 29z 1 12 5 0.

D 5 12

13(1) 2 11(23) 2 29(2) 1 D 5 0

(1, 23, 2)

13x 2 11y 2 29z 1 D 5 0.

5 (13, 211, 229)

2 (23)(3), (23)(22) 2 5(7))

5 (7(3) 2 (22)(24), 5(24)

AB>

3 BC>

5 (23, 7, 24) 3 (5, 22, 3)


b.

c.

d. Since its parallel to the x-axis, its direction vectoris (1, 0, 0).

e. Find a perpendicular vector use the dot product.

f. Since the line is perpendicular to the plane, theline’s directional vector is the normal of the plane.Use the cross product to find the vector.

34. a. This plane will be of the form

To find D, substitute in


b. The direction vector for this line is(we can use this as one of the direction

vectors for the plane), and a point on this line isSo a second direction vector for the

plane will be

So a normal vector for this plane is

The Cartesian equation of this plane has the form

Substitute in to determine D.

The Cartesian equation of this plane is

c. This plane, being parallel to the xy-plane, iscompletely determined by a fixed z-coordinate (thex- and y- coordinates are allowed to be anything atall). Since it passes through the point theequation of this plane is Written in Cartesianform, this is d. Since this plane is to be parallel to

it will have the same normal vector, So this plane will be of theform Since is on this plane, we can substitutethis in to determine the value of D.


e. Since this plane is perpendicular to the yz-plane,it is completely determined by its intersection withthe yz-plane, which will be a line with y-intercept 4 and z-intercept This means that y and z arerelated by because of the y-intercept of 4. We can find the value of m by using the z-intercept of

So y and z are related via and theCartesian equation of the plane is (x is allowed to be anything here.)f. A normal vector, for this plane will beperpendicular to the normal vector for the plane

(A, B, C),

y 2 2z 2 4 5 0.

y 5 2z 1 4,

m 5 2

0 5 m(22) 1 4

22.

y 5 mz 1 4

22.

3x 1 y 2 4z 1 26 5 0.

D 5 26

3(24) 1 2 2 4(4) 1 D 5 0

P(24, 2, 4)

3x 1 y 2 4z 1 D 5 0.

(3, 1, 24).

3x 1 y 2 4z 1 8 5 0,

z 2 3 5 0.

z 5 3.

P(3, 3, 3),

29x 1 27y 1 24z 2 86 5 0.

D 5 286

29(22) 1 27(0) 1 24(6) 1 D 5 0

P(22, 0, 6)

29x 1 27y 1 24z 1 D 5 0.

5 (29, 27, 24)

2 6(25))

2 3(25), 3(22)

2 (22)(2), 6(2)

(3, 25, 2) 3 (6, 22, 25) 5 ((25)(25)

5 (6, 22, 25)

v>

5 (4, 22, 1) 2 P(22, 0, 6)

(4, 22, 1).

(3, 25, 2)

2x 2 4y 1 5z 1 23 5 0.

D 5 23

2(22) 2 4(6) 1 5(1) 1 D 5 0

P(22, 6, 1).

2x 2 4y 1 5z 1 D 5 0.

z 5 5 1 6ty 5 3 1 t, x 5 1,

r>

5 (1, 3, 5) 1 t(0, 1, 6)

AB>

3 BC>

5 (0, 27, 242) 5 (0, 1, 6) 5 n>

BC>

5 (26, 6, 21)

AB>

5 (21, 26, 1)

C(23, 2, 1)B(3, 24, 2),A(4, 2, 1),

tPR r>

5 (1, 3, 5) 1 t(0, 6, 4),

n>

5 (0, 6, 4)

c 5 4b 5 6, a 5 0,

23a 1 4b 2 6c 1 0

(23, 4, 26) ? (a, b, c) 5 0

z 5 5y 5 3, x 5 1 1 t,tPR r

>

5 (1, 3, 5) 1 t(1, 0, 0),

r>

5 r>

0 1 ta>

n>

5 (1, 0, 0)

