MA261-A Calculus III

2006 Fall

Homework 3 Solutions

Due 9/22/2006 8:00AM

9.5 #10 Find the parametric equation and symmetric equation for the line of intersection of theplanes x+ y + z = 1 and x+ z = 0.[Solution]To write down a line equation, we need a directional vector and a point.To get a point, �rst, we assume that z = 0. So, we have

x+ y = 1 and x = 0.

This implies that x = 0 and y = 1. Thus, the point (0; 1; 0) is in the intersection of ourtwo planes, which is a point we need.We can read the normal vector of the plane x+ y+ z = 1 to be (1; 1; 1) and the normal

vector of the plane x+z = 0 to be (1; 0; 1). Since the intersection line lies on both planes,the line must be perpendicular to both normal vectors. So, the directional vector of theline can be (1; 1; 1)� (1; 0; 1) = (1; 0;�1).Therefore, the parametric equation of the line is8<: x = 0 + (1) t = t

y = 1 + (0) t = 1z = 0 + (�1) t = �t

and the symmetric equation of the line should look likex� 01

=y � 10

=z � 0�1 .

But, we cannot divide by 0. Thus, we should write a equation for y individually. So, thesymmetric equation of the line is

x� 01

=z � 0�1 and y = 1.

�9.5 #12 Is the line through (4; 1;�1) and (2; 5; 3) perpendicular to the line through (�3; 2; 0) and

(5; 1; 4)?[Solution]The directional vector for the �rst line is (2� 4; 5� 1; 3� (�1)) = (�2; 4; 4). The

directional vector for the second line is (5� (�3) ; 1� 2; 4� 0) = (8;�1; 4). By dotproduct, since (�2; 4; 4) � (8;�1; 4) = �4 6= 0, these two lines are not perpendicular.

�9.5 #14 (a) Find parametric equations for the line through (5; 1; 0) that is perpendicular to the

plane 2x� y + z = 1.(b) In what points does this line intersect the coordinate planes?[Solution]

1

2

(a) To write down a line equation, we need a directional vector and a point. We have apoint (5; 1; 0). Note that the normal vector (2;�1; 1) is perpendicular to the plane.So, it can serve a directional vector of the line. Thus, the parametric equation is8<: x = 5 + (2) t = 5 + 2t

y = 1 + (�1) t = 1� tz = 0 + (1) t = t

.

(b) When the line intersect xy-plane, the intersection point looks like (x; y; 0). Thus, bythe parametric equation, we have 0 = z = t. This implies that x = 5+2 (0) = 5 andy = 1� (0) = 1. Hence, the line intersects the xy-plane at (5; 1; 0).When the line intersect xz-plane, the intersection point looks like (x; 0; z). Thus, bythe parametric equation, we have 0 = y = 1 � t, that is t = 1. This implies thatx = 5+2 (1) = 7 and z = (1) = 1. Hence, the line intersects the xz-plane at (7; 0; 1).When the line intersect yz-plane, the intersection point looks like (0; y; z). Thus, bythe parametric equation, we have 0 = x = 5 + 2t, that is t = �5

2.. This implies that

y = 1 ���52

�= 7

2and z =

��52

�= �5

2. Hence, the line intersects the xy-plane at�

0; 72;�5

2

�.

�9.5 #30 Find the equation of the plane that passes through the line of intersection of the planes

x� z = 1 and y + 2z = 3 and is perpendicular to the plane x+ y � 2z = 1.[Solution]We can read the directional vector of the plane x�z = 1 is (1; 0;�1) and the directional

vector of the plane y+2z = 3 is (0; 1; 2). The directional vector of the line of intersectionof both planes is perpendicular to both. So, it can be (1; 0;�1) � (0; 1; 2) = (1;�2; 1).Also, by letting z = 0 for both planes, we can have a point in the line of intersection(1; 3; 0).The normal vector of the plane x+ y � 2z = 1 is (1; 1;�2). Since our plane is perpen-

dicular to the plane x+ y � 2z = 1, we know that (1; 1;�2) is on our plane.Now, we have two vectors (1;�2; 1) and (1; 1;�2) on our plane and a point (1; 3; 0).

Thus we have the normal vector of our plane is (1;�2; 1)� (1; 1;�2) = (3; 3; 3). Hence,the equation of the plane is

(x� 1; y � 3; z � 0) � (3; 3; 3) = 0,which is

x+ y + z = 4.

�9.5 #38 Find an equation for the plane consisting of all points that are equidistant from the points

(�4; 2; 1) and (2;�4; 3).[Solution]Assume that (x; y; z) is the point that are equidistant from the points (�4; 2; 1) and

(2;�4; 3). The distance between (x; y; z) and (�4; 2; 1) isq(x� (�4))2 + (y � 2)2 + (z � 1)2

and the distance between (x; y; z) and (2;�4; 3) isq(x� 2)2 + (y � (�4))2 + (z � 3)2.

