ch08

Solutions for Selected Problems

Exercise 8.1

7. Vector equations of planes:

a. Through A(�4, 5, 1), parallel to vectors a�� (�3,

�5, 3) and b�� (2, �1, �5). A vector equation is

r� � (�4, 5, 1) � s(�3, �5, 3) � t(2, �1, �5).

b. Contains two intersecting lines l1: r� � (4, 7, 3)

� t(1, 4, 3) and l2: r� � (�1, �4, 6) � s(�1, �1, 3).

A point on the plane is A(4, 7, 3) and two

directions are a�� (1, 4, 3) and b�� (�1, �1, 3).

A vector equation of the plane is

r� � (4, 7, 3) � s(1, 4, 3) � t(�1, �1, 3).

c. Contains the line r� � (�3, 4, 6) � t(�5, �2, 3)

and the point A(8, 3, 5). A second point on the

plane is B(�3, 4, 6). Two directions are a�� (�5,

�2, 3) and BA�� (11, �1, �1). A vector equation

is r� � (8, 3, 5) � s(�5, �2, 3) � t(11, �1, �1).

d. Contains parallel lines l1: r� � (0, 1, 3) � t(�6,

�3, 6) and l2: r� � (�4, 5, �4) � s(4, 2, �4).

A direction of the parallel lines is a�� (2, 1, �2).

Two points on the plane are A(0, 1, 3) and B(�4, 5,

�4), therefore a second direction is BA�� (4, �4,

7). A vector equation is r� � (0, 1, 3) � s(2, 1, �2)

� t(4, �4, 7).

e. Contains the three points A(2, 6, �5), B(�3, 1,

�4), C(6, �2, 2). Two directions are BA�� (5, 5,

�1) and AC�� (4, �8, 7). Choosing any of the

three points, a vector equation is r� � (2, 6, �5) �

s(5, 5, �1) � t(4, �8, 7).

8. Parametric equations of planes:

a. Through A(7, �5, 2), parallel to vectors a�� (4,

�1, 1) and b�� (�3, 4, 4). Parametric equations

are x � 7 � 4s � 3t, y � �5 � s � 4t, z � 2 � s

� 4t.

b. A plane contains the two intersecting lines

l1: r� � (5, 4, 2) � t(4, �2, 1) and

l2: r� � (7, 4, �7) � s(�3, 1, 4).

Since the plane contains both lines, two directions

of the plane will be a�� (4, �2, 1) and b�� (�3,

1, 4). Any point on l1

or l2

can be used, therefore

parametric equations of the plane arex � 5 � 4t � 3sy � 4 � 2t � sz � 2 � t � 4s.

c. Contains the line r� � (1, 3, �1) � t(2, 2, �5) and

the point A(8, 3, 5). A direction is a�� (2, 2, �5),

a second point is B(3, 5, �6), (when t � 1),

therefore another direction is BA�� (5, �2, 11).

An equation is x � 8 � 5s � 2t, y � 3 � 2s � 2t,

z � 5 � 11s � 5t.

d. Contains two parallel lines l1: r� � (3, 2, 2) � t(�9,

6, �6) and l2: r� � (1, 6, �6) � s(6, �4, 4). Two

points in the plane are A(3, 2, 2) and B(1, 6, �6).

Two directions are a�� (3, �2, 2) and BA��

(2, �4, 8) � 2(1, �2, 4). Parametric equations are

x � 3 � s � 3t, y � 2 � 2s � 2t, z � 2 � 4s � 2t.

e. Contains the three points A(2, 6, 5), B(�3, 1, �4),

C(6, �2, 2). From 7e, parametric equations are

x � 2 � 5s � 4t, y � 6 � 5s � 8t, z � �5 �

s � 7t.

9. a. A plane parallel to the yz-plane has directions

a�� (0, 1, 0) and b�� (0, 0, 1). This plane passes

through the point A(6, 4, 2). A vector equation is

r� � (6, 4, 2) � s(0, 1, 0) � t(0, 0, 1).

b. A plane passes through O(0, 0, 0), A(3, 3, 3), and

B(8, �1, �1). Two directions are OA�� (3, 3, 3)

� 3(1, 1, 1) and OB�� (8, �1, �1). An equation is

r� � s(1, 1, 1) � t(8, �1, �1).

Chapter 8: Equations of Planes 109

Chapter 8 • Equations of Planes

c. A plane contains the x-axis and the point A(�1,

�4, �7). Two points on the x-axis are O(0, 0, 0)

and B(1, 0, 0). Two directions are AO�� (1, 4, 7)

and OB�� (1, 0, 0). An equation is r� � s(1, 0, 0) �

t(1, 4, 7).

10. a. The three points A(2, 3, �1), B(8, 5, �5), and

C(�1, 2, 1) give directions AB�� (6, 2, �4) �

2(3, 1, �2) and AC�� (�3, �1, 2) � �1(3, 1, �2).

Since AB�� 2AC��, A, B, and C are collinear and

three collinear points do not define a unique plane.

b. The point P(8, �7, 5) is on the line r� � (4, 9, �3)

� t(1, �4, 2), (t � 4). Collinear points do not

define a unique plane.

11. The plane contains the line x � 7 � t, y � �2t,

z � �7 � t. A point on the plane will be A(7, 0, �7),

(t � 0), and a direction is a�� (�1, �2, 1) � �1(1,

2, �1). Since the plane does not intersect the z-axis,

the z-axis will be parallel to the plane, hence a second

direction will be b�� (0, 0, 1). A vector equation of

the plane is r� � (7, 0, �7) � s(0, 0, 1) � t(1, 2, �1).

Parametric equations are x � 7 � t, y � 2t, z � �7 �

s � t.

12. A plane has equations r� � (a, b, c) � s(d, e, f) �t(a, b, c). If s � 0 and t � �1, r � (0, 0, 0), hence the plane passes through the origin.

13. a.

A, B, and C are three points on the plane. Let two

directions be

AB�� a�� b�� and

AC�� a�� c��.

With point A, the vector equation of the plane

containing A, B, and C is

r� � a�� s(�a�� b��) � t(�a�� c��)

� (1 � s � t)a�� b��s � c��t

or r� � pa�� sb�� tc�� where p � 1 �s � t

or p � s � t � 1.

b. r� � (1 �s � t)a�� sb�� tc��, 0 � s � 1, 0 � t � 1.

Now if s � t � 0, r� � a��

s � 0, t � 1, r� � c��

s � 1, t� 0, r� � b��

s � t � 1, r� � �a�� b�� c��.

Now �a�� b�� c�� c�� a�� b��

� OC�� CD��, CD�� a�� b��

� OD��.

CD�� a�� b�� AB��, therefore ABCD is a

parallelogram and the fourth vertex D has position

vector OD�� a�� b�� c��. Hence the region in the

plane is all points in and on the parallelogram

whose vertices have position vectors a��, b��, �a�� b��

� c��, and c��.

14. a.

A line l has equation r� � r�0

� td��. r�0

is the position

vector of point R on the line. Q is a point not on l

and has position vector q��. Two directions of the

R

Q

l→

d

→r

→q

O

0

A

B

D

C

→a

→

b→c

�a +

b→

→

�a +

b→

→

c – a + b=�a + b + c

→ → →

→ → →

AB

C

O

→a

→

b

→c

110 Chapter 8: Equations of Planes

plane are d�� and RQ�� r�0

� q��.

The vector equation of this plane will be

r� � r�0

� l (�r�0

� q��) � td��

� (1 � l)r�0

� lq�� td��

r� � kr�0

� lq�� td��, 1 � l � k

or k � l � 1.

b. r� � kr�0

� lq�� td��, k � l � 1

r� � kr�0

� (1 � k)q�� td��, 0 � k � 1.

k � 0 gives the line r� � q�� td��. The line passes

through Q and has direction d��.

k � 1 gives the line r� � r�0

� td��. The line passes

through R and has direction d��.

Therefore the region of the plane determined by

0 � k � 1 is the region between and including the

two parallel lines through R and Q with direction d��.

Exercise 8.2

8. A plane contains the x-axis and the point A(4, �2, 1).

O(0, 0, 0) is on the x-axis, therefore a direction is

OA�� (4, �2, 1). A second direction is the direction

of the x�axis, i � (1, 0, 0). A normal is OA�� i � n��

� (0, 1, 2). P(x, y, z) is a point on the plane, hence

AP�� · n�� 0. The plane has equation

(x � 4, y � 2, z � 1) · (0, 1, 2) � 0y � 2z � 0.

9. A plane contains the intersecting lines

�x �

12

� � �2

y� � �

z �

3

3� and �

x

�

�

3

2� � �

4

y� � �

z �

23

�.

The common point is A(2, 0, �3). Two directions are

a�� (1, 2, 3) and b�� (�3, 4, 2). A normal to the

plane is n�� a�� b�� (�10, �11, 10) � �(10, 11,

�10). The scalar equation is AP�� · n�� 0

(x � 2, y, z � 3) · (10, 11, �10) � 0

10x � 11y � 10z � 50 � 0.

10. a. �1: x � 3y � z � 2 � 0, n��

1� (1, 3, �1).

�2: 2x � 6y � 2z � 8 � 0, n��

2� (2, 6, �2) �

2(1, 3, �1). Since n��2

� 2n��1, the planes are parallel.

Since (2, 6, �2, �8) � 2(1, 3, �1, �4) � 2(1, 3,

�1, �2), the two planes are distinct. Hence the

two planes are parallel and distinct.

b. �1: 2x � y � z � 3 � 0, n��

1� (2, 1, 1).

�2: 6x � 2y � 2z � 9 � 0, n��

2� (6, 2, 2) �

2 (3, 1, 1).

Since n��2

� kn��1, the two planes are distinct and they

intersect.

c. �1: 3x � 3y � z � 2 � 0, n��

1� (3, �3, 1)

�2: 6x � 6y � 2z � 4 � 0, n��

2� (6, �6, 2)

� 2(3, �3, 1). Since n��2

� 2n��1, the planes are

parallel. Now (6, �6, 2, �4) � 2(3, �3, 2, �2).

Therefore �2

� 2�1

and the planes are coincident.

d. �1: 2x � 4y � 2z � 6 � 0, n��

1� (2, �4, 2)

� 2(1, �2, 1).

�2: 3x � 6y � 3z � 9 � 0, n��

2� (3, �6, 3)

� 3(1, �2, 1).

n��2

� �2

3�n��

1, therefore the planes are parallel.

(3, �6, 3, �9) � 3(1, �2, 1, �3)

(2, �4, 2, �6) � 2(1, �2, 1, �3)

�2

� �23

��1, therefore the planes

are coincident.

