ch07

Solutions for Selected Problems

Exercise 7.1

8. b. The vector equation of a line passing through

P�2, �34

�� with direction d�� 23

�, 6� � �23

�(1, 9)

is r� � �2, �34

�� t(1, 9).

9. a. x � �13

� � 2t, y � 3 � �23

�t.

The direction vector d�� (2, ��23

�) � �23

�(3, �1).

A direction vector with integer components is (3, �1). There is no point on this line with integercoordinates.

b. r� � ��13

�, �12

�� t��13

�, �14

��. The direction vector

d�� 13

�, �14

�� 112��4, 3�. A direction vector with

integer components is (4, 3). A point on the line with

integer coordinates is (1, 1) when t � 2.

c. r�� 12

�, 3� � t��12

�, 5�. The direction vector

d�� 12

�, 5� � ��12

� (1, �10). A direction vector

with integer components is (1, �10). A point on the

line with integer coordinates is (�1, 18) when

t � 3.

10. a. l1: x � 1 � 3t , y � 7 � 4t; l

2: x � 2 �4s,

y � �3s. The direction vectors are d��1

� (�3, 4)

and d��2

� (�4, �3). Since d��1

· d��2

� 0, the lines are

perpendicular.

b. l1: r� � (1, 7), � t(�3, 4); l

2: r� � (2, 0) � s(3, �4),

the direction vectors are d��1

� (�3, 4) and d��2

�

(3, �4). Since d��1

� �d��2, the lines are parallel.

Chapter 7: Lines in a Plane 91

Chapter 7 • Lines in a Plane

c. l1: r� � (1, 7) � t(�3, 4); l

2: r� � (2, 0) � s(4,

�3). The direction vectors are d��1

� (�3, 4) and d��2

� (4, �3). Since d��1

· d��2

� 0 and d��1

≠ kd��2, the

lines are neither perpendicular nor parallel.

11. The direction of r� � (1, 8) � t(3, 7) is d�� (3, 7).

A direction perpendicular to d�� is n�� (7, �3). An

equation of a line through (4, 5) with direction n�� is

r� � (4, 5) � t(7, �3).

12. a. x � 6, y � 1 � 7tSince x � 6, this line is parallel to the y-axistherefore does not intersect the y-axis. The lineintersects the x-axis at the point (6, 0).

b. r� � (�5, 10) � t(1, 5)The parametric equations are x � �5 � t,y � 10 � 5t.The line intersects the x-axis when y � 0, t � �2,and x � �7, i.e., the point (�7, 0). The lineintersects the y-axis when x � 0, t � 5, andy � 35, i.e., the point (0, 35).

y

x

35

–35 7 14 35

–21

y

4

–3

–2

4 6x

c. r� � (2, 3) � t(3, �1)The parametric equations are x � 2 � 3t, y � 3 � t. The line intersects the x-axis when y � 0, t � 3,and x � 11, i.e., the point (11, 0). The lineintersects

the y-axis when x � 0, t � ��23

�, and y � �131�, i.e.,

the point �0, �131��.

13. l1: r� � (3, 9) � t(2, 5)

x � 3 � 2t, y � 9 � 5twhen t � �1; x � 1, y � 4 and (1, 4) lies on l

1.

l2: r� � (�5, 6) � u(3, �1)

x � �5 � 3u, y � 6 � u when u � 2; x � 1, y � 4

and (1, 4) lies on l2.

The directions of l1

and l2

are d��1

� (2, 5) and d��2

�

(3, �1). The angle between the direction vectors is the

angle of intersection of these lines.

Now d��1

· d��2

� �d��1��d��

2� cos �

6 � 5 � �29� �10� cos �

cos � � ��29�

1

�10��

� � 87�.

The acute angle between these lines is 87�.

14. a. i) The line r� � (2, �6) � t(3, �4) has direction

d�� (3, �4). The direction of the positive x-axis

is i � (1, 0). The angle the line makes with the

x-axis is α. (a positive rotation about A). To use

the dot product we use direction AB�� (�3, 4).

Therefore AB�� · i � �AB��i� cos α

�3 � 5 cos α

cos α � ��35

�

α � 127�.The line makes an angle of 127� with the x-axis.

ii) The line r� � (6, 1) � t(5, 1) has direction vector

d�� (5, 1). We see from the diagram that α is

the acute angle between direction vectors

(5, 1) and (1, 0) therefore d�� · i � �d��i� cos α

5 � �26� cos α

cos α � ��

5

26��

α � 11�.The line makes an angle of 11� with the x-axis.

y

x–3

5

–2

5

α

→d

P(6,1)

yB

xA

P

�

α

→

d

5

–2

–2

5 10

y

x

92 Chapter 7: Lines in a Plane


b. Consider a line l that intersects the x-axis at

A(a, 0). Choose a point B(x1, y

1), y

1� 0 on l. The

angle of inclination of the line is ∠BAX � α.

The slope of line l is m � �y

x1

1

�

�

0

a� � �

x1

y

�1

a�.

Translate the line to the left a units

therefore A(a, 0) → A'(0, 0)B(x

1, y

1) → B'(x

1� a, y

1).

By definition tan α � �x

1

y

�1

a�

but �x

1

y

�1

a� � m (the slope of l)

therefore tan α � m.Note: if x

1� a, the slope is undefined and

α � 90�.

15. Point A(24, 96) with velocity vector v�� (85, �65)(units in km and km/h).

a. Parametric equations of the highway line are x � 24 � 85t, y � 96 � 65t.

b. The horizontal velocity component is 85 km/h. The

time taken to travel 102 km at 85 km/h is �1

8

0

5

2�

hours (1h 12 min).

c. When t � �1

8

0

5

2�, x � 126, y � 18. The coordinates

of P at that time will be (126, 18).

16. a. Parametric equations of a line are x � x0

� ab1,

y � y0

� bt.

Solving for t gives �x �

ax

0� � t, �y �

b

y0� � t.

Therefore �x �

a

x0� � �

y �

b

y0�.

b. (i) x � 5 � 8t, y � �3 � 5t.

Solving for t gives �x

�

�

8

5� � t and �

y �

5

3� � t.

A symmetric equation is x � ��

5

8� � �

y �

5

3�.

(ii) r� � (0, �4) � t(4, 1). A symmetric equation

is �4

x� � �

y �

1

4�.

c. A direction of the line through A(7, �2),

B(�5, �4) is AB�� (�12, �2) � �2(6, 1).