3

Since they are equidistant, we haveq(x� (�4))2 + (y � 2)2 + (z � 1)2 =

q(x� 2)2 + (y � (�4))2 + (z � 3)2.

It turns out that

(x� (�4))2 + (y � 2)2 + (z � 1)2 = (x� 2)2 + (y � (�4))2 + (z � 3)2 ,

which is

3x� 3y + z = 2.�

9.5 #41 Find parametric equations for the line through the point (0; 1; 2) that is parallel to theplane x+ y + z = 2 and perpendicular to the line x = 1 + t, y = 1� t, z = 2t.[Solution]To write down a parametric equation for a line, we need a directional vector and a

point. We have a point (0; 1; 2).Assume that our directional vector is (a; b; c).Since our line is parallel to the plane x+y+z = 2, we know that (a; b; c) is perpendicular

to the normal vector of x+y+z = 2, which is (1; 1; 1). This gives us that (a; b; c)�(1; 1; 1) =0, or, a+ b+ c = 0.Also, since our line is perpendicular to the line x = 1 + t, y = 1 � t, z = 2t, we know

that (a; b; c) is perpendicular to the direction vector of this line, which is (1;�1; 2). Thisgives us that (a; b; c) � (1;�1; 2) = 0, or, a� b+ 2c = 0.Now, we can solve a; b; c from the two equations we just found to get (a; b; c) = (�3; 1; 2)

(This is just one possible solution by letting c = 2. We can have your own one.) Thus,the parametric equation is 8<: x = 0 + (�3) t = �3t

y = 1 + (1) t = 1 + tz = 2 + (2) t = 2 + 2t

.

[Alternative Solution]As above, we found that our directional vector (a; b; c) is perpendicular to both (1; 1; 1)

and (1;�1; 2). Thus, we can use the cross product to get (a; b; c) = (1; 1; 1)� (1;�1; 2) =(�3; 1; 2).

�9.5 #42 Find parametric equations for the line through the point (0; 1; 2) that is perpendicular to

the line x = 1 + t, y = 1� t, z = 2t and intersects this line.[Solution]To write down a parametric equation for a line, we need a directional vector and a

point. We have a point (0; 1; 2).Since our line is perpendicular to the line x = 1+ t, y = 1� t, z = 2t, we know that the

driectional vector is perpendicular to the direction vector of this line, which is (1;�1; 2).Now, draw a �gure with the given line x = 1 + t, y = 1 � t, z = 2t with any point of

this line (the easiest point of this time to �nd is (1; 1; 0).) Plot (0; 1; 2). Draw a vector ~vfrom (1; 1; 0) to (0; 1; 2). Also, draw a line which is perpendicular to the given line andpasses through (0; 1; 2). Mark the intersection point as P . Does our �gure look like aright-angle triangle we use to derive the projection formula?

4

In this �gure, the last line (perpendicular one) is the line we want. The vector ~w fromP to (0; 1; 2) plays the role of the directional vector. By projection formula discussion, ~wis the orthoginal projection of ~v onto (1;�1; 2) (see Exercise 9.3. #27.) So,

~w = ~v � proj(1;�1;2) ~v = (�1; 0; 2)�(�1; 0; 2) � (1;�1; 2)

j(1;�1; 2)j2(1;�1; 2)

= (�1; 0; 2)� 12(1;�1; 2) =

��32;1

2; 1

�.

With a directional vector and a point, we know that the line equation is8<: x = 0 +��32

�t = �3

2t

y = 1 +�12

�t = 1 + 1

2t

z = 2 + (1) t = 2 + t.

�9.5 #52 Find equations of the planes that are parallel to the plane x+2y� 2z = 1 and two units

away from it.[Solution]To write down a plane equation, we need a normal vector and a point. Since the planes

are parallel to the plane x + 2y � 2z = 1, the normal vector is (1; 2;�2). Also, the unitnormal vector is 1

k(1;2;�2)k (1; 2;�2) =�13; 23;�2

3

�.

By setting y = 0 and z = 0, we have a point (1; 0; 0) on the plane x + 2y � 2z = 1.There are two points in the direction of the normal vector (1; 2;�2) which are 2 unitsaway from it. They can be found by starting from (1; 0; 0) and walking along the direction(or opposite direction) of the normal vector (1; 2;�2). Thus, they are

(1; 0; 0) + 2

�1

3;2

3;�23

�=

�5

3;4

3;�43

�and

(1; 0; 0)� 2�1

3;2

3;�23

�=

�1

3;�43;4

3

�.

So, we know that for these two points the diatance to the point (1; 0; 0) are all 2. Thus,we have our points.For

�53; 43;�4

3

�, we have the equation of the plane is�

x� 53; y � 4

3; z �

��43

��� (1; 2;�2) = 0,

which is,x+ 2y � 2z = 7.