11. a. �: 2x � y � 3z � 24 � 0. Let x � s, z � t.

Therefore y � �24 � 2s � 3t. A vector equation

is r� � (0, �24, 0) � s(1, 2, 0) � t(0, 3, 1).

b. �: 3x � 5z � 15 � 0. Solve for z: z � 3 � �3

5�x.

Let x � 5s, y � t, z � 3� 3s. A vector equation is

r� � (0, 0, 3) � s(5, 0, 3) � t(0, 1, 0).

12. �: 4x � y � z � 10 � 10. A normal to � is

n�� (4, 1, �1).

a. r� � (3, 0, 2) � t(1, �2, 2). A direction of the line

is d�� (1, �2, 2) since d�� · n�� 4 � 2 � 2 � 0.

The line is parallel to the plane. A point on the line

is A(3, 0, 2). Since A satisfies the equation of �,

the point is on the plane, hence the line lies on

the plane.


b. x � �3t, y � �5 � 2t, z � �10t. A direction of

the line is d�� (�3, 2, �10).

d�� · n�� 12 � 2 � 10 � 0, therefore the line

is parallel to the plane. A point on the line is

A(0, �5, 0), (t � 0). Since A does not satisfy the

equation of the plane, the line is not on the plane.

c. �x �

4

1� � �

y

�

�

1

6� � �

1

z�. A direction of the line is

d�� (4, �1, 1).

d�� · n�� 16 � 1 � 1 � 0. The line is not parallel

to the plane, hence it must intersect the plane at

some point.

13. a. �1: 2x � 3y � z � 9 � 0, n��

1� (2, 3, �1).

�2: x � 2y � 4 � 0, n��

2� (1, 2, 0).

The angle between n��1

and n��2

is where

n��1

· n��2

� �n��1��n��

2� cos

2 � 6 � �4 � 9�� 1� �1 � 4� cos

cos � ��14�

8

�5��

� 17.The angle between the planes is approximately 17.

b. �1: x � y � z � 1 � 0, n��

1� (1, �1, �1).

�2: 2x � 3y � z � 4 � 0, n��

2� (2, 3, �1).

n��1

· n��2

� 2 � 3 � 1 � 0, therefore n��1

⊥ n��2

and the

angle between the planes is 90.

14. a. l: x � 0, y � t, z � 2t, d�� (0, 1, 2)

�: 2x � 10y � 5z � 1 � 0, n�� (2, �10, 5)The line intersects the z-axis at (0, 0, 0).

The z-intercept is �1

5�.

Since d�� · n�� 0 � 10 � 10 � 0, the line is parallel

to the plane and since the z-intercept of the line is 0

and of the plane is �1

5�, the line lies below the plane.

b. �: x � 4y � 2z � 7 � 0, n�� (1, 4, �2)

d�� · n�� 0 � 4 � 4 � 0, the line is parallel to the

plane. The z-intercept of the plane is ��

2

7�.

Therefore the line lies above the plane.

15. a. P(x, y, z) is equidistant from A(1, 2, 3) and

B(4, 0, 1) therefore �AP�� BP�� and �AP��2 � �BP��2

therefore (x � 1)2 � (y � 2)2 � (z � 3)2 �

(x � 4)2 � y2 � (z � 1)2. Squaring and collecting

terms gives

�2x � 1 � 4y � 4 � 6z � 9 � �8x � 16 � 2z � 16x � 4y � 4z � 3 � 0.

b. 6x � 4y � 4z � 3 � 0 is a plane that has normal

coincident with AB�� and passes through the

midpoint of AB, i.e., AB�� (3, �2, �2),

n�� (6, �4, �4) � 2(3, �2, �2). n�� 2AB��. The

midpoint of AB is M��5

2�, 1, 2�

6��5

2�� 4(1) � 4(2) � 3 � 15 � 4 � 8 � 3 � 0.

M satisfies 6x � 4y � 4z � 3 � 0 therefore the

plane passes through the midpoint of AB.

16. a. B

CA

P

→

b

→n

→a

→r

→c

O


a��, b��, and c�� are position vectors of points A, B, and

C. ABC defines a plane with two directions

AB�� a�� b��

and AC�� a�� c��.

A normal to the plane is

AB�� AC�� (b�� a��) � (c�� a��)

� b�� c�� b�� a�� a�� c�� a�� a��.

But a�� a�� 0�� hence a normal is n�� b�� c�� b��

� a�� a�� c��. P(x, y, z) is a point on the plane and

r� is the position vector of P. AP�� a�� r�. Since

AP�� is in the plane and n�� is perpendicular to the

plane, AP�� · n�� 0.

(r� � a��) · (b�� c�� b�� a�� a�� c��) � 0

but �b�� a�� a�� b��, �a�� c�� c�� a��,

therefore the scalar equation of the plane through

A, B, and C is

(r� � a��) · (a�� b�� b�� c�� c�� a��) � 0.

b. A(8, 4, �3), B(5, �6, 1), C(�4, 1, 2). The position

vectors of A, B, and C are a�� (8, 4, �3),

b�� (5, �6, 1),

c�� (�4, 1, 2).

b�� c�� (�13, �14, �19)

�b�� a�� (�14, �23, �68)

�a�� c�� (�11, 4, �24),

r� � a�� (x � 8, y � 4, z � 3).From 16a(x � 8, y � 4, z � 3) · (�38, �33, �111) � 0

38x � 33y � 111z � 304 � 132 � 333 � 038x � 33y � 111z � 103 � 0.

17. �: 2x � 3y � kz � 0. The plane intersects the xy-plane when z � 0. Therefore 2x � 3y � 0. Let x � 3t, y � �2t. The line of intersection of the plane in the xy-plane isthe line x � 3t, y � �2t, z � 0. This line passesthrough the origin (i.e., t � 0 gives the point (0, 0, 0)).Since the equation of the line is independent of k, allplanes 2x � 3y � kz � 0, k � R, intersect thexy-plane in this line and the plane rotates about thisline as k varies.

18. The distance from P1(x

1, y

1, z

1) to the plane Ax � By

� Cz � D � 0 is given by

d � .

The distance from the origin to the plane is

d ��A2 �

�D

B��2 � C2��.

If n�� (A, B, C) is a unit vector, �A2 � B�2 � C2� � 1

and d � �D�.

Therefore if the coefficients A, B, and C are thecomponents of a unit normal, �D� will represent thedistance from the origin to the plane.

19. a, b, and c are the x-, y-, and z-intercept of a plane.These intercepts correspond to the points A(a, 0, 0),B(0, b, 0) and C(0, 0, c). Two directions of this plane

are BA�� (a, �b, 0) and CA�� (a, 0, �c).

A normal is BA�� CA�� n�� (bc, ac, ab).

The scalar equation is AP�� · n��

(x � a, y, z) · (bc, ac, ab) � 0bcx � acy � abz � abc � 0

divide by abc: �ax

� � �by

� � �cz

� � 1.

The distance from the origin to this plane is

1d � ––––––––––––––––

��a12� � ��b

12� � �

c�12��

or �1d

� � ��a12� � ��b

12� � �

c�12��

or �d1

2� � �a12� � �

b12� � �

c12�.

20. The scalar equation of the plane with x-, y-, andz-intercepts a, b, and c from question 19 is

�ax

� � �by

� � �cz

� � 1. Developing this equation using the

results of question 16:

�Ax1

� By1

� Cz1

� D��

�A2 � B�2 � C2�


Let the position vectors of the three intercepts be

a�� (a, 0, 0), b�� (0, b, 0), c�� (0, 0, c).

Now a�� b�� (0, 0, ab), b�� c�� (bc, 0, 0),

c�� a�� (0, ac, 0).

a�� b�� b�� c�� c�� a�� (bc, ac, ab)

r� � a�� (x � a, y, z)

and (r� � a��) · (a�� b�� b�� c�� c�� a��) � 0

(x � a, y, z) · (bc, ac, ab) � 0

bcx � acy � abz � abc � 0

Divide by abc: �ax

� � �by

� � �cz

� � 1 � 0

or �ax

� � �by

� � �cz

� � 1

represents the equation of the plane having a, b, and c as

x-, y-, and z-intercepts respectively.

Exercise 8.3

4. a. The plane r� � (6, �4, 3) � s(�2, 4, 7) � t(�7, 6, �3) has parametric equationsx � 6 � 2s � 7ty � �4 � 4s � 6tz � 3 � 7s � 3t.

(i) To intersect the x-axis, y � 0, z � 0 therefore 4s � 6t � 4 ➀7s � 3t � �3 ➁

➀ � 2 � ➁: 9s � �1

s � ��19

�, t � �2207�

x � 6 � �29

� � �12470

�

� �162 �

267

� 140�

x � �2287�.

The plane intersects the x-axis at ��2287�, 0, 0�.


(ii) To intersect the y-axis, x � z � 0 therefore2s � 7t � 6 ➀7s � 3t � �3 ➁

3 � ➀ � 7 � ➁: 55s � �3,

s � ��535�,

t � �4585�, and y � �

5565�.

The plane intersects the y-axis at �0, �5565�, 0�.

(iii) To intersect the z-axis, x � y � 0 therefore2s � 7t � 6 ➀2s � 3t � 2 ➁

➀ � ➁: 4x � 4, t � 1, s � ��12

�, and z � ��72

�.

The plane intersects the z-axis at (0, 0, ��72

�).

b. (i) For an intersection with the xy-plane, z � 0.

Therefore 3 � 7s � 3t � 0, t � 1 � �73

�s.

Let s � 3k then t � 1 � 7k. Substitute for x andy, hencex � 6 � 6k � 7 � 49kx � �1 � 55k;y � �4 � 12k � 6 � 42ky � 2 � 54k.The intersection with the xy-plane is the line

r� � (�1, 2, 0) � k(�55, 54, 0).

(ii) An intersection with yz-plane, x � 0. Therefore

6 � 2s � 7t � 0, s � 3 � �72

�t.

Let t � 2p, s � 3 � 7p. Substitute for y and zy � �4 � 12 � 28p � 12py � 8 � 16p;z � 3 � 21 � 49p � 6pz � 24 � 55p.

The intersection with the yz-plane is the liner� � (0, 8, 24) � p(0, 16, 55).

(iii)An intersection with the xz-plane, y � 0.

Therefore �4 � 4s � 6t � 0, s � 1 � �32

�t.

Let t � �2u then s � 1 � 3u. Substitute for x and zx � 6 � 2 � 6u � 14ux � 4 � 8u;z � 3 � 7 � 21u � 6u z � 10 � 27u.The intersection with the xz-plane is the liner� � (4, 0, 10) � u(8, 0, 27).