A symmetric equation of this line is

�x �

6

7� � �

y �

1

2�.

17. a. An equation of the line passing through A(7, 3)

with direction vector d�� (2, �5) is r� �

(7, 3) � t(2, �5). If t � �1, r�1

� (5, 8) then

P(5, 8) is on the line. If t � 5, r�2

� (17, �22),

then Q(17, �22) is on the line.

b. The line segment PQ is defined by the parametricequation x � 7 � 2t, y � 3 � 5t, for �1 t 5.

18. a. A direction vector of the line is PQ�� OP�� OQ��

� �p�� q��.

The vector equation of the line, passing through P

with direction PQ��, is r� � p�� t(�p�� q��)

r� � (1� t)p�� tq��.

y

Q

P

xO

y

xα α

x

y

' '

y

αx

A(a, 0)

y

α

A(a, 0)x

b. R is at P when t � 0 and at Q when t � 1.Therefore R is between P and Q for 0 < t < 1.

c. When t � 2, r� � �p�� 2q��

OS�� 2q��, SR�� p��

OPSR is a parallelogramPQ � QRR divides PQ in the ratio 2:�1.

d. If t � �1

2�, R will be the midpoint of PQ. Therefore

for t > �1

2�, R will be closer to Q than to P.

19. a. b.

l1: r�

1� (5, 2) � t(�3, 6), d��

1� (�1, 2)

l2: r�

2� (5, 2) � u(11, 2), d��

2� (11, 2).

Lines intersect at A(5, 2).

Let l3

represent the line bisecting the angle between

l1

and l2

and let the angle between l1

and l3, and l

2

and l3, be �. Let the direction vector of l

3be

m�� (m1, m

2).

m�� · d��2

� �m��d��2� cos �

11m1

� 2m��2

� �m�� 125� cos �

cos � � �11

5

m

�1

5�

�

�m�2

�

m

�2

�.

m�� · d��1

� �m��d��1� cos �

�m1

� 2m2

� �m��5� cos �

cos � � ��m

�1

5�

�

�m�2

��

m2

�.

Equating cos � gives

�11

5

m

�1

�

5��m�2

�

m

�2

� � ��m

�1

5�

�

�m�2

��

m2

�

11m1

� 2m2

� �5m1

� 10m2

16m1

� 8m2

m1

� �12

� m2·

Choose m2

� 2. Therefore m1

� 1 and the

direction of l3

is m�� (1, 2).

The vector equation of l3

is r� � (5, 2) � s(1, 2). To determine the bisector of the other angle weuse directions d��

1and �d��

2which gives equation

��1

5

1

�

m1

5�

�

�m��

2

�

m2

�� m

�1

5�

�

�m�2

��

m2

�

�11m1

� 2m2

� �5m1

� 10m2

6m1

� �12m2

m1

� �2m2.

Choose m2

� �1. Therefore m1

� 2 and the

direction of this line is (2, �1). The vector

equation of the second angle bisector is

l4: r� � (5, 2) � v(2, �1).

c. The direction of the two lines are (1, 2) and (2, �1). Since (1, 2) · (2, �1) � 0, the two lines are perpendicular.

Exercise 7.2

9. a. 5x � 3y � 15 � 0.

Solve for y: y � �53

� � 5.

Let x � 3t, y � 5t � 5

A vector equation is r� � (0, 5) � t(3, 5).Scalar equations are x � 3t

y � 5 � 5t.

A symmetric equation is �3x

� � �y �

55

�.

→d

� �

→m

A(5, 2)

2

→d1

1

4

2

→d2�

3

l

l

l l

PQ

S

R

O


b. �4x � 6y � 9 � 0.

Solve for y: y � �23

�x � �32

�.

Let x � 3t, y � 2t � �32

�.

A vector equation is r� � �0, ��32

�� t(3, 2).

Parametric equations are x � 3t

y � 2t � �32

�.

A symmetric equation is � .

10. Let P be a point not on the line l and D be on l, the footof the perpendicular from P. Choose a point A on lother than D. Now PAD is a right triangle where PA isthe hypotenuse. Therefore PA is the longest side andPA � PD. Therefore the shortest distance from a pointto a line is the perpendicular distance from the point tothe line.

11. The distance d from Q(3, �2) to each of the followinglines:

a. 3x �2y � 6 � 0

d ��Ax

�1

�

A2

B

�

y1

B��

2�C�

�

� �9 �

�

4

1

�

3�

6�

d � ��

7

13��.

b. �x �

23

� � �y �

74

� therefore 7x � 21 � 2y � 8

7x � 2y � 13 � 0

and d � �21

��

494

�

�

4�13

�

d � ��

12

53��.

c. r� � (�3, �7) � t��15

�, �16

��. A direction vector

m�� (6, 5) therefore a normal n�� (5, �6). A pointon the line is P(�3, �7).

Now d � ��PQ�

��

n��· n��

� �(6, 5)

�·

n�

(

�

5

�� 6)

�

� 0

therefore the point is on the line.

d. The distance from Q to x � �5 is 8.

12. Equation of the line is 6x � 3y � 10 � 0.

a. Point (4, 7)

d � �24

��

3

2

6

1

�

�

9�10

�

� �3�

35

5�� · �

��

5�5�

�

d � �7�

3

5��.

x = –5

x = –5 –5

3

2

2

Q (3, –2)

P

D

A

l

y � �32

�

�2

x�3


b. Point (4, �8)

d � ��24 �

3

2

�4

5�� 10��

� �3�

10

5��

d � �2�

3

5��.

c. Point (0, 5)

d � �15

3��

5�10

�

� �3�

5

5��

d � ��

35�

�.

d. Point �5, ��230��

d � �30 �

3�20

5��10

�

d � 0.

13. a. i) Given two lines l1

and l2. Let the direction

vector of l1

be d��1

� (a, b). Since l1�l

2, the

direction vector d��2

of l2

is a multiple of d��1. Let

d��2

� (ka, kb). Now the normal of l1

is n��1

�

(b, �a) and of l2

is n��2

– (kb, �ka). But n��2

�

(kb, �ka) � k(b, �a) � kn��1

therefore n��1�n��

2.

ii) Given two lines l1

and l2

having normals n��1

and

n��2. Let the normal of l

1be n��

1� (A, B) since

n��1�n��

2, n��

2� kn��

1, and n��

2� (kA, kB). The

direction of l1

and l2

will be m��1

� (B, �A) and

m��2

� (kB, �kA). Since m��2

� (kB, �kA) �

k(B, �A) � km��1. m��

2�m��

1and the lines l

1and l

2

are parallel, therefore two lines in a plane are

parallel if and only if their normals are parallel.

b. If two lines in a plane are perpendicular, their

normals are perpendicular. Let two lines l1

and l2

have direction m��1

and m��2. Let m��

1� (a, b). Since

l1

⊥ l2, m��

1· m��

2� 0, therefore m��

2� (�kb, ka).

