Ch07
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Chapter 7 • Lines in a Plane Solutions for Selected Problems
c. l1: r (1, 7) t(3, 4); l2: r (2, 0) s(4,
3). The direction vectors are d1 (3, 4) and d2 (4, 3). Since d · d 0 and d ≠ kd , the
Exercise 7.1
1
2
1
2
lines are neither perpendicular nor parallel. 8. b. The vector equation of a line passing through
3 2 2 P 2, with direction d , 6 (1, 9) 4 3 3
3 is r 2, t(1, 9). 4 1 3
(3, 7). 11. The direction of r (1, 8) t(3, 7) is d 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).
2 3
9. a. x 2t, y 3 t.
12. a. x 6, y 1 7t
2 2 The direction vector d (2, ) (3, 1). 3 3 A direction vector with integer components is (3, 1). There is no point on this line with integer coordinates.
3 2 3 4 1 1 1 d , 4, 3. A direction vector with 3 4 12
Since x 6, this line is parallel to the y-axis therefore does not intersect the y-axis. The line intersects the x-axis at the point (6, 0). y
4
1 1 1 1 b. r , t , . The direction vector
integer components is (4, 3). A point on the line with integer coordinates is (1, 1) when t 2.
2 1
1 2
c. r , 3 t , 5 . The direction vector
1 1 d , 5 (1, 10). A direction vector 2 2 with integer components is (1, 10). A point on the line with integer coordinates is (1, 18) when t 3.
–3
4
x
6
–2
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 line intersects the y-axis when x 0, t 5, and y 35, i.e., the point (0, 35). y
10. a. l1: x 1 3t , y 7 4t; l2: x 2 4s,
y 3s. The direction vectors are d1 (3, 4) and d (4, 3). Since d · d 0, the lines are 2
1
35
2
perpendicular. b. l1: r (1, 7), t(3, 4); l2: r (2, 0) s(3, 4),
the direction vectors are d1 (3, 4) and d2 (3, 4). Since d d , the lines are parallel. 1
Chapter 7: Lines in a Plane
2
–35
7 14
35
x
–21
91
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 line intersects
2 11 the y-axis when x 0, t , and y , i.e., 3 3 11 the point 0, . 3
14. a. i) The line r (2, 6) t(3, 4) has direction
d (3, 4). The direction of the positive x-axis is ˆi (1, 0). The angle the line makes with the x-axis is α. (a positive rotation about A). To use (3, 4). the dot product we use direction AB · ˆi AB ˆi cos α Therefore AB 3 5 cos α 3 cos α 5
y
α 127. The line makes an angle of 127 with the x-axis. 5 y B
x
–2
5
α
10
–2
A
x
→
d
P 13. l1: r (3, 9) t(2, 5) x 3 2t, y 9 5t when t 1; x 1, y 4 and (1, 4) lies on l1. 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 d1 (2, 5) and d2 (3, 1). The angle between the direction vectors is the angle of intersection of these lines. Now d · d d d cos 1
2
1
2
6 5 29 10 cos
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 · ˆi dˆi cos α 5 26 cos α 5 cos α 26 α 11. The line makes an angle of 11 with the x-axis.
1 cos 29 10
y 5
87. →
The acute angle between these lines is 87.
) d
P(6,1 α
–3
5 –2
92
Chapter 7: Lines in a Plane
x
b. (i) x 5 8t, y 3 5t.
b. Consider a line l that intersects the x-axis at
A(a, 0). Choose a point B(x1, y1), y1 0 on l. The angle of inclination of the line is ∠BAX α.
5 y3 A symmetric equation is x . 8 5
y
y
(ii) r (0, 4) t(4, 1). A symmetric equation
α
α
x
x A(a, 0)
A(a, 0)
(12, 2) 2(6, 1). B(5, 4) is AB A symmetric equation of this line is
Translate the line to the left a units
x7 y2 . 6 1
therefore A(a, 0) → A'(0, 0) B(x1, y1) → B'(x1 a, y1).