5. A line r� � (6, 10, 1) � t(3, 4, �1) has parametricequations x � 6 � 3ty � 10 � 4tz � 1 � t.

a. The line meets the xy-plane when z � 0. Thereforet � 1 and x � 9, y � 14. The point is (9, 14, 0).

b. The line meets the xz-plane at y � 0. Therefore

t � ��52

� and x � ��32

�, z � �72

�.

The point is ��32

�, 0, �72

��.

c. The line meets the yz-plane at x � 0. Thereforet � �2 and y � 2, z � 3. The point is (0, 2, 3).

6. a. A line parallel to the x-axis will intersect a planeperpendicular to the x-axis in one point.

b. A line parallel to the y-axis could intersect a planeparallel to the y-axis in an infinite number of pointsor in no points.

c. A line perpendicular to the z-axis could intersect aplane parallel to the z-axis is in one point, aninfinite number of points, or in no points.

7. The plane 3x � 2y � 7z � 31 � 0 has normal n�� (3, �2, 7), which is the direction of the line throughthe origin. An equation of the line is x � 3t, y � �2t,z � 7t. Solving with the plane gives

9t � 4t � 49t � 31 � 0, t � �12

�

and the point of intersection is ��32

�, �1, �72

��.

8. The plane 4x � 2y � 5z � 18 � 0 has normaln�� (4, �2, 5). A line through (6, �2, �2) with direction n�� has equation x � 6 � 4t, y � �2 � 2t,z � �2 � 5t. Solve with the plane4(6 � 4t) � 2(�2 � 2t) � 5(�2 � 5t) � 18 � 0

24 � 16t � 4 � 4t � 10 � 25t � 18 � 045t � �36

t � ��45

�.

The point of intersection is ��154�, ��

25

�, �6�.

9. a. 12x � 3y � 4z � 12 � 0. The x-, y-, andz-intercepts are 1, 4, and 3 respectively.

b. x � 2y � z � 5 � 0. The x-, y-, and z-intercepts are

5, ��52

�, and �5 respectively.

z

y

5

x

z

y

x

1

3

4


c. 2x � y � z � 8 � 0. The x-, y-, and z-interceptsare �4, 8, and �8 respectively.

d. 4x � y � 2z � 16 � 0. The x-, y-, and z-interceptsare 4, �16, and 8 respectively.

10. a. x � y � 4 � 0. The x-intercept is 4. They-intercept is 4. The intersection with thexy-plane:

x � u, y � 4 � u, z � 0xz-plane: x � 4, y � 0, z � syz-plane: x � 0, y � 4, z � t.

b. x � 3 � 0. The x-intercept is 3. The intersection with thexy-plane: x � 3, y � t, z � 0 xz-plane: x � 3, y � 0, z � s.

c. 2y � 1 � 0. The y-intercept is ��12

�. The intersection with the

xy-plane: x � t, y � ��12

�, z � 0

yz-plane: x � 0, y � ��12

�, z � s.

z

y

x

3

z

y

x

4

4

8

�16

4

z

y

x

x

y

z

�4

8

�8


d. 3x � z � 6 � 0. The x-intercept is 2. The z-intercept is 6. The intersection with the xy-plane: x � 2, y � t, z � 0 xz-plane: x � u, y � 0, z � 6 � 3uyz-plane: x � 0, y � s, z � 6.

e. y � 2z � 0. The y-intercept is 0. The z-intercept is 0. The intersection with the xy- and xz-plane is the x-axis: x � t, y � 0, z � 0; the intersection withthe yz-plane is x � 0, y � 2u, z � u.

f. x � y � z � 0. Thex x-, y-, and z-intercepts are 0.The intersection with thexy-plane: x � t, y � �t, z � 0 xz-plane: x � s, y � 0, z � syz-plane: x � 0, y � u, z � u.

11. Given the line l: �x �

3k

� � �y �

24

� � �z �

16

� and the

plane �: x � 4y � 5z � 5 � 0. The parametricequations of the line are: x � k � 3t, y � �4 � 2t,z � �6 � t. Substitute into the equation of the plane:k � 3t � 16 � 8t � 30 � 5t � 5 � 0

0t � 9 � k.

a. No value of k will give a unique value to t. Notethat the direction vector of the line is d�� (3, 2, 1),the normal of the plane is n�� (1, �4, 5) andn�� · d�� 3 � 8 � 5 � 0. Since the direction of theline is perpendicular to the normal, the line isparallel to the plane.

b. If k � 9, t � R and there will be an infinite numberof points. In this case the line is on the plane.

c. If k � 9, there will be no points of intersection.

12. See exercise 8.2, questions 19 and 20.

z

y

x

z

y

x

z

x

y

6

2

z

y

x

12

� –


Exercise 8.4

4. a. 3x � 7y � z � 12x � y � 2z � � 3The augmented matrix is

�.

b. �4x � 3y � 2z � 42y � 5z � 5The augmented matrix is

�.

c. x � 4z � 16y � 8z � �2The augmented matrix is

�.

d. 5y � 2z � 6x � 43z � 5y � 2x � �4The augmented matrix, with the coefficients of x, y, and z in the first, second, and third columnsrespectively, is

�.

5. a. � represents the system

x � 4z � 9,y � 6z � 4.

b. � represents the system

8x � 2y � 3z � �6,2x � 6y � 6z � 9.

c. � represents the system

5x � 10z � 8,3y � 4z � 6.

d. � represents the system

x � 4z � 0,y � 9z � 0.

6. a. The augmented matrix of the system is

�. R2

� R1 �

R1

� 2 � R2 �.

The final matrix corresponds to the equationsx � 15z � 10 or x � 10 � 15z

y � 4z � �3 y � �3 � 4z.The parametric equations of the line of intersectionresult when z is set equal to t. They are x � 10 �15t, y � �3 � 4t, z � t. The vector equation isr� � (10, �3, 0) � t(�15, 4, 1).

b. The augmented matrix of the system is

�. �2 � R1

� R2 �.

The final matrix corresponds to the equations

x � 4y � 3z � 5, 9y � �10, y � ��

910�.

Substituting into the first equation and setting

z � t, x � �490� � 3t � 5, x � �

59

� � 3t.

The parametric equation of the line of intersection

is x � �59

� � 3t, y � ��

910�, z � t. The vector

equation is r� � ��59

�, ��

910�, 0� � t(�3, 0, 1).

c. The augmented matrix of the system is

�. R1

� 2 � R2 �.


2x � 8y � 2z � 7, 4z � 1, z � �14

�.

Substituting into the first equation and letting y � t

2x � 9t � �12

� � 7, x � �143� � 4t.

The vector equation of the line of intersection is

r� � ��143�, 0, �

14

�� t(�4, 1, 0).

71

24

80

20

73

2�1

84

21

5�10

30

�49

10

50

36

�41

12

10�3

15�4

01

10

4�3

7�4

21

10

41

73

23

11

00

49

01

10

�86

�10�4

03

50

�69

3�6

�2�6

82

94

4�6

01

10

4�4

�23

55

6�2

16�2

4�8

01

10

�45

2�5

32

�40

12�3

1�2

�71

31


d. The augmented matrix of the system is

�.

3 � R1

� 4 � R2 �.

The final matrix corresponds to the equations 4x � 8y �3z � 6, �5z � 10, z � �2. Substituting into the first equation and lettingy � t, 4x � 8t � 6 � 6, x � 2t. The vector equation of the line of intersection is

r� � (0, 0, �2) � t(2, 1, 0).

e. The augmented matrix of the system is

�.

2 � R1

� 3 � R2 �.

R2

� (�5) �.

R1

� 2 � R2 �.

The final matrix corresponds to the equations 3x � 21, y � 3z � �8, x � 7.

Substituting z � t into the second equation,y � �8 � 3t. The vector equation of the line ofintersection is r� � (7, �8, 0) � t(0, 3, 1).

f. The augmented matrix of the system is

�.

5 � R1

� 3 � R2 �.

R2

� 46 �.

R1

� 8 � R2 �.

R1

� 3 �.

The final matrix corresponds to the equations 2x � z � 3, y � 0. Substituting x � t into the first equation,2t � z � 3, z � �3 � 2t. The vector equation of the line of intersection is

r� � (0, 0, �3) � t(1, 0, 2).

7. a. �.

�.

�.

z � �11554

�, y � �11056

�, x � ��

32053�.

The three planes intersect at the point

��32053�, �

11056

�, �11554

��.

b. � 4 � R1

� R2 �.3 � R

1� R

3

22 � R2

� 35 � R3 �.

The three lines in R2

are not concurrent.

c. �R

1� 2 � R

2 48 � R

1� 3 � R

3 5 � R

1� 6 � R

4

�10�34�5104

�4�12�19

10

315333

6000

102215

�9

�441

�5

3�6�7

2

6385

522

�111

8350

�100

52217

83522

�100

52

�2

832

�14

�3

520

154

�2�715

6130

2009 � R

2� 13 � R

3

5202

�2�7�6

6139

200

3 � R1

� R2

R1

� R2

5�5

3

�214

65

�3

262

�30

�10

01

20

90

�30

01

60

90

�30

81

60

90

�30

846

60

915

�3�5

8�2

610

21�8

0�3

01

30

5�3

�6�3

21

30

�540

�615

2�5

30

5�10

�6�9

23

32

610

�3�5

�80

40

6�2

�31

�86

4�3


�11 � R2

� 5 � R3

R2

� 5 � R4

�31 � R

3� 37 � R

4

�.

The four planes are not concurrent.

8. �1: A

1x � B

1y � C

1z � D

1� 0.

�2: A

2x � B

2y � C

2z � D

2� 0.

�1

and �2

are two nonparallel planes in space. Now

A1x � B

1y � C

1z � D

1� k(A

2x � B

2y � C

2z � D

2)

� 0 ➀

can be written as (A1

� kA2)x � (B

1� kB

2)y �

(C1

� kC2)z � D

1� kD

2� 0, which is of the form of

a plane. Any point P1(x

1, y

1, z

1) that satisfies the

equation of �1

and also �2

will be on the line of

intersection of �1

and �2; i.e., A

1x

1� B

1y

1� C

1z

1�

D1

� 0 and A2x

1� B

2y

1� C

2z

1� D

2� 0. Also P

1

satisfies ➀ since substituting gives

L.S. � A1x

1� B

1y

1� C

1z

1� D

1� k(A

2x

1� B

2y

1�

C2z

1� D

2)

� 0 � k(0)

� 0

� R.S.

Therefore all members of the family of planes

represented by ➀ will also pass through the line of

intersection of �1

and �2.