For�13;�4

3; 43

�, we have the equation of the plane is�

x� 13; y �

��43

�; z � 4

3

�� (1; 2;�2) = 0,

which is,x+ 2y � 2z = �5.

�

5

9.5 #54 Find the distance between the skew lines with parametric equations x = 1+ t, y = 1+6t,z = 2t and x = 1 + 2s, y = 5 + 15s, z = �2 + 6s.[Solution](This is similar to Example 10 in the textbook page 673.)First, the normal vector is (1; 6; 2)� (2; 15; 6) = (6;�2; 3).Second, by putting s = 0 in the equations of the second line, we get the point (1; 5;�2).

Thus, the equation of the plane that our second line lies on it is

(x� 1; y � 5; z + 2) � (6;�2; 3) = 0,

that is,

6x� 2y + 3z + 10 = 0.

Now, by putting t = 0 in the equations of the �rst line, we get the point (1; 1; 0). So thedistance from (1; 1; 0) to the plane 6x� 2y + 3z + 10 = 0 is

D =j6 (1)� 2 (1) + 3 (0) + 10jq

62 + (�2)2 + 32=14

7= 2.

�

9.6 #4 Let f (x; y) = ln (x+ y � 1).(a) Evaluate f (1; 1).(b) Evaluate f (e; 1).(c) Find and sketch the domain of f .(d) Find the range of f .[Solution](a) f (1; 1) = ln (1 + 1� 1) = ln 1 = 0.(b) f (e; 1) = ln (e+ 1� 1) = ln e = 1.(c) To use the natural logarithm function, we have to have x + y � 1 > 0. Thus, the

domain is f(x; y) j x+ y > 1g.(d) It is easy to see that x+ y � 1 can be equal to any positive real number. Thus, the

range of f is R.

�

9.6 #12 Sketch the graph of

f (x; y) = cosx

[Solution]Thr graph is

6

­1.0

­4­2

1.0

­4

­0.5

­2 0.0

2

0.5

0

xy0

z

442

�

9.6 #18 Use traces to sketch the graph of the function f (x; y) = x2 � y2.[Solution]For a �x x = k, we have traces z = k2 � y2.

­5 ­4 ­3 ­2 ­1 1 2 3 4 5

­4

­2

2

4

y

z

For a �x y = k, we have traces z = x2 � k2.

­5 ­4 ­3 ­2 ­1 1 2 3 4 5

­4

­2

2

4

y

z

For a �x z = k, we have traces k2 = x2 � y2.

7

­5 ­4 ­3 ­2 ­1 1 2 3 4 5

­5

­4

­3

­2

­1

1

2

3

4

5

x

y

The graph is

­4

­2

­204

x 2­10

000

y2

z

10

4

­4

20

­2

�9.6 #22 Classify the surface

4y2 + z2 � x� 16y � 4z + 20 = 0by comparing with one of the standard forms in Table. Then sketch its graph.[Solution]

4y2 + z2 � x� 16y � 4z + 20 = 0)

x = 4y2 + z2 � 16y � 4z + 20)

x = 4y2 � 16y + 16 + z2 � 4z + 4)

x = 4 (y � 2)2 + (z � 2)2

8

)x

4= (y � 2)2 + (z � 2)

2

4By comparing to the Table, the surface is a Elliptic Paraboloid at the top (0; 2; 2). Thegraph is

2015­20 5

x10

­2

20

0

64

2

4

6

y

�9.6 #32 Show that the curve of intersection of the surfaces x2+2y2� z2+3x = 1 and 2x2+4y2�

2z2 � 5y = 0 lies in a plane.[Solution]The �rst surface can be written as

x2 + 3x+

�3

2

�2+ 2y2 � z2 = 1 +

�3

2

�2) �

x+3

2

�2+ 2y2 � z2 = 13

4) �

x+ 32

�22

+ y2 � z2

2=13

8.

So, it is a hyperboloid of one sheet.The second surface can be written as

x2 + 2y2 � 52y � z2 = 0

)x2 + 2y2 � 5

2y + 2

�5

8

�2� z2 = 2

�5

8

�2)

x2 + 2

�y � 5

8

�2� z2 = 50

64)

x2

2+

�y � 5

8

�2� z

2

2=25

16.

9

So, it is also a hyperboloid of one sheet.For an intersection point (x; y; z), it satis�es both x2 + 2y2 � z2 + 3x � 1 = 0 and

x2 + 2y2 � 52y � z2 = 0. Thus, we substract the second equation from the �rst equation

and get 3x+ 52y = 1, that is, 6x+5y = 2. So, an intersection point must lie on the plane

6x+ 5y = 2.�

9.6 #34 Find an equation for the surface consisting of all points P for which the distance from Pto the x-axis is twice the distance from P to the yz-plane. Identify the surface.[Solution]