A normal to l1

is (�b, a) and to l2

is (ka, kb). Now

(�b, a) · (ka, kb) � �kab � kab � 0, therefore the

normals are perpendicular.

If the normals of two lines are perpendicular, then

the two lines are perpendicular.

Let two lines l1

and l2

have normals n��1

� (A, B)

and n��2. Since n��

1⊥ n��

2, n��

1· n��

2� 0 and

n��2

� (kB, � kA). The normal to l1

is n��1

� (A, B),

therefore a direction is m��1

� (B, �A). Similarly a

direction of l2

will be m��2

� (kA · kB). Now

m��1

· m��2

� 0, therefore m��1

⊥ m��2

and the two lines

are perpendicular.

Therefore two lines in a plane are perpendicular if

and only if their normals are perpendicular.

14. a.

Given a line l intersecting the x-axis at A. The

angle of inclination of l is ∠BAX � α. Let the

direction vector of l be AB�� m�� (m1, m

2).

In ∆ABC, AE � m1

� �m�� cos αand EB � m

2� �m�� sin α.

Therefore a direction of the line is (cos α, sin α)

and a normal is (sin α, �cos α). The scalar

equation of a line is Ax � By � C � 0 where

(A, B) is a normal therefore the equation is

x sin α � y cos α � C � 0.

A

y

B

xE

l

→m

α


b. 2x � 4y � 9 � 0 has normal (1, 2) therefore adirection is (2, �1).

tan � � �m

m2

1

�

� ��12

�

� � 153�.The angle of inclination is 153�.

c. The equation will be x sin 120�� y cos 120� � D � 0.

sin 120� � ��

23�

�, cos 120�� 12

�

therefore we have �3�x � y � 2D � 0.

Now (6, �4) is on the line. 6�3� � 4 � 2D � 0,

2D � 4 � 6�3�and the equation is �3�x � y � 4 � 6�3� � 0.

15. a.

AN�� (x � 2, y � 2), BN�� (x � 8, y � 10)

AN�� · BN�� 0

(x � 2, y � 2) · (x � 8, y � 10) � 0

(x � 2)(x � 8) � (y � 2)(y � 10) � 0

x2 � 10x � 16 � y2 � 12y � 20 � 0

x2 � y2 � 10x � 12y � 36 � 0

(x2 � 10x � 25)� 25 � (y2 � 12y � 36)� 36 � 36� 0

(x � 5)2 � (y � 6)2 � 25.

Hence N lies on a circle with centre (5, 6) andradius 5.

b. The midpoint of AB is (5, 6), which is the centre ofthe circle. Since both A and B are on the circle, ABis a diameter of the circle.

16. a.

n�� is a normal to the line l. P(x, y) is a point on the

line and OP�� is the position vector of P. Rotate the

line about P until it passes through the origin. Now

the line l and the position vector OP�� are coincident.

Since n�� ⊥ l , it will now be perpendicular to OP��

and n�� · OP�� 0.

b. If the line goes through the origin then OP�� is a

direction vector of the line. n�� is normal to the line

therefore n�� · OP�� 0.

If n�� · OP�� 0 then n�� is perpendicular to OP��. But n��

is a normal to the line hence is perpendicular to l.

Since n�� is perpendicular to both OP�� and the line,

OP�� and the line are parallel. But P is a point on the

line, hence they are coincident and the line passes

through the origin.

Exercise 7.3

7. Given the points A(2, 3, �2) and B(4, �1, 5). The

midpoint of AB is M�3, 1, �32

��. The line passes through

C(0, �1, 1) and M, therefore a direction is

CM � �3, 2, �12

��.

Using direction vector (6, 4, 1) and point C(0, �1, 1),the parametric equations arex � 6t, y � �1 � 4t, z � 1 � t.

→n

P(x, y)

l

O

ly

5

–5 5x

–5

N(x, y)

A(2, 2)

B(8, 10)


8. A line through the origin and parallel to AB;

A(4, 3, 1), B(�2, �4, 3) has direction BA��

(6, 7, �2) and symmetric equation

�6x

� � �7y

� � ��

z2�.

9. a. line l1: r� � (1, 0, 3) � t(3, �6, 3) with direction

d��1

� 3(1, �2, 1) and l2: r� � (2, �2, 5) �

t(2, �4, 2) with direction d��2

� 2(1, �2, 1) since

d��1

� �32

�d��2, the lines are parallel. The symmetric

equation of l1

is

�x �

11

� � ��

y

2� � �

z �

13

�.

(2, �2, 5) is on l2. Check to see if it is on l

1.

�2 �

11

� � ��

�

22� � �

5 �

13

�.

Therefore the lines are parallel and distinct.

b. l1: r� � (2, �1, 4) � s(3, 0, 6); d��

1� 3(1, 0, 2).

l2: r� � (�3, 0, 1) � t(2, 0, 2); d��

2� 2(1, 0, 1).

Since d��1

� kd��2, the lines are not parallel nor the

same line.

c. l1: r� � (1, �1, 1) � s(6, 2, 0); d��

1� 2(3, 1, 0).

l2: r� � (�5, �3, 1) � t(�9, �3, 0);

d��2

� �3 (3, 1, 0).

Since d��1

� ��23

�d��2, the lines are parallel.

Symmetric equation of l1

is �x �

31

� � �y �

11

�; z � 1.

Check to see if (�5, �3, 1) is on l1.

��5

3� 1� � �

�31� 1� � �2; z � 1.

Since (�5, �3, 1) lies on l1

and l1

is parallel to l2,

l1

and l2

are the same line.