Note: If k � 0, we get the plane �1; however, no value

of k gives �2.

b. The equation of the family of planes passingthrough the line of intersection of the given planesis 3x � 4y � 7z � 2 � k(2x � 3y � 4) � 0. Sincethe required plane contains the origin, then (0, 0, 0)must satisfy the equation. Therefore, substituting(x, y, z) � (0, 0, 0), we get�2 � 4k � 0,

k � ��12

�.

Substituting into the family

3x � 4y � 7z � 2 � �12

�(2x � 3y � 4) � 0

4x � 5y � 14z � 0. The equation of the required plane is4x � 5y � 14z � 0.

c. The line with equation x � 2y � 3z can be written

as �6x

� � �3y

� � �32

�. Therefore a direction is d�� (6, 3, 2).

The equation of the family of planes passingthrough the intersection of the given planes is4x � 3y � 5z � 10 � k(4x � y � 3z � 15) � 0(4 � 4k)x � (�3 �k)y � (�5 � 3k)z � 10 � 15k � 0.

Each plane of the family has normal n�� (4 � 4k,

�3 � k, �5 � 3k). Since the line is parallel to the

required plane, d�� ⊥ n�� and d�� · n�� 0,(6, 3, 2) · (4 � 4k, �3 � k, �5 � 3k) � 0

24 � 24k � 9 � 3k � 10 � 6k � 015k � �5

k � ��13

�.

Substituting, we get

4x � 3y � 5z � 10 � �13

�(4x � y � 3z � 15) � 012x � 9y � 15z � 30 � 4x � y � 3z � 15 � 0

8x � 8y � 12z � 15 � 0Therefore, the equation of the required plane is8x � 8y � 12z � 15 � 0.

9. Given the plane �: r� � (�2, 1, 3) � s(5, �2, �2) �

t(�1, 0, 1). Two directions of � are a�� (5, �2, �2)

and b�� (�1, 0, 1). A normal to this plane is a�� b��

� (�2, �3, �2) � �1(2, 3, 2). Since the line l: r� �

(9, �1, �5) � p(2, �2, 2) is on the required plane, a

second direction will be (2, �2, 2) � 2(1, �1, 1).

10�34349570

�4�12

370

31500

6000

10�34349277

�4�12

3731

31500

6000

�1 � R3

R4

� (�2)

10�34

�349�554

�4�12�37�62

31500

6000


A normal to the required plane is (2, 3, 2) � (1, �1, 1)

� (5, 0, �5) � 5(1, 0, �1).

A point on the required plane is any point on the given

line (9, �1, �5).

The required plane has equation

(x � 9, y � 1, z � 5) · (1, 0, �1) � 0x � z � 14 � 0.

Exercise 8.5

5. The matrix forms of the given systems are:

a. �b. �c.

� �

6. The systems of equations from the given matrices are:a. x � 8

y � �6z � 3

b. x � 6z � 4y � 5z � �50z � 0

c. x � 0y � 00z � 1

7. The augmented matrix of the given system is

�.

Now

�.

1125

�45

4�4

�18

�60

42

200

R1

� 2 � R2

R3

� 4 � R1

11�7�1

44

�2

�6�318

218

121510

306

450

024

08

�6

�302

153

�210

512

�3

1�5

2

�21

�5

531


Interchange

�.

�.


4z � �25, z � ��

425�

14y � 6z � �152x � 6y � 4z � 11

Substituting gives 14y � 6��

425�� 15,

y � ��145�

2x � 6��

415�� 4 ��

�

425�� 11, x � �

247�.

The planes intersect in the point

��247�, �

�

415�, �

�

425��.

8. a. The augmented matrix of the system is

�.

Now

�.

R2

� R3

�.

The final matrix corresponds to the equations�4z � �16, z � 4

5y � z � 19, 5y � 4 � 19, y � 3x � 2y � z � 12, x � 6 � 4 � 12, x � 2The three planes intersect in the point (2, 3, 4). Thesolution is unique.

1219

�16

11

�4

250

100

121935

115

255

100

2 � R1

� R2

3 � R1

� R3

1251

11

�2

2�1

1

123

11�15�25

4�6

4

�6140

200

R2

� 3R

3� (�1)

11�45

25

4�18�4

�6420

200R

2and R

3

b. The augmented matrix of the system is

�.

Now

�.

There are no values satisfying the equation from R3,

i.e., 0z � 1. Therefore the three planes do not

intersect. Note that the normals,

n��1

� �12

�n��2

� �13

�n��3, are collinear and the three planes

are parallel and distinct.

c. The augmented matrix of the system is

�.

Now

�.

�.


x � y � z � 5z � 2

and 0x � 0y � 0z � 0. Substituting z � 2 into the first equation and lettingy � t gives x � t � 2 � 5, x � 7 � t. The threeplanes intersect in a line with vector equation

r� � (7, 0, 2) � t(�1, 1, 0). There are an infinitenumber of solutions.

d. The augmented matrix of the system is

�.

Now

�.

Each row represents the same equation, x � 2y �3z � 1, hence the three planes are coincident andthere are an infinite number of solutions.

e. The augmented matrix of the system is

�.

Now

�.

The third row corresponds to the equation 0x � 0y� 0z � 1 or 0z � 1. There are no values for thevariable that will satisfy the equation, thereforethere are no solutions to the system of equations.The planes x � y � 2z � 2 and 3x � 3y � 6z � 5are parallel and distinct; the plane x � y � 2z � 5intersects these two planes.

f. The augmented matrix of the system is

�.

Now

�.

There is no solution to the system. The two planes2x � 6y � 10z � 18 and x � 3y � 5z � 9 arecoincident. The other plane, x � 3y � 5z � 10, isdistinct and parallel to the coincident planes.

g. The augmented matrix of the system is

�.

Now

�.

�.

�9�2

0

�210

�320

100

R2

� 7R

2� R

3

9�14�14

�277

�31414

100

R2

� R1

R3

� 2 � R1

9�5

4

�253

�3118

112

1020

500

300

100

2 � R1

�R2

R2

� 2 � R3

10189

5105

363

121

2�3

1

240

120

100

R1

� R2

3 � R1

� R3

255

2�2

6

1�1

3

113

111

�3�3�3

�2�2�2

111

R1

� (�1)R

2� 4

�241

6�12�3

4�8�2

�241

520

�110

100

100

R2

� 2R

2� 2 � R

3

542

�121

100

100

2 � R1

� R2

R1

� R3

563

�1�4�2

121

121

411

200

�100

100

2 � R1

� R2

3 � R1

� R3

47

11

246

�1�2�3

123


The final matrix corresponds to the equations x � 3y � 2z � 9

2y � z � �20z � 0, z � 2t.

Substitute into the second equation2y � 2t � �2, y � �1 � t and x � 3(�1 � t) �2(2t) � 9

x � 3 � 3t � 4t � 9, x � 6 � t.

There is an infinite number of solutions. The three

planes intersect in a line with equation r� � (6, �1,

0) � t(1, �1, 2).

h. The augmented matrix of the system is

�.

Now

�.

�.

There is no solution. Since no two planes areparallel, their intersection forms a triangular prism.

i. The augmented matrix is �.

Now

�.

R2

� 5 � R3

�.

R3

� (�18) �

.

The final matrix corresponds to the equationsz � 0, 5y � z � 0, 5y � 0, y � 0

2x � y � z � 0, 2x � 0, x � 0There is a unique solution. The three planesintersect in a single point, the origin, (0, 0, 0).

9. The three planes �1: x � 2y � z � 0

�2: x � 9y � 5z � 0

�3: kx � y � z � 0.

We find the line of intersection between �1

and �2.

Subtracting these equations gives 11y � 4z � 0,

y� �141z�.

Let z � 11t, then y � 4t. Substitute to find x:x � 2(4t) � (11t) � 0, x � 19t. �

1and �

2intersect in

a line with equation x � 19t, y � 4t, z � 11t. For theplanes to intersect in a line, this line must lie on �

3.

Therefore k(19t) � 4t � 11t � 019kt � �7t

k � ��

197�, t � 0.

The planes intersect in a line when k � ��179�. The

equation of this line is r� � t(19, 4, 11).

Review Exercise

1. a. The line with equation x � z, y � 0 has direction

vector d�� (1, 0, 1). A plane perpendicular to the

x-axis has normal n�� (1, 0, 0). If the plane

contains the line then the direction of the line will

be perpendicular to the normal, i.e., d�� · n�� 0. But

d�� · n�� (1, 0, 1) · (1, 0, 0) � 1 ≠ 0. Therefore a

plane perpendicular to the x-axis cannot contain

the line x � z, y � 0.

b. A plane parallel to the yz-coordinate plane will

have normal parallel to the x-axis, n�� (1, 0, 0).

Equation of the plane passing through A(�4, 0, 5)

with normal n�� is (x � 4, y, z � 5) · (1, 0, 0) � 0;

x � 4 � 0 is a plane parallel to the yz-coordinate

plane and containing the point (�4, 0, 5).

2. a. The plane passes through A(�1, �1, 2) and is

parallel to the plane r� � (2, �1, 0) � s(5, 4, 2) �

t(0, 0, 1). Two directions of the plane are (5, 4, 2)

and (0, 0, 1). The vector equation of the plane is

r� � (�1, �1, 2) � s(5, 4, 2) � t(0, 0, 1).

Parametric equations are

x � �1 � 5s,

y � �1 � 4s,

z � 2 � 2s � t.

000

171

150

200

000

17

�18

150

200

000

17

�5

15

�1

200

R1

� 2 � R2

3 � R1

� 2 � R3

000

1�3

4

1�2

2

213

64

�1

230

110

100

R2

� 2R

2� R

3

68

�9

26

�6

12

�2

100

R1

� R2

3 � R1

� R3

6�227

2�412

1�1

5

113


b. The plane passes through A(1, 1, 0) and B(�2, 0,

3) and is parallel to the y-axis. The direction of the

y-axis is i � (0, 1, 0). A second direction is BA��

(3, 1, �3). A vector equation of this plane is r� �

(1, 1, 0) � s(0, 1, 0) � t(3, 1, �3). Parametric

equations are x � 1 � 3t, y � 1 � s � t, z � �3t.

c. The plane has x-, y-, and z-intercepts �2, �3, and

4 respectively. Therefore three points that the plane

passes through are A(�2, 0, 0), B(0, �3, 0), and

C(0, 0, 4). Two directions of the plane are AB��

(2, �3, 0) and AC�� (2, 0, 4) � 2(1, 0, 2). A

vector equation of the plane is r� � (0, 0, 4) �

s(2, �3, 0) � t(1, 0, 2) and parametric equations

are x � 2s � t, y � �3s, z � 4 � 2t.

d. The plane contains the point A(1, 1, 1) and the line

�3x

� � �4y

� � �5z

�. Since the plane contains the line, the

direction of the line, (3, 4, 5), is also a direction of

the plane. A point on the line is B(0, 0, 0), hence a

second direction is BA�� (1, 1, 1). A vector

equation of the plane is r� � s(1, 1, 1) � t(3, 4, 5)

and parametric equations are x � s � 3t, y � s �

4t, z � s � 5t.

e. The plane contains the two intersecting lines

r� � (3, �1, 2) � s(4, 0, 1) and r� � (3, �1, 2) �

t(4, 0, 2). Since the plane contains these lines, the

direction of the lines, (4, 0, 1) and (2, 0, 1), will be

the direction of the plane. A point on both lines is

(3, �1, 2). A vector equation of the plane is r� �

(3, �1, 2) � s(4, 0, 1) � t(2, 0, 1) and parametric

equations are x � 3 � 4s � 2t, y � �1, z �

2 � s � t.

