Math 250 Review Problems for Exam 1

1. Find the point of intersection of the lines L1:!x = !t,!!y = !t,!!z = 5t + 6 ,

and L2:!x = t + 2,!!y = !t,!!z = !t .

Solution: (1, 1, 1).

It is probably best to first rewrite the second line as

L2:!x = s + 2,!!y = !s,!!z = !s so that we can distinguish the parameters for each

line. Then we get!t = s + 2 and ! t = !s . So, t = s = !1 .

Don’t be misled here into thinking that a common point will always occur for the

same value of the parameter in each line. For example,

What if the second line had been L2:!x = 2t ! 3,!!y = t !1,!!z = 3t ! 5 ?

Then rewriting this asL2:!x = 2s ! 3,!!y = s !1,!!z = 3s ! 5 , we get the

equations!t = 2s ! 3 and ! t = s !1 . Solving these, we get t = !1 and s = 2 . And

again, the point of intersection is (1, 1, 1).

2. (a). Find the equation of the plane that contains the line

L1:!x = t + 3,!!y = !t +1,!!z = 3t !1,

and is parallel to the lineL2:!x = t + 2,!!y = t,!!z = 2t +1 .

Solution: A normal to the plane isn = i ! j + 3k( ) " i + j + 2k( ) = !5i + j + 2k .

So, the equation of the plane is5x ! y ! 2z = 16 .

(b). Suppose that there is a plane that contains the line

L1:!x = t + 3,!!y = !t +1,!!z = 3t !1 and is perpendicular to

the line L2:!x = kt + 2,!!y = t,!!z = 2t +1. What is the value of k?

What is the equation of the plane?

Solution: The vectors (1, –1, 3) must be perpendicular to (k, 1, 2).

So, we get k – 1 + 6 = 0 or k = –5.

The equation of the plane is5x ! y ! 2z = 16 .

3. What is the cosine of the angle between the

linesx ! 3

1=y ! 2

2=z ! 4

1 and

x ! 3

2=y ! 2

1=z ! 4

1?

Solution:56

.

cos! =i + 2 j + k( ) " 2i + j + k( )

i + 2 j + k 2i + j + k=

5

6 # 6=5

6.

4. Find the area of the triangle whose vertices

are: A :!(1,!0,!3),!!B :!(0,!2,!3),!!C :!(!1,!5,!!2) .

Solution:126

2.

Since, AC!

" AB!

=

i j k

#2 5 #5

#1 2 0

= 10i + 5 j + k , the Area

is1

210i + 5 j + k =

1

2100 + 25 +1 =

126

2.

5. Find a unit vector that is perpendicular to both 2i + j ! 3k and i + k .

Solution:1

3 3i ! 5 j ! k( ) .

Letv = 2i + j ! 3k( ) " i + k( ) = i ! 5 j ! k .

Thus the answer isv

v=

1

3 3i ! 5 j ! k( ) .

6. Suppose that v,!u,!w are vectors in R3.Write expressions for each of the following

using dot and cross products.

(a). A vector that is perpendicular to both v and u.

(b). The vector projection of v onto u.

(c). A vector in the direction of v but having the same length as u.

Solution: (a). v ! u (b). v !uu

2

"

#$

%

&' u (c).

u

v

!

"#$

%&v .

7. Suppose that v,!u,!w,!x are vectors in R3. For each of the following expressions,

determine it represents a scalar [i.e., a number], a vector, or is simply nonsense.

(a).v ! (u " w) (b).v ! u + w( ) (c).v ! (u "w)

(d).v ! (u "w)x[ ] (e). (v !u) " (v !w) (f). v ! u( ) " x ! w( ) .

Solution: a and f represent numbers, b and d represent vectors, c and e are nonsense.

8. Is it true or false that for any two n ! n matrices A and

B,det(A + B) = det(A) + det(B) ?

Solution: False – for example letA = B =1 0

0 1

!

"#

$

%& .

9. Find the point P that lies on the line L: x = 2t + 3,!!y = 3t + 5,!!z = t + 4 that is closet

to the point A: (10, 9, 6).

Solution: The point is (7, 11, 6). Suppose that P has coordinates (a, b, c). Then

a = 2t + 3,!b = 3t + 5,!c = t + 4 for some number t. Since the line must be

perpendicular to the vector PA, we get 2,!3,!1( ) ! 2t " 7,!3t " 4,!t " 2( ) = 0 .

So we get, 4t !14 + 9t !12 + t ! 2 = 0"14t = 28 , and so t = 2.

Thus the point is (7, 11, 6). The distance from the point P to the line L is 13 .

10. Suppose that the angle between the vectors v and u is 60 degrees.

Given that v = 3 and u = 4 , what is the value ofv ! v + u( ) ?

Solution: 15.

v ! v + u( ) = v !v + v !u = v

2

+ v u cos60!

== 9 + 6 = 15 .

11. 4 3

1 1

!"#

$%&

'1

= ? Solution:1 !3!1 4

"#$

%&'

.

12. What is the equation of the sphere having the line segment from (1, 4, 2) to (5, 2, 6) as

a diameter?