10. a. x � t, y � 2, z � �1: perpendicular to the yz-planepassing through (0, 2, �1),

b. x � 0, y � 1 � t, z � 1 � t: a line in the yz-planehaving y-intercept 2 and z-intercept 2.

c. x � �5, y � 2 � t, z � 2 � t represents a line inthe plane x � �5, a plane parallel to the yz-plane.In this yz-plane the line has equation y � z, a linepassing through (�5, s, s) for all s � R.

z

x

y

l

�5

(�5, s, s)

z

y

x

2

2

z

x

y

x = t, y = 2, z = –1


11. a. If a line in R3 has one direction number zero it will

be parallel to one of the coordinate planes; i.e., if

d�� (a, b, 0), the line is parallel to the xy-plane,

d�� (a, 0, c), the line is parallel to the xz-plane,

d�� (0, b, c), the line is parallel to the yz-plane.

b. If a line in R3 has two direction numbers zero, theline will be perpendicular to one of the coordinateplanes; i.e., if

d�� (a, 0, 0), the line is perpendicular to the yz-plane,

d�� (0, b, 0), the line is perpendicular to the xz-plane,

d�� (0, 0, c), the line is perpendicular to the xy-plane.

12. A line l1

passes through the point A(�6, 4, 2) and is

perpendicular to both

l1: �

�

x4� � �

y �

�610

� � �z �

32

� and

l2: �

x �

35

� � �y �

25

� � �z �

45

�.

A direction of l1

is d��1

� (4, 6, �3)

and of l2

is d��2

� (3, 2, 4). The direction of l1d��

1is

perpendicular to both l1

and l2. Therefore

d�� d��1

d��2

� (30, �25, �10) � 5(6, �5, �2). The

symmetric equation of a line through A(�6, 4, 3)

having direction (6, �5, �2) is �x �

66

� � �y�

�

54

�

� �z�

�

23

�.

13. a. The equation of a line l1

passing through C(0, 0, 2)

having direction vector d�� (3, 1, 6) is

r�� (0, 0, 2) � t(3, 1, 6). The parametric

equations are x � 3t, y � t, z � 2 � 6t. Check to see if A(�9, �3, �16) is on l byequating components 3t � �9, t � �3, 2 � 6t � �16t � �3 t � �3

Since t � �3 generates the point A, A(�9, �3,�16) is on the line l.Check B(6, 2, 14):3t � 6, t � 2, 2 � 6t � 14t � 2 t � 2

Since t � 2 generates the point B, B(6, 2, 14) is onthe line l.


b. The line segment AB is the set of points on the line x � 3t, y � t, z � 2 � 6t, for �3 t 2.

14. A line l has equation �x �

311

� � �y�

�

18

� � �z �

14

�.

The parametric equation of l is x � 11 � 3t,

y � �8 � t, z � 4 � t. A direction of l is d�� (3, �1, 1).

Let m be the required line passing through A(4, 5, 5)

and intersecting l at T. Since T is on l, represent its

coordinates as T(11 � 3t, �8 � t, 4 � t). Now a direction

of line m is AT�� (7 � 3t, �13 � t, �1 � t). Since l and

m are perpendicular, d�� · AT�� 021 � 9t � 13 � t � 1 � t � 011t � �33.

t � �3. Now AT�� (�2, �10, �4). Line m passes

through A(4, 5, 5) and has a direction (1, 5, 2). An

equation of m is r� � (4, 5, 5) � s(1, 5, 2).

15. a.

l is a line passing through P and having direction

vector d��. Q is a point, not on the line and FQ is the

distance from Q to the line. FQ�� ⊥ l. � is the angle

between PQ�� and d��.

From ∆PFQ, �FQ�� PQ�� sin �

� �PQ�� sin � ��d

d

�

�

�

��

�

� ��PQ��

��d

d�

�

�

�

�� sin ��.

Fl

Q

P �

→

d

T l

m

A(4, 5, 5)

d = (3, �1, 1)→

But �PQ�� d�� PQ��d�� sin �

therefore �FQ�� PQ��

�d��

�d��

�.

b. The distance from Q(1, �2, �3) to the line

r� � (3, 1, 0) � t(1, 1, 2). A point on the line is

P(3, 1, 0), therefore PQ�� (�2, �3, �3) and

d�� (1, 1, 2). PQ�� d�� (�3, 1, 1)

�PQ�� d�� 9 � 1�� 1� � �11�,

�d�� 1 � 1�� 4� � �6�.

Therefore the distance from Q

to the line is ��

1

6�1�

� � ��

666��.

c.

Two lines l1: r� � (�2, 2, 1) � t(7, 3, �4),

l2: r� � (2, �1, �2) � u(u, 3, �4).

The distance between two parallel lines is theperpendicular distance from a point on one of thelines to the other line.

A(�2, 2, 1) is a point on l1. The direction of l

2is

d��2

� (7, 3, �4) and a point on l2

is B(2, �1, �2).

Now BA�� (�4, 3, 3), BA�� d�� (�21, 5, �33).

Therefore �AF�� BA��

�d�

��d��

�

� ��

15

7

5

4�5�

�.

The distance between the two parallel lines is

��157545

�.

Exercise 7.4

4. a. r� � (�2, 0, �3) � t(5, 1, 3);

r� � (5, 8, �6) � u(�1, 2, �3).The parametric equations are:x � �2 � 5t x � 5 � uy � t y � 8 � 2uz � �3 � 3t z � �6 � 3uEquating components and rearranging gives:�2 � 5t� 5 � u 5t � u � 7 ➀

t � 8 � 2u t � 2u � 8 ➁�3 � 3t� �6 � 3u 3t � 3u � �3 ➂➀ � ➂ � 3: 4t � 8, t � 2, u � �3From ➁: t � 2u � 2 �2(�3) � 8.u � �3, t � 2 satisfies all three equations.Therefore the two lines intersect at the point (8, 2, 3).

b. line 1 line 2x � 1 � t x � 3 � 2uy � 1 � 2t y � 5 � 4uz � 1 � 3t z � �5 � 6uEquating components and rearranging terms:

1 � t � 3 � 2u t � 2u � 2 ➀1 � 2t � 5 � 4u 2t � 4u � 4 ➁1 � 3t � �5 � 6u 3t � 6u � 6 ➂

Each of equations ➀, ➁, and ➂ are equivalent.

Note that d��1

� (1, 2, �3), d��2

� (�2, �4, 6) �

�2(1, 2, �3) � �2d��1

therefore the lines are

parallel. Also a point on line 1 is (1, 1, 1) and it is

also on line 2 (u � 1) therefore the two lines are

coincident.

c. l1: r� � (2, �1, 0) � t(1, 2, �3);

l2: r� � (�1, 1, 2) � u(�2, 1, 1).