3. a. The plane passes through A(1, 7, 9) and has normal

n�� (1, 3, 5). The scalar equation is AP�� · n�� 0,

(x � 1, y � 7, z � 9) · (1, 3, 5) � 0

x � 3y � 5z � 67 � 0.

b. The plane passes through the points A(3, 2, 3),

B(�4, 1, 2), and C(�1, 3, 2). Two directions of the

plane are CA�� (4, �1, 1) and BC�� (3, 2, 0). A

normal to the plane is CA�� BC�� (�2, 3, 11).

The scalar equation is

(x � 3, y � 2, z � 3) · (2, �3, �11) � 02x � 3y � 11z � 33 � 0.

c. The plane passes through the point A(0, 0, 6) and

parallel to the plane y � z � 5. The family of

planes parallel to y � z � 5 is y � z � D. Since

A(0, 0, 6) lies on this family, substituting gives

0 � 6 � D, D � 6. The required plane has

equation y � z � 6 or y � z � 6 � 0.

d. The plane contains the point A(3, �3, 0) and the

line x � 2, y � 3 � t, z � �4 � 2t. The direction

of the line, d�� (0, 1, �2), is also a direction of the

plane. A point on the line, B(2, 3, �4), gives a

second direction BA�� (1, �6, 4). A normal to the

plane is d�� BA�� (�8, �2, �1). The equation of

the plane is (x � 3, y � 3, z) · (8, 2, 1) � 0

8x � 2y � z � 18 � 0.

e. The plane contains the line r� � (2, 1, 7) � s(0, 1,

0). Therefore a point it passes through is A(2, 1, 7)

and a direction is a�� (0, 1, 0). Since it is parallel

to the line r� � (3, 0, 4) � t(2, �1, 0), a second

direction is b�� (2, �1, 0). A normal to the plane

is a�� b�� (0, 0, �2). The equation of the plane is

(x � 1, y � 1, z � 7) · (0, 0, 1) � 0; z � 7 � 0.

f. The plane contains the points A(6, 1, 0) and B(3, 0,

2). One direction is BA�� (3, 1, �2). It is also

parallel to the z-axis, therefore a second direction is

k � (0, 0, 1). A normal to the plane will be k � BA��

� (1, �3, 0) and the scalar equation is (x � 6,

y � 1, z) · (1, �3, 0) � 0, x � 3y � 3 � 0.

4. Given the planes �1: 3x � ky � z � 6 � 0 with

normal n��1

� (3, k, 1) and �2: 6x � (1 � k)y � 2z � 9

� 0 with normal n��2

� (6, 1 � k, 2).


a. If the planes are parallel, the normals will be scalar

multiples and n��1

� an��2.

Therefore (3, k, 1) � a(6, 1 � k, 2)

3 � 6a, k � a(1 � k), 1 � 2a

a � �12

� a � �12

�

Since a � �12

�, k � �12

�(1 � k)

2k � 1 � k

k � �13

�.

The planes are parallel for k � �13

�.

b. If the planes are perpendicular, their normals will

be perpendicular. Since n��1

⊥ n��2, n��

1· n��

2� 0,

(3, k, 1) · (6, 1 � k, 2) � 0

18 � k � k2 � 2 � 0

k2 � k � 20 � 0

(k � 5)(k � 4) � 0

k � 5 or k � �4.

The planes are perpendicular for k � 5 or �4.

5. A plane contains the parallel lines

l1: x � 1, �

y �

43

� � �32

� and l2: x � 5, �

y �

25

� � �z �

13

�.

A point on l1

is A(1, 3, 0) and on l2

is B(5, �5, 3).

A and B are also on the required plane, hence one

direction of the plane is AB�� (4, �8, 3). Since both

lines are on the plane, a second direction of the plane

is the direction of the line, d�� (0, 2, 1). A normal to

the plane is d�� AB�� (14, 4, �8) � 2(7, 2, �4).

The scalar equation of the required plane is

(x � 1, y � 3, z) · (7, 2, �4) � 0

7x � 2y � 4z � 13 � 0.

6. Since the required plane is perpendicular to �: x � 2y

� z � 3 � 0, the normal n�� (1, 2, �1) will be a

direction vector. Since it passes through the origin,

O(0, 0, 0), and A(2, �3, 2), a second direction is

OA�� (2, �3, 2). A vector equation of the required

plane is r� � s(1, 2, �1) � t(2, �3, 2).

7. A plane is parallel to vectors a�� 6k � (0, 0, 6) and

b�� i � 2j � 3k � (1, 2, �3). a�� and b�� will be two

directions of the plane and a normal will be a�� b��

(�12, 6, 0) � �6(2, �1, 0). The plane passes through

A(1, 2, 3), therefore the scalar equation will be

(x � 1, y � 2, z � 3) · (2, �1, 0) � 0

2x � y � 0.

8. A line passes through the origin, O(0, 0, 0), and the

point A(1, �3, 2). Since the line is perpendicular to

the plane, a normal will be OA�� (1, �3, 2). The

plane passes through A, therefore (x � 1, y � 3,

z � 2) · (1, �3, 2) � 0 and the scalar equation of the

plane is x � 3y � 2z � 14 � 0.

9. Two lines, l1: �

x �

21

� � �y �

31

� � �z�

�

11

�, direction d��1

� (2, 3, �1) and l2: �

x�

�

11

� � �y �

51

� � �z �

41

�, direction

d��2

� (�1, 5, 4) intersect at the point A(1, 1, 1). A

normal to the plane containing l1

and l2

is d��1

� d��2

�

(17, �7, 13). Now (x � 1, y � 1, z � 1) · (17, �7,

13) � 0 and the scalar equation of the plane

containing the intersecting lines l1

and l2

is 17x � 7y

� 13z � 23 � 0.

10. The line r� � (�4, �3, �1) � t(2, 8, 3) passesthrough the point A(2, 21, 8) (t � 3 will give the pointA). A point and two non-collinear directions define aunique plane. Since A is on the given line, only onedirection is known, hence the equation of a planecannot be determined.

11. The distance from a point P1(x

1, y

1, z

1) to a plane

Ax � By � Cz � D � 0 is given by

d ��Ax

�1

�

A

B

2

y

�

1�

B�2

C

�

z1

C

�

2�D�

�.


a. The distance from the point P1(7, 7, �7) to the

plane by �z � 5 � 0 is

d �

D � ��5437��.

b. Point P1(3, 2, 1) and the plane �: 3x � 2y � z �

10. The distance from P1

to � is

d ��9 �

�4

9

�

�

1

4��

1�10�

��

414��.

c. The line l: r� � (1, 3, 2) � t(1, 2, �1) has direction

d��1

� (1, 2, �1). The plane �: y � 2z � 5 has

normal n�� (0, 1, 2). Since d��1

· n�� 0, the line is

parallel to the plane. The distance between the line

l and the plane � will be the distance from a point

on l, A(1, 3, 2) to the plane.

Therefore d � ��3

�

�

1

4

�

�

4�

5��

�2

5��.

The distance between the line and the plane is ��2

5��.

d. The plane �1: x � 2y � 5z � 10 � 0 has normal

n��1

� (1, 2, �5) and the plane �2: 2x � 4y � 1�z

� 17 � 0 has normal n��2

� (2, 4, �10) � 2(1, 2,

�5). Since n��2

� 2n��1, the planes are parallel, hence

the distance between the planes is the distance

from a point on �1, say A(10, 0, 0), and �

2.

d ��4

�2

�

0

1

�

6�

1

�

7�

100��

2�1 �

3

4�� 25��

2�

3

30��.�

The distance between the planes is �2�

3

30��.

12. The scalar equation of the plane having x-intercept�1, y-intercept 2, and z-intercept 3 is

��

x1� � �

2y

� � �3z

� � 1 or 6x � 3y � 2z � 6 � 0.

The distance from A(1, �2, �2) to this plane is

d � �6

�

�

36

6

�

�

9�

4

�

�

4�

6� � �

272�.

13. A normal to the plane �: 4x � 2y � 5z � 9 � 0 is

n�� (4, �2, 5). An equation of a line through the

origin with direction n�� will be x � 4t, y � �2t, z �

5t. Substituting into � gives 16t � 4t � 25t � 9 � 0,

t � �495� � �

15

�. The normal through the origin intersects

� at the point ��45

�, ��

52�, 1�.

14. The x-, y-, and z-intercepts of the plane �: 4x �5y � z � 20 � 0 are �5, �4, and 20 respectively.

z

x

y

20

�4

�5

�0(7) � 6(7) � 1(�7) � 5��

�02 � 6�2 � (��1)2�


15. a. 2x � y � z � 3 � 0.The x-, y-, and z-intercepts are �

32

�, 3, and 3,respectively.

b. 3y � 4z � 24 � 0.The y-intercept is �8, the z-intercept is 6.

c. 3z � 9 � 0. The z-intercept is �3, and the plane is parallel tothe xy-plane.

d. r� � (4, �5, 0) � s(�12, 9, 8) � t(8, �7, �8).

Two directions are a�� (�12, 9, 8)

and b�� (8, �7, �8).

A normal is n�� a�� b�� (�16, �32, 12)

� �4(4, 8, �3)(x � 4, y � 5, z) · (4, 8, �3) � 0

4x � 8y � 3z � 24 � 0x-, y-, and z-intercepts are �6, �3, and 8respectively.

16. The line l: x � �5 � 3t, y � 3 � 4t, z � 1 � 5t

passes through the point A(�5, 3, 1) and has direction

d�� (�3, �4, 5). The plane �: 2x � y � 2z � 5 � 0

has normal n�� (2, 1, 2). Since d�� · n�� 6 � 4 � 10

� 0, d�� ⊥ n��, hence the line is parallel to the plane.

Since 2(�5) � 3 � 2(1) � 5 � 0, the point A is on

the plane. Since a point of the line is on the plane

and the line is parallel to the plane, the line lies

on the plane.