Solution: x ! 3( )2

+ y ! 3( )2

+ z ! 4( )2

= 36 .

13. If a ! b = 3i + 4 j + 2k anda = 2i + j + ck ,

(a). What is the value of c?

Solution: c = –5.

(b). What is the vector b?

Solution:b = j ! 2k .

14. Suppose thatdet

a b c

d e f

g h i

!

"

##

$

%

&&= 12 . What is the value ofdet

g h i

d e f

a b c

!

"

##

$

%

&&

?

Solution: –12.

[Switching any two rows of a matrix reverses the sign of its determinant.]

15. The line through the points (1, 2, 3) and (5,!b,!c) is parallel to the

linex ! 3

2=y + 7

6=z !1

3. What are the values of b and c?

Solution:b = 14,!c = 9 .

The vector 4,!b ! 2,!c ! 3( ) must be parallel to (2,!6,!3) .

16. Show that lim(x,y)!(0,0)

x2+ y

4

x + y2( )2

does not exist.

Solution:

Approaching along the line x = 0 , the limit would be 1.

Approaching along the parabola x = y2 , the limit would be 0.5.

17. The line L is tangent to a circle with center at (1, 2, 0) at the point (3, 4, 0). What is the

equation of L?

Hint: Since z = 0 for both points, this line lies in the xy-plane and you can write its

equation in the ‘old fashioned way’ i.e., in the formax + by = c .

Solution: x + y = 7 .

Suppose that (x,!y,!z) is some point on the line. The vector (2, 2, 0) is perpendicular

to (x ! 3,!y ! 4,!z) . So we get2(x ! 3) + 2(y ! 4) = 0" x + y = 7 .

18. For the vectors below,a = (1,!3,!6),!b = (2,!4,!!2),!!c = (1,!3,!!1),!!e = (2, 4,!5) .

What is the value d? [Picture is not to any scale!]

Solution: d = (2, 6, 8). Just note thatd = a + b + c ! e .

19. If the angle between v and u is 60 degrees, then v !u

v ! u = ?

Solution: 12

.

v !u

v ! u= cos60

!

=1

2.

20. If v lies in the plane determined by a and b, then v ! a " b( )=______. Solution: 0.

21. Let M denote the parallelogram determined by the vectors a and b.

Suppose that the diagonals of M are perpendicular to one another.

Show that a and b must have the same length.

Solution: The diagonals of M area + b and a ! b . If these are perpendicular,

then (a + b) ! (a " b) = 0# a !a " a !b + b !a " b !b = 0 . This reduces to ,

a !a " b !b = 0# a2

= b2

, and so a = b .

22. (a). Suppose that the vectors a,!b,!c,!and d all lie in a common plane.

What is the value of a ! b( ) ! c ! d( ) ?

Solution: 0.

(b). If a !b = a ! c where a, b and c are non-zero vectors, does it follow that b = c?

Solution: No.

23. What is the line of intersection of the plane through (1, 2, 3) and normal to the vector

(1, 5, 4) and the xy-plane?

Solution: x + 5y = 23 .

The plane through (1, 2, 3) and normal to the vector (1, 5, 4) has

equation x + 5y + 4z = 23 . This plane meets the xy-plane whenever z = 0. Thus the

equation of the line is most easily obtained by just setting z = 0 in x + 5y + 4z = 23 .

A longer, but acceptable, solution would be to notice that the points

(3, 4, 0) and (8, 3, 0) lie on the line of intersection. The equation of the line through

these two points is x = 5t + 3,!!y = !t + 4 . This could be rewritten as x + 5y = 23 .

24. If the line x = 3t +1,!y = !2t + 3,!z = !4t !1 intersects the

line x = 2t ! 6,!y = t + 3,!z = 4t + c , then what is the value of c?

Solution: c = –5 – and the point of intersection is (–2, 5, 3).

Rewrite the second line as x = 2s ! 6,!y = s + 3,!z = 4s + c to distingiuish its

parameter from that of the first line. Then solving 3t +1 = 2s ! 6 and ! 2t + 3 = s + 3 ,

we get t = !1 and s = 2 . So the two lines meet when the z-coordinate for t = –1 and

s = 2 are the same. That is, 3 = 8 + c. So, c = –5.

25. Suppose thata !x = 22 and a " x = #i + 2 j # k , where a = (1, 3, 5).

Then find the value of x.

Solution: x = (1, 2, 3)

26. Suppose that the vector u = (x, y,!6) is parallel to the vector v = (1,!2,!4) and u is

perpendicular to the vector w = (2,!1,!k) . What is the value of k?

Solution: k = – 1.

27. Find the equation of the plane tangent to the sphere (x !1)2 + (y ! 2)2 + (z ! 3)2 = 29 at

the point (5, 4, 6).

Solution: A vector normal to the plane is n = 4i + 2 j + 3k .

So the equation of the plane is 4x + 2y + 3z = 46 .

Review Problems From Your Text:

1.3 #11, 26, 33 1.5 #19 2.1 #21, 23, 24, 26, 27

Review Exercises (Page 88 – 90) #3, 7, 12, 15, 18(a), 43, 45.