The parametric equations are:x � 2 � t x � �1 � 2uy � �1 � 2t y � 1 � uz � �3t z � 2 � uEquating components and rearranging gives:t � 2u � �3 ➀2t � u � 2 ➁3t � u � �2 ➂➁ � ➂: 5t � 0, t � 0 and u � �2.

From ➀ t � 2u � 0 � 4 � �4 � �3, the lines do

not intersect. Since d��1

� (1, 2, �3), d2

�

(�2, 1, 1) and d��1

≠ kd��2, the lines are not parallel.

Therefore the lines are skew.

A(�2, 2, 1)

F

l

l

B(2, �1, �2)

1

2



d. l1: (x, y, z) � (1 � t, 2 � t, �t);

l2: (x, y, z) � (3 � 2u, 4 � 2u, �1 � 2u).

Equating components and rearranging gives:t � 2u � 2t � 2u � 2t � 2u � 1

there is no solution to this system of equation hence

the lines do not intersect. d��1

� (1, 1, �1),

d��2

� (�2, �2, 2) � �2(1, 1, �1).

Since d��1

� �2d��2, the lines are parallel. Therefore

the lines are parallel and distinct.

e. l1: �

x �

43

� � �y �

12

�0 � z � 2;

l2: �

x�

�

32

� � �y �

21

� � �z�

�

12

�

The parametric equations are:x � 3 � 4t x � 2 � 3uy � 2 � t y � �1 � 2uz � 2 � t z � 2 � uEquating components and rearranging gives:4t � 3u � �1 ➀

t � 2u � �3 ➁t � u � 0 ➂

➂ – ➁: 3u � 3, u � 1, t � �1.Substitution into ➀: 4t � 3u � �4 � 3 � �1.u � 1, t � �1 satisfies all three equations.Therefore the two lines intersect at the point (�1, 1, 1).

5. a. l1: r� � (1, �1, 1) � t(3, 2, 1);

l2: r� � (�2, �3, 0) � u(1, 2, 3).

Equating components and rearranging gives:3t � u � �3 ➀

2t � 2u � �2 ➁t � 3u � �1 ➂

➀ � ➁ � 2: 2t � �2, t � �1, u � 0.t � �1, u � 0 also satisfies ➂. Therefore the twolines intersect at A(�2, �3, 0).

b. The direction of the two lines are d��1

� (3, 2, 1) and

d��2

� (1, 2, 3).

d��1

d��2

� (4, �8, 4).

A direction perpendicular to both given lines is

(1, �2, 1). The equation of the line passing through

A(�2, �3, 0) with direction (1, �2, 1) is

r� � (�2, �3, 0) � s(1, �2, 1).

6. l1: r� � (4, 7, �1) � t(4, �8, �4),

l2: r� � (1, 5, 4) � u(�1, 2, 3).

Equating components and rearranging gives:4t � u � �3 ➀

8t � 2u � �2 ➁4t � 3u � �5 ➂

➀ � ➁ ÷ 2: 8t � �4, t � ��12

�, u � �1 which also

satisfies ➂ therefore the two lines intersect at

(2, 3, 1). The directions are d��1

� (1, 2, �1) and

d��2

� (�1, 2, 3).

d��1

· d��2

� �1 � 4 �3 � 0. Therefore the two lines

intersect at right angles at the point (2, 3, 1).

7. x � 24 � 7t, y � 4 � t, z � �20 � 5t.For the x-intercept, both y � 0 and z � 0 which is truefor t � �4. Therefore the x-intercept is 24 � 7(�4) ��4.For the y-intercept, both x � 0 and z � 0 for the samet, which is not possible, therefore there is no y-intercept.Similarly there is no z-intercept.

8. Given the line 10x � 4y � 101 � 0 and a point A(3, �4). A normal to the line is n�� (5, 2) which is adirection for a line perpendicular to 10x � 4y � 101 � 0. A vector perpendicular to n�� is (2, �5).The equation of the line perpendicular to 10x �4y � 101 � 0 and passing through A(3, �4) is(2, �5) · (x � 3, y � 4) � 0

2x � 5y � 26 � 0.Now solving 10x � 4y � 101 ➀

2x � 5y � 261 ➁➀ � 5 ➁: 29y � �29

y � �1, x � �221�.

The point of intersection is ��221�, �1�.

9. Three lines l1, l

2, and l

3are in the same plane. Possible

intersections are:

a. All intersect in a common point.

1l

2l

3l

b. Pairs of lines intersect, but there is no commonintersection.

c. One line intersects two distinct parallel lines.

d. One line intersects two coincident lines.

e. The three lines are coincident.

10. Three lines l1, l

2, l

3are in space. The possible

intersections, in addition to those of question 9, are:

a. A line intersecting two skew lines.

l1

and l2

are skew. l3

intersects l1

and l2

at A and B.

b. Two parallel lines and the third intersecting one ofthe two.

l1�l

2, l

3intersects l

2at A.

11.

Given line l: r� � (7, �13, 8) � t(1, 2, �2), pointA(�5, �4, 2).The line m, through A, intersects l at right angles. LetT be the point of intersection of l and m. Since T is onl, we represent its coordinates by T(7 � t, �13 � 2t,8 � 2t).

AT�� (12 � t, �9 � 2t, 6 � 2t) is a direction of m.

Since l and m are perpendicular, their directions are

perpendicular, hence d�� · AT�� 0.(1, 2, �2) · (12 � t, �9 � 2t, 6 � 2t) � 0

12 � t � 18 � 4t �12 � 4t � 09t � 18

t � 2.

Now AT�� (14, �5, 2) and the equation of m is

r� � (�5, �4, 2) � s(14, �5, 2). The coordinates of

the point of intersection are T(9, �9, 4).

12. The line r� � (0, 5, 3) � t(1, �3, �2) with parametric

equations x � t, y � 5 � 3t, z � 3 � 2t, intersects the

sphere x2 � y2 � z2 � 6. (The centre of the sphere is

C(0, 0, 0) and its radius is �6�.) Substituting for x, y,

and z into the equation of the sphere:t2 � (5 � 3t)2 � (3 � 2t)2 � 6

t2 � 25 � 30t � 9t2 � 9 � 12t � 4t2 � 614t2 � 42t � 28 � 0

t2 � 3t � 2 � 0(t � 2)(t � 1) � 0

t � 2 or � 1

T

l

m

B (7, �13, 8)

d=(1, 2, �2)

A (�5, �4, 2)→

A

l3

l1

l2

A

B

l2

l1

l3

l1 l2 l3, , and

l 1

l 2 and l 3

l 3

l 2

l 1

l 1

l 2

l 3


therefore the line intersects the sphere at A(1, 2, 1)

and B(2, �1, �1). The midpoint of AB is ��32

�, �12

�, 0�which is not the centre of the sphere, therefore AB isnot a diameter of the sphere.