17. The plane �1: 2x � 6y � 4z � 3 � 0 has normal

n��1

� (2, �6, 4) � 2(1, �3, 2) and the plane �2:

3x � 9y � 6z � k � 0 has normal n��2� (3, �9, 6)�

3(1,�3, 2).

Since n��1

� �23

�n��2, the planes are parallel.

3�1

� 2�2

� 0, therefore 9 � 2k � 0, k � �92

�.

z

y

x

�3

– 6

8

�3

z

y

x

z

y

x

�8

6

3

3

32–

z

x

y


a. Since the planes are parallel, they will not intersect

for k � �92

�.

b. The planes will never intersect in a line.

c. If k � �92

�, the two planes are coincident, hence

intersect in a plane.

18. A plane passes through the points A(1, 0, 2),

B(�1, 1, 0), and has a direction a�� (�1, 1, 1).

a. A second direction is BA�� (2, �1, 2) and a

normal to the plane is a�� BA�� (3, 4, �1).

Now (x � 1, y, z � 2) · (3, 4, �1) � 03x � 4y � z � 1 � 0.

The scalar equation of the plane is 3x � 4y � z � 1 � 0.

b. A line through Q(0, 3, 3) perpendicular to the plane

has direction n�� (3, 4, �1).

An equation of this line is

r� � (0, 3, 3) � t(3, 4, �1).

c. The parametric equations of the line are x � 3t,y � 3 � 4t, z � 3 � t. Solve by substituting thesevalues into the equation of the plane.3(3t) � 4(3 � 4t) � (3 � t) � 1 � 0

t � ��143�. The perpendicular through Q intersects

the plane at A��1132

�, �2133�, �

4133��.

d. The distance from A to the plane is given by

d �

� ��

�26�

d � 0.The distance from A to the plane is 0, implies thatA is on the plane.

19. Plane �1: x � 2y � 7z � 3 � 0 has normal n��

1�

(1, 2, �7), and plane �2: x � 5y � 4z � 1 � 0 has

normal n��2

� (1, �5, 4). A direction of the line of

intersection of the two planes is n��1

� n��2

�

(�17, �11, �7) � �1(27, 11, 7). A plane through

A(3, 0, �4) perpendicular to the line of intersection of

�1

and �2

has n��1

� n��2

as a normal. Therefore

(x � 3, y, z � 4) · (27, 11, 7) � 027x � 11y � 7z � 53 � 0

is the equation of the plane through A(3, 0, �4) and perpendicular to the line of intersection of thegiven planes.

20. a. The family of planes passing through the line ofintersection of the planes x � y � z � 1 � 0 and 2x � 3y � z � 2 � 0 is x � y � z � 1 �k(2x � 3y � z � 2) � 0. To find the particularmember that passes through the origin set (x, y, z) � (0, 0, 0). Now �1 � 2k � 0

k � �12

�.

The particular plane is

x � y � z � 1 � �12

�(2x � 3y � z � 2) � 0

2x � 2y � 2z � 2 � 2x � 3y � z � 2 � 04x � y � z � 0.

b. A normal to the plane 4x � y � z � 0 is n��1

�

(4, �1, 1) and a normal to the plane x � z � 0 is

n��2

� (1, 0, �1). The angle between the planes

is the angle between the normals. Therefore

n��1

· n��2

� � n��1�� n��

2� cos

4 � 1 � �16 � 1� � 1� �2� cos

cos � �36

� � �12

�

� 60and the angle between the planes is 60.

�36 � 92 � 43 �13��

13

��3��1132

�� 4��2133��

4133�� 1�

��9 � 16� � 1�


21. Plane �1: r� � (4, 0, 3) � t(�8, 1, �9) � u(�1, 5, 7)

has directions a��1

� (�8, 1, �9), b��1

� (�1, 5, 7) and

normal n��1

� a��1

� b��1

� (52, 65, �39) � 13(4, 5, �3).

The scalar equation is (x � 4, y, z � 3) · (4, 5, �3) � 04x � 5y � 3z � 7 � 0.

Plane �2: r� � (�14, 12, �1) � p(1, 1, 3) � q(�2, 1, �1),

has directions a��2

� (1, 1, 3), b��2

� (�2, 1, �1) and

normal n��2

� a��2

� b��2

� (�4, �5, 3) � �1(4, 5, �3).

The scalar equation is

(x � 14, y � 12, z � 1) · (4, 5, �3) � 04x � 5y � 3z � 56 � 60 � 3 � 0

4x � 5y � 3z � 7 � 0Since the scalar equations of both planes is the same, theplanes are coincident.

22. a. x � 5y � 8 ➀5x � 7y � �8 ➁

5 � ➀ � ➁: 32y � 48

y � �32

�, x � �12

�.

The two lines in R2 intersect in the point ��12

�, �32

��.

b. �1: 2x � 2y � 4z � 5, n��

1� (2, �2, 4) � 2(1, �1, 2).

�2: x � y � 2z � 2, n��

2� (1, �1, 2).

Since n��1

� 2 n��2, the planes are parallel. Since

�1

� 2 � 2

� 1 � 0 the two planes will be distinct.

c. �1: 3x � 2y � 4z � �1 ➀

�2: 2x � y � z � �3 ➁

➀ � 2 � ➁: 7x � 6z � �7

x � �1 � �67

�z

Let z � 7t, x � �1 � 6t. Substitute into ➁�2 � 12t � y � 7t � �3

y � 1 � 5t.The two planes intersect in a line with equation

r� � (�1, 1, 0) � t(6, 5, 7).

d. x � 2y � 3z � 11 ➀2x � y � 7 ➁

3x � 6y � 8z � 32 ➂2 � ➀ � ➁: 3y � 6z � 15, y � 2z � 5.3 � ➀ � ➂: �z � 1, z � �1, y � 3, x � 2.The three planes intersect at the point (2, 3, �1).


e. �1: x � y � 3z � 4, n��

1� (1, �1, 3).

�2: x � y � 2z � 2, n��

2� (1, 1, 2).

�3: 3x � y � 7z � 9, n��

3� (3, 1, 7).

Since no two normals are collinear, no two planesare parallel.➀ � ➁: 2x � 5z � 6 ➃

➀ � ➂: 4x � 10z � 13 ➄2 � ➃ � ➄: 0z � �1

There is no solution and the planes intersect to forma triangular prism.

f. �1: x � 3y � 3z � 8

�2: x � y � 3z � 4

�3: 2x � 6y � 6z � 16

Since 2 � �1

� �3

� 0, �1

and �3

are coincident.

�1

� �2: 4y � 4, y � 1 Substitute into �

2

x � 1 � 3z � 4

x � 5 � 3z.

Let z � t, x � 5 � 3t.

The planes intersect in the line with equation

r� � (5, 1, 0) � t(�3, 0, 1).

g. The augmented matrix of the system is

�.

Now

�.

�.

The last row corresponds to the equation0z � 0. Let z � �5t (to avoid fractions)

then 5y � 3z � �10, y � �2 � �35

�z, y � �2 � 3t.

x � 2y � z � �3, x � 4 � 6t � 5t � � 3,x � 1 � t.Letting t � 0, y � � 2, x � 1, z � 0r� � (1, �2, 0) � t(�1, 3, �5).

�3�10

0

130

250

100

�1 � R2

R2

� R3

�310

�10

1�3

3

2�5

5

100

R1

� R2

2 � R1

� R2

�3�13

4

14

�1

27

�1

112

h. �1: 3x � 3z � 12.

�2: 2x � 2z � 8.

�3: x � z � 4.

The three planes are coincident with the planex � z � 4.

i. �1: x � y � z � � 3.

�2: x � 2y � 2z � �4.

�3: 2x � 2y � 2z � �5.

�1

and �3

are parallel and distinct. �2

intersects

both �1

and �3

in two parallel lines.

Chapter 8 Test

1. a. Two planes, with normals satisfying n��1

· n��2

� 0,will be perpendicular to each other and intersect ina line.

b. Two planes, with normals satisfying n��1

� n��2

� 0,will be parallel.

c. Three planes, with normals satisfying n��1

� n��2

· n��3

� 0, will be parallel to each other.

2. The plane �: 4x � y � z � 10 � 0 has normaln�� (4, 1, �1)

a. The line l: x � �3t, y � �5 � 2t, z � �10t has

direction d�� (�3, 2, �10). Since d�� · n�� 12

� 2 � 10 � 0, the line is parallel to the plane. A

point on the line, A(0, �5, 0), (t � 0) does not

satisfy the equation of the plane, therefore the line

does not coincide with the plane.

b. The line �x �

42

� � �y �

12

� � ��

z1� has direction

d�� (4, 1, �1) and passes through the point A(2, 2,

0). Since A satisfies the plane, and d�� n��, the line

intersects the plane, at right angles, at the point

A(2, 2, 0).

3. Three planes intersect in a point A.

Three planes intersect in a line.

The three planes are coincident, thus intersect in a plane.

Two planes are coincident and the third planeintersects them in a line.

�3

�2

�1

line of intersection

�2

�3

�1

line ofintersection

A


4. The plane r� � (0, 0, 5) � s(4, 1, 0) � t(2, 0, 2) hasparametric equation x � 4s � 2ty � sz � 5 � 2t.

a. For an intersection with the x-axis, y � z � 0,

therefore s � 0, t � ��

25� and x � �5 and the point is

(�5, 0, 0).

b. An intersection with the xz-coordinate plane, y � 0,and the line of intersection, will be r� � (0, 0, 5) �t(1, 0, 1).

5. The line x � y, z � 0 has direction d�� (1, 1, 0) and

passes through the origin, O(0, 0, 0). The required plane

passes through A(2, �5, �4), therefore a second

direction of the plane is OA�� (2, �5, �4) and a normal

is d�� OA�� (�4, 4, �7) � �1(4, �4, 7). The scalar

equation of the plane is 4x � 4y � 7z � 0.

6. Given the system of equations x � 2y � z � �3 ➀x � 7y � 4z � �13 ➁

2x � y � z � 4 ➂➀ � ➂: 3x � y � 1 ➃

➁ � 4 � ➂: 9x � 3y � 3, 3x � y � 1.Let x � t, y � 1 � 3t and from ➀t � 2 � 6t � z � �3

z � �5 � 5tThe three planes intersect in a line with equationr� � (0, 1, �5) � t(1, �3, 5).

7. a. The distance from the origin, O(0, 0, 0) to the plane �:3x � 2y � z � 14 � 0 is

d � ��9

��

1

4

4

�� 1�

� � ��

14

14�� 14�.

b. The distance from the point P(10, 10, 10) to the plane3x � 2y � z � 14 � 0 is

d � � ��26

14��.