17. a. An equation of the line passing through A(7, 3)
y
1
α
α
with direction vector d (2, 5) is r (7, 3) t(2, 5). If t 1, r (5, 8) then
'
'
x
x y4 is . 4 1 c. A direction of the line through A(7, 2),
y1 0 y1 The slope of line l is m . x1 a x1 a
y
x5 y3 Solving for t gives t and t. 8 5
x
P(5, 8) is on the line. If t 5, r2 (17, 22), then Q(17, 22) is on the line. b. The line segment PQ is defined by the parametric
y1 By definition tan α x1 a y1 but m (the slope of l) x1 a therefore tan α m. Note: if x1 a, the slope is undefined and α 90.
equation x 7 2t, y 3 5t, for 1 t 5. OP OQ 18. a. A direction vector of the line is PQ p q. The vector equation of the line, passing through P , is r p t(p q) with direction PQ r (1 t)p tq.
15. Point A(24, 96) with velocity vector v (85, 65)
y
(units in km and km/h). a. Parametric equations of the highway line are
x 24 85t, y 96 65t.
Q
b. The horizontal velocity component is 85 km/h. The
102 time taken to travel 102 km at 85 km/h is 85 hours (1h 12 min). 102 c. When t , x 126, y 18. The coordinates 85 of P at that time will be (126, 18).
P O
x
16. a. Parametric equations of a line are x x0 ab1,
y y0 bt.
xx yy Solving for t gives 0 t, 0 t. a b xx y y0 Therefore 0 . a b
Chapter 7: Lines in a Plane
93
· d m d cos m 1 1
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.
5 cos m1 2m2 m
c. When t 2, r p 2q 2q, SR p OS OPSR is a parallelogram PQ QR R divides PQ in the ratio 2:1.
m1 2m2 cos . 5m Equating cos gives S
11m1 2m2
Q
m1 2m2
55m 5m
R
11m1 2m2 5m1 10m2
P
16m1 8m2 1 m1 m2· 2
O
1 d. If t , R will be the midpoint of PQ. Therefore 2 1 for t > , R will be closer to Q than to P. 2 19. a. b.
l1 →
m →
d2
→
The vector equation of l3 is r (5, 2) s(1, 2). To determine the bisector of the other angle we use directions d1 and d2 which gives equation
11m1 2m2 5m1 10m2 6m1 12m2
A(5, 2)
d 2
m1 2m2
55m 5m
→
l4
3
11m1 2m2
l3
d1
Choose m2 2. Therefore m1 1 and the (1, 2). direction of l is m
m1 2m2. Choose m2 1. Therefore m1 2 and the
l2
direction of this line is (2, 1). The vector
l1: r1 (5, 2) t(3, 6), d1 (1, 2) l : r (5, 2) u(11, 2), d (11, 2). 2
2
2
Lines intersect at A(5, 2).
equation of the second angle bisector is l : r (5, 2) v(2, 1). 4
c. The direction of the two lines are (1, 2) and
(2, 1). Since (1, 2) · (2, 1) 0, the two lines are perpendicular.
Let l3 represent the line bisecting the angle between l1 and l2 and let the angle between l1 and l3, and l2
Exercise 7.2
and l3, be . Let the direction vector of l3 be (m , m ). m
9. a. 5x 3y 15 0.
1
2
· d m d cos m 2 2 m 125 11m1 2m cos 2 11m1 2m2 cos . 55 m
94
Chapter 7: Lines in a Plane
5 Solve for y: y 5. 3 Let x 3t, y 5t 5 A vector equation is r (0, 5) t(3, 5). Scalar equations are x 3t y 5 5t. x y5 A symmetric equation is . 3 5
x3 2
b. 4x 6y 9 0.
y4 7
b. therefore 7x 21 2y 8
2 3 Solve for y: y x . 3 2 3 Let x 3t, y 2t . 2
7x 2y 13 0 21 4 13 and d 49 4
3 A vector equation is r 0, t(3, 2). 2 Parametric equations are x 3t
12 d . 53
1 1 c. r (3, 7) t , . A direction vector
3 y 2t . 2
5 6 (6, 5) therefore a normal n (5, 6). A point m on the line is P(3, 7).