P(1, 3, 5),

x 2 1

265

x 2 3

2135

x 2 5

14

z 5 5 1 14ty 5 3 2 13t, x 5 1 2 6t,tPR r

>

5 (1, 3, 5) 1 t(26, 213, 14),

r>

5 r>

0 1 ta>

RS>

5 (26, 213, 14)

P(1, 3, 5)

x 2 1

285

x 2 3

65

x 2 5

22

z 5 5 2 2ty 5 3 1 6t, x 5 1 2 8t,tPR r

>

5 (1, 3, 5) 1 t(28, 6, 22),

r>

5 r>

0 1 ta>

PQ>

5 (28, 6, 22)

Q(27, 9, 3)P(1, 3, 5),

x 2 1

225

y 2 3

245

z 2 5

210

z 5 5 2 10ty 5 3 2 4t, x 5 1 2 2t,

















or Similarly










5 (3, 213, 2)

2 (2)(21), (2)1 2 (0)5)

2 (23)1, (23)5

(2, 0, 23) 3 (5, 1, 21) 5 ((0)(21)

tPRr>

5 (4, 23, 5) 1 s(2, 0, 23) 1 t(5, 1, 21),

(4, 23, 5)

(5, 1, 21)(2, 0, 23)

2x 2 7y 1 3z 5 0.

D 5 0.3(0) 1 D 5 0,2 (0) 2 7(0) 1

2x 2 7y 1 3z 1 D 5 0,

5 (21, 27, 3)

2 (2)5, (2)2 2 (1)1)

(2, 1, 3) 3 (1, 2, 5) 5 ((1)5 2 (3)2, (3)1

tPRr>

5 s(2, 1, 3) 1 t(1, 2, 5),

(3, 3, 8) 2 (2, 1, 3) 5 (1, 2, 5)

(2, 1, 3) 1 1(1, 2, 5) 5 (3, 3, 8)

(2, 1, 3) 1 0(1, 2, 5) 5 (2, 1, 3)

6x 1 4y 1 3z 5 12.

x2

1y3

1z4

5 1.

c 5 4.b 5 30

a1

0

b1

4

c5 1

0

a1

3

b1

0

c5 1a 5 2.

2

a1

0

b1

0

c5 1

xa

1yb

1zc

5 1

(1, 21, 212)

8(212) 2 28 5 227 2 0,2(1) 2 3(21) 1

2x 2 3y 1 8z 2 28 5 0.

2x 2 3y 1 8z 2 28 5 0.

D 5 228.

28 1 D 5 0,2(1) 2 3(2) 1 8(4) 1 D 5 0.

2x 2 3y 1 8z 1 D 5 0,

(2, 23, 8)

5 (2, 23, 8)

2 (1)0, (1)2 2 (22)3)

AB>

3 AC>

5 ((22)0 2 (21)2, (21)3

tPRz 5 4 2 s,

y 5 2 2 2s 1 2t,x 5 1 1 s 1 3t,

tPRt(3, 2, 0),r>

5 (1, 2, 4) 1 s(1, 22, 21) 1

5 (3, 2, 0)

AC>

5 (4, 4, 4) 2 (1, 2, 4)

5 (1, 22, 21)

AB>

5 (2, 0, 3) 2 (1, 2, 4)

AC>

AB>

25x 1 y 1 7z 1 18 5 0.

D 5 18

25(2) 1 (21) 1 7(21) 1 D 5 0

25x 1 y 1 7z 1 D 5 0.

5 (25, 1, 7)

2 (1)(1), 3(1) 2 1(2), 1(1) 2 3(22))

(A, B, C) 5 (1, 22, 1) 3 (3, 1, 2) 5 ((22)(2)

(A, B, C)

(1, 22, 1).x 2 2y 1 z 5 6,


chapter 8 equations of lines and planes

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