�30 � 20 � 10 � 14��

�14�


c. A plane Ax � By � Cz � D � 0 divides R3 into

three regions. All points P1(x

1, y

1, z

1) satisfying the

inequality Ax1

� By1

� C1

� D 0 lie on the

same side of the plane. Those satisfying Ax1

� By1

� D 0 lie on the other side of the plane, and

those satisfying Ax1

� By1

� Cz1

� D � 0 lie on

the plane. Since the sign of Ax1

� By1

� Cz1

� D

is positive for P and negative for the origin, P does

not lie on the same side of the plane as the origin.

Cumulative Review Chapters 4–8

1. Choose two unit vectors, â � ��1

2��, �

�1

2��, 0�

and b� ��13�

�, ��13�

�, �13

��.

Now â � b � ��16�

�, ��16�

�, 0�

�â � b� � ��16

� � �16��

�1

3�� 1.

Therefore the cross product of unit vectors is notnecessarily a unit vector.

2. Question as posed in first printing of textbook ismeaningless. Use (u�� v��) � v��.

3. ∆ABC has coordinates A(2, 4), B(0, 0), C(� 2, 1). To

find the cos ∠ABC, we need BA�� (2, 4) and BC��

(�2, 1). Now BA�� · BC�� 4 � 4 � 0. Therefore

BA�� ⊥ BC��, ∠ABC � 90 and cos ∠ABC � 0.

4. The vector (0, 8) is a linear combination of (2, 4) and(�2, 1). Therefore (0, 8) � m(2, 4) � n(�2, 1).Equating components 2m � 2n � 0 ➀and 4m � n � 8 ➁Solve for m and n: ➀ � 2 � ➁: 5m � 8,

m � �8

5�, n � �

8

5�

and (0, 8) � �85

�(2, 4) � �85

�(�2, 1).

5. Given four points A(2k, 0, 0), B(0, 2k, 0), C(0, 0, 2k),and D(2l, 2l, 2l).The midpoint of AB is W(k, k, 0)The midpoint of BC is X(0, k, k)The midpoint of CD is Y(l, l, l � k)

and of DA is Z(l � k, l, l).

Now WX�� (�k, 0, k) � �k(1, 0, �1)

ZY�� (�k, o, k) � �k(1, 0, �1)

and WX�� ZY��.

Since WX�� ZY��, W, X, Y, and Z are four points of a

parallelogram, hence the four points W, X, Y, and Z are

coplanar.

6.

In ∆ABC, let BP�� PC�� a��,

AQ�� 5b��, QP�� 2b��.

Extend BQ to meet AC at R. BQR is a straight line.

Let BQ�� md��, QR�� (1 � m)d��, AC�� c��, therefore

AR�� kc�� and RC�� (1 � k)c��.

In ∆BQP: md�� a�� 2b��.

In ∆BRC: 2a�� d�� (1 � k)c��,

therefore md�� 1

2�d��

1

2�(1 � k)c�� 2 b��.

In ∆AQR: 5b�� kc�� (1 � m)d��

and 2b�� 2

5�kc��

2

5�(1 � m)d��.

Now md�� 1

2�d��

1

2�(1 � k)c��

2

5�kc��

25

�(1 � m)d��

d�� m � �1

2� � �

2

5��

2

5�m� � c��

1

2� � �

1

2�k � �

2

5�k�

��7

5�m � �

1

9

0�� d��

1

2� � �

1

9

0�k�c��.

Since c�� and d�� are linearly independent,

�7

5�m � �

1

9

0� � 0 and �

1

2� � �

1

9

0�k � 0

m � �1

9

4�, k � �

5

9�.

Therefore, if k � �5

9�, BQR is a straight line.

7.

Place the polygon in the Cartesian plane so that P1

is

at the origin and P1P

2�� is along the positive x-axis. The

interior angles of the polygon as 150.

∠AP2P

1� 90, therefore ∠AP

2P

3� 60 and ∠AP

3P

2

� 30. Let the magnitude of each side of the polygon

be 2, therefore in ∠AP2P

3, P

2P

3� 2, AP

2� 1, and

AP3

� �3�. Now P2P

3�� (�3�, 1). Similarly in

∆BP4P

3, P

3P

4� 2, P

3B � �3�, BP

4� 1, and P

3P

4��

(1, �3�).

Similarly, we have the following:

P1P

2�� (2, 0), P

4P

5�� (0, 2), P

5P

6�� (�1, �3�),

P6P

7�� (��3�, 1).

A

B

Y

P1 P2

P3

P4

P5

P6

P7

X

A

B CP→

a→a

→kc

R

(l – k) c→

b

→

2→dm

Q

→

b5

(l – m) d→


a. P1P

3�� y��, P

1P

2�� x��.

In ∆P1P

2P

3, P

2P

3�� x�� y��.

b. P1P

4�� P

1P

2�� P

2P

3�� P

3P

4�� mx�� ny��.

Now x�� P1P

2�� (2, 0), y�� P

1P

2�� P

2P

3��

� (2, 0) � (�3�, 1)

y�� (2 � �3�, 1).

P1P

4�� (2, 0) � (�3�, 1) � (1, �3�) � m(2, 0) �

n(2 � �3�, 1).

Equating components:

2 � �3� � 1 � 2m � (2 � �3�)n ➀

1 � �3� � n ➁

Substitute in ➁:

3 � �3� � 2m � (2 � �3�)(1 � �3�)

3 � �3� � 2m � 2 � 3�3� � 3

2m � �2�3� � 2

m � �1 � �3�

therefore P1P

4�� (�1 � �3�)x�� (1 � �3�)y��.

c. P3P

4�� P

3P

4�� P

4P

5�� P

5P

6�� P

6P

7�� mx�� ny��

(1, �3�) � (0, 2) � (�1, �3�) � (��3�, 1) �

m(2, 0) � n(2 � �3�, 1).

(��3�, 2�3� � 3) � m(2, 0) � n(2 � �3�, 1).

Equating components:

2m � (2 � �3�)n � ��3� ➀

n � 2�3� � 3 ➁

Substitute in ➀:

2m � (2 � �3�)(3 � 2�3�) � ��3�

2m � 6 � 7�3� � 6 � ��3�

2m � �12 �8�3�

m � �6 �4�3�

therefore P3P

7�� (�6 �4�3�)x�� (3 � 2�3�)y��.

8. a��, b��, and c�� are three linearly independent vectors. If

u�� 3a�� 2b�� c��, v�� 2a�� 4c��, and w�� a�� 3b��

� kc�� are coplanar then one of u��, v��, or w�� can be

written as a linear combination of the other two.

Say w�� pu�� qv��.

Now �a�� 3b�� kc�� p(3a�� 2b�� c��) �

q(�2a�� 4c)

(�1 � 3p � 2q)a�� (3 � 2p)b�� (k � p � 4q)c�� 0��

Since a��, b��, and c�� are linearly independent

�1 � 3p � 2q � 0 ➀, 3 � 2p � 0 ➁, and

k � p � 4q � 0. ➂

From ➁, p � �3

2�.

Substituting in ➀, �1 � �9

2� � 2q � 0

q � �1

4

1�.

Substituting for p and q in ➂: k � �3

2� � 11 � 0,

k � �1

2

9�.

If k � �1

2

9� then u��, v��, and w�� will be coplanar.

9.

ABCD is a parallelogram with

DA�� CB�� a��

DC�� AB�� b��.

Diagonals DB and AC intersect at E and E divides

AC in the ration m:n.

Since E divides AC in the ratio m:n,

DE�� m �

n

n� a��

m

m

� n� b��.

A B

CD

E→a

→a

→

b

→

b

m

n


Also BE�� m �

n

n� BA��

m

m

� n� BC��

EB�� m �

n

n� AB��

m

m

� n� CB��

EB�� m �

n

n� b��

m �

n

n� a��.

Since D, E, and B are collinear, DE�� kEB��.

�m �

n

n� a��

m

m

� n� b�� k�m �

n

n� b�� k�m

m

� n� a��

��m �

n

n� � �

m

k

�

m

n�� a�� m

m

� n� � �

m

k

�

n

n�� b�� 0.

But a�� and b�� are linearly independent, therefore

�m �

n

n� � �

mk�

mn

� � 0 and �m

m

� n� � �

m

k

�

n

n� � 0

n � km � 0 ➀ m � kn � 0 ➁

➀ � ➁: m � n � k(m � n) � 0 and k � 1.

Since k � 1, DE�� EB�� and E is the midpoint of DB.

Substitute k � 1 into ➀: n � m � 0, m � n. Since

m � n, E divides AC in the ration m:m � 1:1 and E is

the midpoint of AC. Therefore the diagonals of a

parallelogram bisect each other.

10.

ABCD is a quadrilateral with AB�DC. AC and DBintersect at M. A line through M parallel to AB meetsAD and BC at P and Q respectively. Draw RMSperpendicular to AB. Since AB�DC, RMS will also beperpendicular to DC. In ∆ABC, MQ�AB. Therefore

�CC

QB� � �

CC

MA� � �

MAB

Q�.

Since SR is an altitude of ∆ABC, M divides SR in thesame ratio.

Therefore �CC

QB� � �

CC

MA� � �

MAB

Q� � �

SSMR� ➀

A R B

P

DS

MQ

C


Similarly in ∆ADB, PM�AB

and �DD

PA� � �

DD

MB� � �

PAMB� � �

SSMR�. ➁

From ➀ and ➁ �MAB

Q� � �

PAMB� �both equal �

SSMR��

and MQ � PM

and M is the midpoint of PQ.

11.

ABC is an isosceles triangle with AB � AC. Apexangle BAC is bisected by DA therefore ∠BAD �∠CAD. We are to show that AD ⊥ BC. In ∆ABD and∆ADC,

AB � AC∠BAD � ∠CAD.

DA is common, therefore ∆ABD � ∆ACDand ∠ACD � ∠ADC � x

∠BDC � 180therefore 2x � 180

x � 90and AD ⊥ BC, hence the bisector of the apex angle ofan isosceles triangle is perpendicular to the base.

12. Two lines l1: �

x �

18

� � �y �

34

� �z �

12

� has direction

d��1

� (1, 3, 1), and l2: (x, y, z) � (3, 3, 3) � t(4, �1, �1)

has direction d��2

� (4, �1, �1).

a. Since d��1

· d��2

� 0, the two lines are perpendicular.

A

B D C

✓ ✓

b. From l1: x� �8 � s, y � �4 � 3s, z � 2 � s

l2: x � 3 � 4t, y � 3 � t, z � 3 � t.