13.

Two lines: l1: r� � (2, �16, 19) � t(1, 1, �4);

l2: r� � (14, 19, �2) � u(�2, 1, 2). A line l

3passes

through the origin and intersects l2

at A and l1

at B.

Since B is on l1, we represent its coordinates as

B(2 � t, �16 � t, 19 � 4t).

Similarly the coordinates of A will be

A(14 � 2u, 19 � u, �2 � 2u).

Since O, A, and B are collinear

OB�� kOA��

(2 � t, �16 � t, 19 � 4t) � k(14 � 2u, 19 � u, �2 � 2u).Equating components: 2 � t � 14k � 2ku ➀

�16 � t � 19k � ku ➁19 � 4t � �2k � 2ku ➂

We solve by first eliminating ku:➀ � ➂: 21 � 3t � 12k ➃

➀ � 2 ➁: �30 � 3t � 52k ➄add: �9 � 64k

k � ��694�.

Substitute in ➃: 21 � 3t � 12k7 � t � 4k

t � 7 � 4 · �694�

t � �11261

�.

With t � �11261

�, we can find the coordinates of B hence

we have OB�� 11563

�, ��

11635�, �

�

11680��.

A direction for l3

is (17, �15, �20) and the equation

of l3

is r� � s(17, �15, �20).

NOTE: To determine the equation of l3

we require a

direction. Once t � �11261

� is established, a

direction is evident and further substitutions to determine u are not required. Uponsubstitution, one would find that u � 41.

14. a.

A line l with equation Ax � By � C � 0. N is on l

so that ON ⊥ l. The normal to l, n�� (A, B), is a

direction of the line m along ON. The equation of

m is r� � t(A, B); x � At, y � Bt.

Substituting in l: A2t � B2t � C � 0

t � �A2

�

�

C

B2�.

The coordinates of N are ��A2

�

�

AC

B2�, �

A

�2 �

BC

B2��.

b. ON�� A2

�

�

AC

B2�, �

A

�2 �

BC

B2��

�ON�� A

(

2

A

C2

2

�

�B

B2

2

)

C2

2

��

C

(

2

A

(A2

2

��

B

B2)

2)�

� ��A2

C

�

2

B2�

�ON�� A

�2

C

�

�

B�2��.

y

N

x

m

l

O

z A

y

xl3

B

l1

l2

O


15. Two skew lines, l1: (x, y, z) � (0, �1, 0) � s(1, 2, 1);

l2: (x, y, z) � (�2, 2, 0) � t(2, �1, 2). Let the line l

intersect l1

at A, l2

at B so that l ⊥ l1

and l ⊥ l2. Let the

coordinates of the intersection points be

A(s, �1 � 2s, s) and B(�2 � 2t, 2 � t, 2t).

BA�� (2 � s � 2t, �3 � 2s � t, s � 2t).

BA�� is perpendicular to l1

therefore BA�� · d��1

� 0

2 � s � 2t � 6 � 4s � 2t � s � 2t � 0

6s � 2t � 4, 3s � t � 2 ➀

BA�� is perpendicular to l2, therefore BA�� · d��

2� 0

4 � 2s � 4t � 3 � 2s � t � 2s � 4t � 02s � 9t � �7 ➁9 � ➀ � ➁: 25s � 25s � 1, t � 1.The coordinates of the points of intersection are A(1, 1, 1) and B(0, 1, 2).

16. The distance between two skew lines r� � OP�� td��1,

r� � OQ�� sd��2

is given by ��PQ�

�n

�

��

·

�n��

� where n�� d��1

d��2.

a. r� � (0, �2, 6) � t(2, 1, �1);

r� � (0, �5, 0) � s(�1, 1, 2).

The two points P(0, �2, 6), Q(0, �5, 0)

and directions d��1

� (2, 1, �1), d��2

� (�1, 1, 2).

Now PQ�� (0, �3, �6), n�� d��1

d��2

� (3, �3, 3)

��PQ�

�n

�

��

·

�n��

� � ��9

3

�

�

1

3�

8��

�

3

3�� 3�.

The distance between the lines is �3�.

b. x � 6, y � �4 � t, z � t; x � �2s, y � 5,

z � 3 � s. Two points P(6, �4, 0) and Q(0, 5, 3).

d��1� (0, �1, 1), d��

2� (�2, 0, 1).

Now PQ�� (�6, 9, 3),

n�� d��1

d��2

� (�1, �2, �2).

��PQ�

�n

�

��

·

�n��

� � ��

�6

1

�

�

18

4�

�

�

6

4�

��

1

3

8� � 6.

The distance between the lines is 6.

Review Exercise

2. a. A line through A(3, 9), B(�4, 2) has a direction AB��

� (�7, �7). An equation is r� � (3, 9) � t(1, 1).

b. A line passes through A(�5, �3) and is parallel to r�

� (4, 0) � t(0, 5). A direction is (0, 1) and an

equation is r� � (�5, �3) � s(0, 1).

c. A line passes through A(0, �3) and is perpendicular

to l: 2x � 5y � 6 � 0. A normal to l is (2, �5)

which is a direction of the required line. An

equation will be r� � (0, �3) � s(2, �5).

3. a. A line passes through A(�9, 8) and has slope

��23

�. A direction is (3, �2) and parametric

equations are x � �9 � 3s, y � 8 � 2s.

b. A line passes through A(3, �2) and is perpendicular

to l: r� � (4, �1) � t(3, 2). A direction of l is (3, 2);

a direction perpendicular to (3, 2) is (2, �3). An

equation of the line is x � 3 � 2s, y � �2 � 3s.

c. A line through A(4, 0), B(0, �2) has direction

AB�� (�4, �2)

� �2(2, 1).The line has equation x � 4 � 2t, y � t.

4. a. The line passes through A(2, 0, �3), B(�3, 2, �2).

A direction is AB�� (5, �2, �1) and a vector

equation is r� � (2, 0, �3) � t(5, �2, �1).