3
y 2 x A symmetric equation is . 3 2 10. Let P be a point not on the line l and D be on l, the foot
of the perpendicular from P. Choose a point A on l other than D. Now PAD is a right triangle where PA is the hypotenuse. Therefore PA is the longest side and PA PD. Therefore the shortest distance from a point to a line is the perpendicular distance from the point to the line. P
· n PQ Now d n (6, 5) · (5 6) n 0 therefore the point is on the line. d. The distance from Q to x 5 is 8.
l
x = –5
A
2 2
D x = –5
–5
11. The distance d from Q(3, 2) to each of the following
3
lines:
Q (3, –2)
a. 3x 2y 6 0
Ax1 By1 C d 2 A B2 946 13 7 d . 13
12. Equation of the line is 6x 3y 10 0. a. Point (4, 7)
24 21 10 d 36 9
5 35 · 35 5 75 d . 3
Chapter 7: Lines in a Plane
95
b. Point (4, 8)
b. If two lines in a plane are perpendicular, their
normals are perpendicular. Let two lines l1 and l2 and m . Let m (a, b). Since have direction m
24 24 10
d 35
1
2
1
· m 0, therefore m (kb, ka). l1 ⊥ l2, m 1 2 2 A normal to l1 is (b, a) and to l2 is (ka, kb). Now
10 35
(b, a) · (ka, kb) kab kab 0, therefore the normals are perpendicular.
25 d .
If the normals of two lines are perpendicular, then
3
the two lines are perpendicular. Let two lines l1 and l2 have normals n1 (A, B) and n . Since n ⊥ n , n · n 0 and
c. Point (0, 5)
15 10 d 35
2
1
2
1
2
n2 (kB, kA). The normal to l1 is n1 (A, B), (B, A). Similarly a therefore a direction is m 1
(kA · kB). Now direction of l2 will be m 2 · m 0, therefore m ⊥ m and the two lines m 1 2 1 2
5
35
are perpendicular.
5 d 3.
Therefore two lines in a plane are perpendicular if 20 3
d. Point 5,
and only if their normals are perpendicular.
y
14. a.
l
30 20 10
d 35 d 0.
B →
m
13. a. i) Given two lines l1 and l2. Let the direction
A
α
x E
vector of l1 be d1 (a, b). Since l1l2, the direction vector d of l is a multiple of d . Let 2
2
1
d2 (ka, kb). Now the normal of l1 is n1 (b, a) and of l is n – (kb, ka). But n 2
2
2
(kb, ka) k(b, a) kn1 therefore n1n2. ii) Given two lines l1 and l2 having normals n1 and
angle of inclination of l is ∠BAX α. Let the m (m , m ). direction vector of l be AB 1
2
n2. Let the normal of l1 be n1 (A, B) since n n , n kn , and n (kA, kB). The
cos α In ∆ABC, AE m1 m sin α. and EB m m
(B, A) and direction of l1 and l2 will be m 1 m (kB, kA). Since m (kB, kA)
Therefore a direction of the line is (cos α, sin α)
. m m and the lines l and l k(B, A) km 1 2 1 1 2
equation of a line is Ax By C 0 where
are parallel, therefore two lines in a plane are
(A, B) is a normal therefore the equation is
parallel if and only if their normals are parallel.
x sin α y cos α C 0.
1
2
2
1
2
2
96
Given a line l intersecting the x-axis at A. The
Chapter 7: Lines in a Plane
2
2
and a normal is (sin α, cos α). The scalar
b. 2x 4y 9 0 has normal (1, 2) therefore a
16. a.
direction is (2, 1).
