ch07
Transcript of 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
Chapter 7: Lines in a Plane 93
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
94 Chapter 7: Lines in a Plane
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
Chapter 7: Lines in a Plane 95
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
96 Chapter 7: Lines in a Plane
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)
Chapter 7: Lines in a Plane 97
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
98 Chapter 7: Lines in a Plane
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.
Chapter 7: Lines in a Plane 99
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
100 Chapter 7: Lines in a Plane
Chapter 7: Lines in a Plane 101
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
102 Chapter 7: Lines in a Plane
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
Chapter 7: Lines in a Plane 103
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
�
104 Chapter 7: Lines in a Plane
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.
Chapter 7: Lines in a Plane 105
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)
106 Chapter 7: Lines in a Plane
Chapter 7: Lines in a Plane 107
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
108 Chapter 7: Lines in a Plane