Equating components and rearranging gives thefollowing equationss � 4t � 11 ➀3s � t � 7 ➁

s � t � 1 ➂➁ � ➂: 2s � 6, s � 3 and t � �2. Substitute in➀: 3 � 4(�2) � 11 � R.S. The lines intersect atthe point (�5, 5, 5).

13. Given four points: O(0, 0, 0), P(1, �1, 3), Q(�1, �2,

�5), and R(�5, �1, 1). Now OP�� (1, �1, 3), OQ��

(�1, �2, 5), OR�� (�5, �1, 1). OP�� OQ��

(1, �8, �3) and OP�� OQ�� · OR�� 5 � 8 �3 � 0,

therefore OP�� , OQ��, and OR�� are coplanar, hence O(0,

0, 0) lies on the plane that passes through P, Q, and R.

14. A plane �, passes through P(6, �1, 1), has z-intercept�4 therefore passes through the point A(0, 0, �4),

and is parallel to the line �x �

32

� � �y �

31

� � ��

z1�.

Two directions of � will be AP�� (6, �1, 3) and the

direction of the line,

d�� (3, 3, �1). A normal to the plane is d�� AP��

(14, �21, �21) � 7(2, �3, �3). The scalar equation

of the plane is (x, y, z � 4) · (2, �3, �3) � 0

or 2x � 3y � 3z � 12 � 0.

15. The coordinates of a point on the line (x, y, z) � (�3,4, 3) � t(�1, 1, 0) is A(�3 � t, 4 � t, 3) and on theline (x, y, z) � (3, 6, �3) � s(1, 2, �2) is B(3 � s,6 � 2s, �3 � 2s).

AB�� (6 � t � s, 2 � t � 2s, �6 � 2s)

AB�� is parallel to m�� (2, �1, 3), therefore AB�� km��

and 6 � t � s � 2k ➀2 � t � 2s � �k ➁

� 6 � 2s � 3k ➂We solve for s and t:➀ � 2 � ➁: 10 � t � 5s � 0 ➃3 � ➁ � ➂: � 3t � 4s � 0 ➄

3 � ➃ � ➄: 30 � 11s � 0, s � ��3101�, t � �

43

�s � ��4101�.

The points are A��171�, �

141�, 3� and B ��

131�, �

161�, �

2171��.

16. The sphere (x � 1)2 � (y � 2)2 � (z � 3)2 � 9 has

centre C(1, 2, 3). The plane tangent to the sphere at

A(2, 4, 5), a point at one end of a diameter, will have

CA�� (1, 2, 2) as normal. Therefore (x � 2, y � 4, z � 5) · (1, 2, 2) � 0

x � 2y � 2z � 20 � 0 is therequired plane.

17. The line l: x � �1 � t, y � 3 � 2t, z � �t interestseach of the following planes.

a. �1: x � y � z � 2 � 0.

Substituting for x, y, and z:�1 � t � 3 � 2t � t � 2 � 0

0t � 2.

There is no intersection. The direction of the line is

d��1

� (1, 2, �1), a normal to the plane is n��1

�

(1, �1, �1), d��1

· n��1

� 0, hence the line is parallel

to the plane and distinct from the plane.

b. �2: �4x � y � 2z � 7 � 0.

Substituting for x, y, and z:4 � 4t � 3 � 2t � 2t � 7 � 00t � 0, t � R.

Note that the plane has normal n��2

� (�4, 1, �2)

and d�� · n��2

� 0, the line is parallel to the plane; in

fact, the line is on the plane. The intersection will be

r� � (�1, 3, 0) � t(1, 2, �1).

c. �3: x � 4y � 3z � 7 � 0.

Substituting for x, y, and z:�1 � t � 12 � 8t � 3t � 7 � 0

12t � �18

t � ��32

�.

The line intersects the plane at the point ��52

�, 0, �32

��.


18. Given the planes �1: 4x � 2y � z � 7

and �2: x � 2y � 3z � 3.

Solve to find the line of intersection.

Add �1

� �2: 5x � 2z � 10

x � 2 � �25

�z.

Let z � 5t, x � 2 � 2t.From �

2: 2 � 2t � 2y � 15t � 3

y � �12

� � �123�t.

The parametric equation of the line of intersection is

x � 2 � 2t, y � �12

� � �123�t, z � 5t.

For the intersection with the xy-plane, z � 0. Therefore

t � 0 and the point of intersection is �2, �12

�, 0�.

19. A plane �1

passes through A(2, 0, 2), B(2, 1, 1), and

C(2, 2, 4). Two directions of � are

AB�� (0, 1, �1)

and AC�� (0, 2, 2) � 2(0, 1, 1).

A normal to the plane will be n�� (0, 1, � 1) � (0, 1, 1)

� (2, 0, 0)

� 2(1, 0, 0).The equation of the plane � is (x � 2, y, z � 2) · (1, 0, 0) � 0

x � 2 � 0.A line l through P(3, 2, 1), Q(1, 3, 4) has direction

QP�� (2, �1, �3) and parametric equations

x � 3 � 2t, y � 2 � t, z � 1 � 3t. Solving the line

with the plane gives 3 � 2t � 2 � 0

t � � �12

�.

The coordinates of the point of intersection of the line

with the plane is �2, �52

�, �52

��.

20. a. To determine the line of intersection of the two planes

�1: 3x � y � 4z � �6 and

�2: x � 2y � z � 5, we solve.

2 � �1

� �2: 7x � 7z � �7, x � z � �1.

Let z � t, x � �1 � t substitute into �1

�3 � 3t � y � 4t � �6y � 3 � t

The parametric equations of the two planes arex � �1 � t, y � 3 � t, z � t.

b. To intersect the xy-plane, z � 0 therefore t � 0 andthe point is A(�1, 3, 0).To intersect the xz-plane, y � 0 therefore t � �3and the point is B(2, 0, �3).To intersect the yz-plane, x � 0 therefore t � �1and the point is C(0, 2, �1).

c. The distance between the xy- and xz-intercepts is

�AB��. AB�� (3, �3, �3), �AB�� 3�3�.

21. Since Q is the reflection of P(�7, �3, 0) in the plane

�: 3x � y � z � 12, PQ will be perpendicular to the

plane and the plane will bisect PQ. Let this midpoint

be R. A normal to the plane is n�� (3, �1, 1). The

line passing through PQ will have direction n�� and

parametric equations x � �7 � 3t,

y � �3 � t,

z � t.

Solving with the plane gives�21 � 9t � 3 � t � t � 12

t � �3101�.

This gives the coordinates of the midpoint of PQ,

R��1131�, ��

6131�, �

3101��.

�

Q

R

→n

P (�7, �3, 0)


Let Q have coordinates (a, b, c). Since R is themidpoint of PQ,

�a �

27

� � �1131�, �

b �

23

� � ��6131�, �

2c

� � �3101�

a � �11013

�, b � ��9131�, c � �

6101�

and the coordinates of Q will be ��11013

�, ��9131�, �

6101��.

22. Two planes 3x � 4y � 9z � 0 and 2y � 9z � 0intersect in a line. From the second plane we have

y � �92

�z. Let z � 2t, then y � 9t, substituting in the

first plane gives 3x � 36t � 18t � 0x � 6t.

Parametric equations of the line of intersection are

x � 6t, y � 9t, z � 2t. A direction of this line is

d�� (6, 9, 2). Now �d�� 36 � 8�1 � 4�

� �121�

�d�� 11.

A unit vector along d�� is d � ��161�, �

191�, �

121��.

A vector of length 44 that lies on this line ofintersection will be

44d � 44 ��161�, �

191�, �

121��

44d � (24, 36, 8).

23. The line through P(a, 0, a) with direction

d��1

� (�1, 2, �1) has equation l1: x � a � t

y � 2t

z � a � t.

l1

intersects the plane �: 3x � 5y � 2z � 0 at Q.

Solving l1

and � gives

3a � 3t � 10t � 2a � 2t � 0

5t � �5a

t � �a.

The point Q has coordinates (2a, �2a, 2a).

The line through P(a, 0, a) with direction

d��2

� (�3, 2, �1) has equation

l2: x � a � 3s

y � 2s

z � a � s.

l2

intersects � at R. Solving l2

and � gives

3a � 9s � 10s � 2a � 2s � 0

s � 5a.

The point R has coordinates (�14a, 10a, �4a)

RQ�� (16a, �12a, 6a).

Since RQ�� 3 we have

�(16a)2� � (��12a)2 �� (6a)2� � 3

256a2 � 144a2 � 36a2 � 9

a2 � �4936�

a � � ��

3436��

� � �2�

3109��.

The distance between Q and R will be 3 if

a � �2�

3109�� or a � �

2��

1309�

�.

24. Two lines L1: (x, y, z) � (2, 0, 0) � t(1, 2, �1)

L2

: (x, y, z) � (3, 2, 3) � s(a, b, 1).

To determine the intersection of L1

and L2

we equatecomponents then solve.x � 2 � t � 3 � sa, t � sa � 1 ➀y � 2t � 2 � sb, 2t � sb � 2 ➁z � �t � 3 � s, t � s � �3 ➂➂ � ➀: s � sa � �4

s(1 � a) � �4

s � �1

�

�

4a

�

Substitute in ➂: t �1

�

�

4a

� � �3

t � �1 �

4a

� � �3

t � �11

�

�

3aa

�

Substitute for s and t into ➁:

�21

�

�

6aa

� � �1

4�

ba

� � 2

2 � 6a � 4b � 2 � 2a8a � 4b

a � �12

�b

L1

and L2

will intersect whenever a � �12

�b.



25. x � 2y � 3z � 1 ➀2x � 5y � 4z � 1 ➁

3x � 5y � z � 3 ➂2 � ➀ � ➁: �y � 10z � 1

3 � ➀ � ➂: �8z � 0z � 0y � �1

Substitute into ➀: x � 2 � 1, x � 3.The solution to the system is x � 3, y � �1, z � 0.

26. �2x � y � z � k � 1 ➀kx � z � 0 ➁y � kz � 0 ➂

➀ � ➂: �2x � z � kz � k � 1 ➃2 � ➁ � k � ➃: 2z � kz � k2z � k(k � 1)

(k2 � k � 2)z � � k(k � 1)(k � 2)(k � 1)z � �k(k � 1).

a. (i) If k � 2, 0z � �4 and there will be no solution.

(ii) If k � 2, k � �1, z � ��k �

k2

� and the system

will have exactly one solution.

(iii) If k � �1, 0z � 0 and there will be an infinitenumber of solutions.

b. Since 0z � 0, let z � t, back substituting will givefrom ➂: y � z � 0, y � t, and from ➁: � x � z �0, x � t. The solution set is (x, y, z) � (t, t, t),which is the equation of a line passing through theorigin with direction (1, 1, 1).

ch08

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