→
n
m2 tan m1 P(x, y)
1 2
O
153. The angle of inclination is 153.
l
c. The equation will be x sin 120 y cos 120 D 0.
3 1 sin 120 2, cos 120
n is a normal to the line l. P(x, y) is a point on the is the position vector of P. Rotate the line and OP
therefore we have 3x y 2D 0.
line about P until it passes through the origin. Now are coincident. the line l and the position vector OP
2
Now (6, 4) is on the line. 63 4 2D 0, 2D 4 63 and the equation is 3x y 4 63 0.
Since n ⊥ l, it will now be perpendicular to OP 0. and n · OP is a b. If the line goes through the origin then OP
15. a.
direction vector of the line. n is normal to the line 0. therefore n · OP
B(8, 10)
y l
5
0 then n is perpendicular to OP . But n If n · OP N(x, y)
is a normal to the line hence is perpendicular to l. and the line, Since n is perpendicular to both OP
A(2, 2) 5
–5
x
and the line are parallel. But P is a point on the OP line, hence they are coincident and the line passes
–5
(x 2, y 2), BN (x 8, y 10) AN · BN 0 AN (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) and radius 5.
through the origin.
Exercise 7.3 7. Given the points A(2, 3, 2) and B(4, 1, 5). The
3 midpoint of AB is M 3, 1, . The line passes through 2 C(0, 1, 1) and M, therefore a direction is
1 CM 3, 2, . 2 Using direction vector (6, 4, 1) and point C(0, 1, 1), the parametric equations are x 6t, y 1 4t, z 1 t.
b. The midpoint of AB is (5, 6), which is the centre of
the circle. Since both A and B are on the circle, AB is a diameter of the circle.
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
10. a. x t, y 2, z 1: perpendicular to the yz-plane
passing through (0, 2, 1),
(6, 7, 2) and symmetric equation
z
x y z . 6 7 2 9. a. line l1: r (1, 0, 3) t(3, 6, 3) with direction
d1 3(1, 2, 1) and l2: r (2, 2, 5) t(2, 4, 2) with direction d 2(1, 2, 1) since 2
y
x x = t, y = 2, z = –1
3 d1 d2, the lines are parallel. The symmetric 2 equation of l1 is y x1 z3 . 1 1 2
b. x 0, y 1 t, z 1 t: a line in the yz-plane
having y-intercept 2 and z-intercept 2. z
(2, 2, 5) is on l2. Check to see if it is on l1. 2
21 2 53 . 1 2 1 Therefore the lines are parallel and distinct.
2
y
3(1, 0, 2). b. l1: r (2, 1, 4) s(3, 0, 6); d 1
l2: r (3, 0, 1) t(2, 0, 2); d2 2(1, 0, 1). Since d kd , the lines are not parallel nor the 1
x
2
same line. 2(3, 1, 0). c. l1: r (1, 1, 1) s(6, 2, 0); d 1 l2: r (5, 3, 1) t(9, 3, 0); d 3 (3, 1, 0).
c. x 5, y 2 t, z 2 t represents a line in
the plane x 5, a plane parallel to the yz-plane. In this yz-plane the line has equation y z, a line passing through (5, s, s) for all s R.
2
z
2 Since d1 d2, the lines are parallel. 3 x1 y1 Symmetric equation of l1 is ; z 1. 3 1
(5, s, s) 5
Check to see if (5, 3, 1) is on l1.
l
5 1 3 1 2; z 1. 3 1
y
Since (5, 3, 1) lies on l1 and l1 is parallel to l2, l1 and l2 are the same line.