A

B

l

l1

l2

�


b. An x-intercept of �7 and a y-intercept of 4 means

the line passes through A(�7, 0, 0), B(0, 4, 0) and

a direction is AB�� (7, 4, 0). A vector equation is

r� � (�7, 0, 0) � t(7, 4, 0).

c. A line l, parallel to �x �

45

� � �y�

�

22

� � �z �

56

�,

passing through (0, 6, 0), has direction (4, �2, 5).

An equation of l is r� � (0, 6, 0) � t(4, �2, 5).

5. a. A line l passes through the origin and is parallel to

the line �x�

�

31

� � �y�

�

22

� � z � 3. A direction of

l is (3, 2, �1). Parametric equations for l are:

x � 3t, y � 2t, z � �t.

b. The line passes through A(6, �4, 5) and is parallelto the y-axis. A direction is (0, 1, 0). Parametricequations are x � 6, y � �4 � t, z � 5.

c. A line with z-intercept �3 with direction vector (1, �3, 6) passes through the point (0, 0, �3).Parametric equations are x � t, y � �3t,z � �3 � 6t.

6. a. A line l passes through A(�1, �2) and is parallelto 3x � 4y � 5 � 0. Line l will have equation 3x � 4y � c � 0. Since A � l, 3(�1) � 4(�2) � c � 0, c � �5 and the scalar equation is 3x � 4y � 5 � 0.

b. A line l passes through A(�7, 3) and isperpendicular to x � 2 � t, y � �3 � 2t. A normalof l is (1, 2) hence an equation is (1, 2) · (x � 7,y � 3) � 0, x � 2y � 1 � 0.

c. A line perpendicular to x � 4y � 1 � 0 will have a

direction vector d�� (1, 4). A vector perpendicular

to d�� is (4, �1). Therefore the equation of a line

through the origin with normal (4, �1) is

4x � y � 0.

7. a. A line l through A(6, 4, 0) and parallel to a line

through B(�2, 0, 4), C(3, �2, 1) has direction

BC�� (5, �2, �3) and parametric equations x � 6

� 5t, y � 4 � 2t, z � �3t.

b. Since (�4, m, n) � l, 6 � 5t � �4, t � �2; 4 � 2t � m, m � 8;

�3t � n, n � 6.

8. a. l1: r� � (2, 3) � t(�3, 1), d��

1� (�3, 1)

l2: r� � (�1, 4) � u(6, �2), d��

2� (6, �2)

� �2(�3, 1). Since d��2

� �2d��1, l

1and l

2are

parallel. The point (�1, 4) is a point on l1(t � 1),

therefore the two lines are coincident.

b. l1: x � 1 � 2t, y � �3 � t; d��

1� (2, �1).

l2: x � u, y � �

13

� � 2u; d��2

� (1, 2).

Since d��1

· d��2

� 0, the two lines are perpendicular.

c. l1: �

x �

21

� � �y �

14

�, z � 1; d��1

� (2, 1, 0).

l2: x � 4t, y � 1 � 2t, z � 6; d��

2� (4, 2, 0)

� 2(2, 1, 0). Since d��2

� 2d��1, l

1and l

2are parallel.

Since points on l1

are of the form (a, b, 1) and on l2

of the form (p, q, 6), there are no points common to

the two lines; hence the lines are parallel and

distinct.

d. l1: (x, y, z) � (1, 7, 2) � t(�1, �1, 1),

d��1

� (�1, �1, 1).

l2: (x, y, z) � (�3, 0, 1) � u(2, �2, �2),

d��2

� (2, �2, �2) � 2(1, �1, �1). Since d��1

� kd��2,

d��1

· d��2

� 0, the two lines are neither parallel nor

perpendicular.

9. Parametric equations of the line are x � �4 � 2t,y � 6 � t, z � �2 � 4t. If it meets the xy-plane, z � 0,

t � �12

�, and the point is (�3, �121�, 0).

If it meets the xz-plane, y � 0, t � 6, and the point is(8, 0, 22). If it meets the yz-plane, x � 0, t � 2, and the point is (0, 4, 6).

10. a. The symmetric equations of the line are

�x�

�

12

� � �y �

53

� and the scalar equation is

5x � y � 13 � 0.


b. The line 5x � 2y � 10 � 0 has normal (5, �2). A

direction is (2, 5) and a point on the line is (0, 5).

A vector equation is r� � (0, 5) � t(2, 5).

c. The line y � �34

�x � �12

� has slope �34

� and a direction

(4, 3). A point on the line is (2, 2) and a vector

equation is r� � (2, 2) � t(4, 3).

11. The parametric equations of the line are x � 12 � 3t,y � �8 � 4t, z � �4 � 2t.

a. Intersection with:xy-plane, z � 0, t � 2, the point is (6, 0, 0)xz-plane, y � 0, t � 2, the point is (6, 0, 0)yz-plane, x � 0, t � 4, the point is (0, 8, 4).

b. The x-intercept is 6 and is the only intercept.

c.

12. a. The line �x �

53

� � �y �

26

� � �z�

�

11

� has direction

d�� (5, 2, �1), �d�� 30�; direction cosines

cos � � ��

5

30��, cos � � �

�2

30��, cos δ � �

��

3

1

0��;

and direction angles � � 24�, � � 69�, δ � 101�.

b. The line x � 1 � 8t, y � 2 � t, z � 4 � 4t has

direction d�� (8, �1, �4), �d�� 9; direction

cosines cos � � �89

�, cos � � ��91�, cos δ � �

�94�;

and direction angles � � 27�, � � 96�, δ � 116�.

c. The line r� � (�7, 0, 0) � t(4, 1, 0) has direction

d�� (4, 1, 0), �d�� 17�; direction cosines

cos � � ��

417��, cos � � �

�117��, cos δ � 0;

and direction angles α � 14�, � � 76�, δ � 90�.

13. a. The parametric equation of the two lines are:l1: x� 4t, y � 3t, z � 2 � 4t and l

2: x � �4 � 4u,

y � 1 � u, z � �2u.Equating components and rearranging4t � 4u � �4 ➀3t � u � 1 ➁

4t � 2u � �2 ➂2 � ➁ � ➂: 10t � 0, t � 0, u � �1. Substitute in ➀: 4(0) � 4(�1) � �4 whichverifies. Therefore the lines intersect at (0, 0, 2).

b. Two lines x � t, y � 1 � 2t, z � 3 � t and

x � �3, y � �6 � 2u, z � 3 � 6u.Equating components gives t � �3.