98
Chapter 7: Lines in a Plane
x
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, the
line will be perpendicular to one of the coordinate planes; 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.
b. The line segment AB is the set of points on the line
x 3t, y t, z 2 6t, for 3 t 2. x 11 3
y8 1
z4 1
14. A line l has equation .
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 (7 3t, 13 t, 1 t). Since l and of line m is AT 0 m are perpendicular, d · AT 21 9t 13 t 1 t 0 11t 33. l
T
12. A line l1 passes through the point A(6, 4, 2) and is
perpendicular to both y 10 z2 x l1: and 6 3 4 x5 y5 z5 l2: . 3 2 4 A direction of l1 is d1 (4, 6, 3) and of l2 is d2 (3, 2, 4). The direction of l1d1 is perpendicular to both l1 and l2. Therefore d d d (30, 25, 10) 5(6, 5, 2). The 1
2
symmetric equation of a line through A(6, 4, 3) x6 y4 having direction (6, 5, 2) is 6 5 z3 . 2
→
d = (3, 1, 1) A(4, 5, 5)
m
(2, 10, 4). Line m passes t 3. Now AT 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. →
d P
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 by equating components 3t 9, t 3, 2 6t 16 t 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 14 t 2 t2 Since t 2 generates the point B, B(6, 2, 14) is on the line l.
l
F
Q
l is a line passing through P and having direction vector d. Q is a point, not on the line and FQ is the ⊥ l. is the angle distance from Q to the line. FQ and d. between PQ PQ sin From ∆PFQ, FQ
d sin PQ d d sin PQ . d
Chapter 7: Lines in a Plane
99
d PQ d sin But PQ
Exercise 7.4
d PQ . therefore FQ d
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 5u yt y 8 2u z 3 3t z 6 3u Equating 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 ➂ ➀ ➂ 3: 4t 8, t 2, u 3 From 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. The distance from Q(1, 2, 3) to the line
r (3, 1, 0) t(1, 1, 2). A point on the line is (2, 3, 3) and P(3, 1, 0), therefore PQ d (3, 1, 1) d (1, 1, 2). PQ d PQ 91 1 11 ,
d 11 4 6. Therefore the distance from Q
66 11 to the line is . 6 6
b. line 1
c.
l1 A(2, 2, 1) l2 F
B(2, 1, 2)
Two lines l1: r (2, 2, 1) t(7, 3, 4), l : r (2, 1, 2) u(u, 3, 4). 2
The distance between two parallel lines is the perpendicular distance from a point on one of the lines to the other line. A(2, 2, 1) is a point on l1. The direction of l2 is d (7, 3, 4) and a point on l is B(2, 1, 2). 2
2
(4, 3, 3), BA d (21, 5, 33). Now BA d BA Therefore AF d
1555 . 74 The distance between the two parallel lines is
1555 . 74
line 2 x1t x 3 2u y 1 2t y 5 4u z 1 3t z 5 6u Equating components and rearranging terms: 1 t 3 2u t 2u 2 ➀ 1 2t 5 4u 2t 4u 4 ➁ 1 3t 5 6u 3t 6u 6 ➂ Each of equations ➀, ➁, and ➂ are equivalent. Note that d (1, 2, 3), d (2, 4, 6) 1
2
2(1, 2, 3) 2d1 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: x2t x 1 2u y 1 2t y1u z 3t z2u Equating components and rearranging gives: t 2u 3 ➀ 2t u 2 ➁ 3t u 2 ➂ ➁ ➂: 5t 0, t 0 and u 2.
From ➀ t 2u 0 4 4 3, the lines do not intersect. Since d (1, 2, 3), d 1
Therefore the lines are skew.
100 Chapter 7: Lines in a Plane
2
(2, 1, 1) and d1 ≠ kd2, the lines are not parallel.
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 2 t 2u 2 t 2u 1 there is no solution to this system of equation hence the lines do not intersect. d (1, 1, 1), 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 ➁ 4t 3u 5 ➂ 1 ➀ ➁ ÷ 2: 8t 4, t , u 1 which also 2 satisfies ➂ therefore the two lines intersect at
d2 (2, 2, 2) 2(1, 1, 1). Since d 2d , the lines are parallel. Therefore
(2, 3, 1). The directions are d1 (1, 2, 1) and d (1, 2, 3).
the lines are parallel and distinct.
d1 · d2 1 4 3 0. Therefore the two lines
1
x3 4
2
y2 1
e. l1: 0 z 2;
x2 y1 z 2 l2: 3 2 1 The parametric equations are: x 3 4t x 2 3u y2t y 1 2u z2t z 2u Equating components and rearranging gives: 4t 3u 1 ➀ t 2u 3 ➁ tu0 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 ➁ t 3u 1 ➂ ➀ ➁ 2: 2t 2, t 1, u 0. t 1, u 0 also satisfies ➂. Therefore the two lines intersect at A(2, 3, 0).
2
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 true for t 4. Therefore the x-intercept is 24 7(4) 4. For the y-intercept, both x 0 and z 0 for the same t, 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 a direction 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 ➁ ➀ 5 ➁: 29y 29 21 y 1, x . 2 21 The point of intersection is , 1 . 2
9. Three lines l1, l2, and l3 are in the same plane. Possible
intersections are:
a. All intersect in a common point.
l1
(3, 2, 1) and b. The direction of the two lines are d 1 d2 (1, 2, 3). d d (4, 8, 4). 1
2
l2
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).
l3
Chapter 7: Lines in a Plane 101
b. Pairs of lines intersect, but there is no common
b. Two parallel lines and the third intersecting one of
intersection.
the two. l1 l1
l2 l2
A
l3 l3
l1l2, l3 intersects l2 at A.
c. One line intersects two distinct parallel lines. 11.
l1
l
l2
T →
l3
(1,
2,
) 2
d=
B (7, 13, 8)
A (5, 4, 2) m
d. One line intersects two coincident lines.
l1
l 2 and l 3 e. The three lines are coincident.
l1, l2 , and l3 10. Three lines l1, l2, l3 are in space. The possible
intersections, in addition to those of question 9, are:
a. A line intersecting two skew lines. l1
Given line l: r (7, 13, 8) t(1, 2, 2), point A(5, 4, 2). The line m, through A, intersects l at right angles. Let T be the point of intersection of l and m. Since T is on l, we represent its coordinates by T(7 t, 13 2t, 8 2t). (12 t, 9 2t, 6 2t) is a direction of m. AT Since l and m are perpendicular, their directions are 0. perpendicular, hence d · AT (1, 2, 2) · (12 t, 9 2t, 6 2t) 0 12 t 18 4t 12 4t 0 9t 18 t 2. (14, 5, 2) and the equation of m is Now AT 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
l2
equations x t, y 5 3t, z 3 2t, intersects the sphere x2 y2 z2 6. (The centre of the sphere is
B
C(0, 0, 0) and its radius is 6.) Substituting for x, y, A
l3
l1 and l2 are skew. l3 intersects l1 and l2 at A and B. 102 Chapter 7: Lines in a Plane
and z into the equation of the sphere: t2 (5 3t)2 (3 2t)2 6 2 t 25 30t 9t2 9 12t 4t2 6 14t2 42t 28 0 t2 3t 2 0 (t 2)(t 1) 0 t 2 or 1
therefore the line intersects the sphere at A(1, 2, 1)
3 1 and B(2, 1, 1). The midpoint of AB is , , 0 2 2
A direction for l3 is (17, 15, 20) and the equation of l is r s(17, 15, 20). 3
NOTE:
which is not the centre of the sphere, therefore AB is not a diameter of the sphere. 13.
B l1 l2 z
To determine the equation of l3 we require a 121 direction. Once t is established, a 16 direction is evident and further substitutions to determine u are not required. Upon substitution, one would find that u 41.
14. a.
y
m
A l
y
O
N
x l3 x
O
Two lines: l1: r (2, 16, 19) t(1, 1, 4); l : r (14, 19, 2) u(2, 1, 2). A line l passes 2
3
through the origin and intersects l2 at A and l1 at B.
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
Since B is on l1, we represent its coordinates as B(2 t, 16 t, 19 4t).
direction of the line m along ON. The equation of m is r t(A, B); x At, y Bt.