�5 � � 6 � 2u, u � �12

�

6 � 3 � 6u, u � ��12

� � �12

�; therefore the two linesdo not intersect.

14. a. P(2, 1, 3), Q(0, �4, 7). The distance between these

points is �QP�� where QP�� (2, 5, �4) and

�QP�� 4 � 25� � 16�

� �45�.

The shortest distance between P and Q is 3�5�.

b. The distance from A(3, 7) to 2x � 3y � 7 � 0 is

given by d ��Ax

�1

�

A2

B

�

y1

B�

�

2�

C��

� ��6 �

�

21

13�

� 7��

d � ��

22

13��.

The shortest distance from the point to the line is

�22�

1313�

�.

z

y

x

6

4

8

(0, 8, 4)



c. From the point A(4, 0, 1) to the line,

r� � (2, �2, 1) � t(1, 2, �1). A direction of the

line is d�� (1, 2, �1) and a point on the line is

P(2, �2, 1).

Now PA�� (2, 2, 0), PA�� d�� (�2, 2, 2)

�PA�� d�� 2�3�, �d�� 6�.

The perpendicular distance from the point A to the

line is ��PA��

�d�

��d��

� � �2

�

�

6�

3�� 2�.

d. From the point A(1, 3, 2) to the line

�x

�

�

1

1� � �

y �

1

3� � �

z �

2

7�. A direction of the line is

d�� (�1, 1, 2). A point on the line is P(1, 3, 7).

Now AP�� (0, 0, 5), AP�� d�� (�5, �5, 0),

�AP�� d�� 5�2�, �d�� 6�.

The perpendicular distance from A to the line is

given by ��PA��

d�

�

d��

5

�

�

6�

2��

5�

3

3��.

15.

Let the foot of the perpendicular be A(�6 � 5t, �7 � 3t, �3 � 4t)

QA�� (�9 � 5t, �9 � 3t, �7 � 4t).

Since QA�� is perpendicular to the line, QA�� · d�� 0

therefore �45 � 25t � 27 � 9t � 28 � 16t � 050t � 100

t � 2.The coordinates of the foot of the perpendicular are(4, �1, 5).

Chapter 7 Test

1. A line through A(9, 2), B(3, 4) has direction

AB�� (�6, 2) � �2(3, �1).

a. A vector equation is r� � (9, 2) � t(3, �1).

b. Parametric equations are x � 9 � 3t, y � 2 � t.

c. Symmetric equations are �x �

3

9� � �

y

�

�

1

2�.

d. The scalar equation is x � 3y � 15 � 0.

2. A line l is perpendicular to 2x � 3y � 18 � 0

therefore its direction is d�� (2, �3). The y-intercept

of (x, y) � (0, 1) � t(�3, 4) is 1. The symmetric

equation of l is �2

x� � �

y

�

�

3

1� and the scalar equation is

3x � 2y � 2 � 0.

3. The line �x �

6

2� � �

y �

3

4� � �

z

�

�

3

2� has direction

d�� (6, 3, �3) � 3(2, 1, �1). Parametric equations

are x � 2 � 2t, y � 4 � t, z � �2 � t. For an

intersection with the xy-plane, z � 0, t � �2, and

x � �2, y � 2. The point in the xy-plane is A(�2, 2, 0).

For an intersection in the yz-plane, x � 0, t � �1, and

y � 3, z � �1. The point in the yz-plane is

B(0, 3, �1). The intersection with the xz-plane is

C(�6, 0, 2).

z C

A

y

B

x

A

l

Q (3, 2, 4)

d=(5, 3, 4)→

4. The line x � y � z � 2 has direction d�� (1, 1, 1).

A point on the line is A(0, 0, 2). P has coordinates

(1, �2, �3) and AP�� (1, �2, �5). AP�� d��

(3, �6, 3).

�AP�� d�� 9 � 36� � 9� � 3�1 � 4�� 1� � 3�6�.

�d�� 3�.

The perpendicular distance from P to the line is

��AP��

�d��

�d��

� � �3

�

�

3�

6�� 3�2�.

5. A line through (0, 0, 0) has direction angles � � 120�,

� � 45�.

Since cos2 � � cos2 � � cos2 � � 1,

cos2 � � cos2 120� � cos2 45� � 1;

cos 120� � ��

21�,

cos � � ��

1

2��

�

2

2��

cos2 � � �1

4� � �

1

2� � 1

cos2 � � �1

4�, cos � � �

1

2� or �

�

21�.

Now a direction is (1, �2, �2�) or (�1, �1, �2�).

Vector equations of the two lines are

r� � (1, �1, �2�)s and r� � (�1, �1, �2�)t.

6. Given l1: x � �2 � 5t, y � t, z� �3 � 3t

and l2: x � 5 � s, y � 8 � 2s, z � �6 � 3s.

Equating components and rearranging:5t � s � 7 ➀t � 2s � 8 ➁3t � 3s � �3 ➂➁ � ➂ � 3: �3s � 9, s � �3, t � 2from ➀: 5(2) �3 � 7, which verifies. The lines intersect at the point (8, 2, 3).

7. Given l1: x � �8 � t, y � �3 � 2t, z � 8 � 3t,

d��1

� (1, �2, 3) and

l2: x � 1 � 2s, y � �1 � s, z � 3s, d��

2� (2, 1, 3).

Since the direction d��1

� kd��2, the lines are not parallel.

Equating components gives:t � 2s � 9 ➀2t � s � �2 ➁3t � 3s � �8 ➂2 ➁ � ➀: 5t � 5, t � 1, s � �4.

From ➂, 3(1) �3(�4) � 3 � 12 � 15 � �8therefore the lines do not intersect. Since the lines are not parallel and do not intersect,the two lines are skew.

b. When t � �2, the coordinates of P1

are (�10, 1, 2).

c.

c. Let the coordinates of P2

be (1 � 2s, �1 � s, 3s).

Now P1P

2�� (11 � 2s, �2 � s, �2 � 3s).

Since P1P

2is perpendicular to l

2P

1P

2�� · d��

2� 0,

therefore 22 � 4s � 2 � s � 6 � 9s � 0

14s � �14

s � �1.

The coordinates of P2

are (�1, �2, �3).

l2

l1

P2

P1 (�10, 1, 2)

→d = (2, 1, 3)2


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