Similarly the coordinates of A will be A(14 2u, 19 u, 2 2u). Since O, A, and B are collinear kOA OB (2 t, 16 t, 19 4t) k(14 2u, 19 u, 2 2u). Equating components: 2 t 14k 2ku ➀ 16 t 19k ku ➁ 19 4t 2k 2ku ➂ We solve by first eliminating ku: ➀ ➂: 21 3t 12k ➃ ➀ 2 ➁: 30 3t 52k ➄ add: 9 64k 9 k . 64 Substitute in ➃: 21 3t 12k 7 t 4k 9 t 7 4 · 64 121 t . 16 121 With t , we can find the coordinates of B hence 16 153 135 180 we have OB , , . 16 16 16
Substituting in l: A2t B2t C 0 C t . 2 A B2 BC AC The coordinates of N are , . 2 2 A B A2 B2
b.
BC AC ON , 2 2 A B A2 B2
ON
A2C2 B2C2 (A2 B2)2
C2(A2 B2) (A2 B2) C2 2 A B2 C
. ON A2 B2
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). (2 s 2t, 3 2s t, s 2t). BA
z 3 s. Two points P(6, 4, 0) and Q(0, 5, 3). d (0, 1, 1), d (2, 0, 1). 1
2
(6, 9, 3), Now PQ n d d (1, 2, 2). 1
2
· n PQ 6 18 6 18 6. 14 4 3 n The distance between the lines is 6.
l1
A
b. x 6, y 4 t, z t; x 2s, y 5,
Review Exercise
2. a. A line through A(3, 9), B(4, 2) has a direction AB
B l2
(7, 7). An equation is r (3, 9) t(1, 1). b. A line passes through A(5, 3) and is parallel to r
l
is perpendicular to l therefore BA · BA d1 0 1 2 s 2t 6 4s 2t s 2t 0 6s 2t 4, 3s t 2 ➀ is perpendicular to l , therefore BA · BA d2 0 2 4 2s 4t 3 2s t 2s 4t 0 2s 9t 7 ➁ 9 ➀ ➁: 25s 25 s 1, t 1. The coordinates of the points of intersection are A(1, 1, 1) and B(0, 1, 2). td , 16. The distance between two skew lines r OP 1 · n PQ
sd is given by where n d d . r OQ 2 1 2 n
r (0, 5, 0) s(1, 1, 2). The two points P(0, 2, 6), Q(0, 5, 0) and directions d (2, 1, 1), d (1, 1, 2).
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
2 . A direction is (3, 2) and parametric 3 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. (4, 2) AB 2(2, 1). The line has equation x 4 2t, y t.
2
(0, 3, 6), n d d (3, 3, 3) Now PQ 1 2
9 18 3 3 . 3 33 n · n PQ
The distance between the lines is 3 .
104 Chapter 7: Lines in a Plane
c. A line passes through A(0, 3) and is perpendicular
c. A line through A(4, 0), B(0, 2) has direction
a. r (0, 2, 6) t(2, 1, 1);
1
(4, 0) t(0, 5). A direction is (0, 1) and an equation is r (5, 3) s(0, 1).
4. a. The line passes through A(2, 0, 3), B(3, 2, 2).
(5, 2, 1) and a vector A direction is AB equation is r (2, 0, 3) t(5, 2, 1).
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 (7, 4, 0). A vector equation is a direction is AB r (7, 0, 0) t(7, 4, 0). x5 y2 z6 c. A line l, parallel to , 4 2 5 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
x1 y2 the line z 3. A direction of 3 2 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 parallel
to the y-axis. A direction is (0, 1, 0). Parametric equations 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 parallel
to 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 is
perpendicular to x 2 t, y 3 2t. A normal of 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 (5, 2, 3) and parametric equations x 6 BC 5t, y 4 2t, z 3t. b. Since (4, m, n) l, 6 5t 4, t 2;
8. a. l1: r (2, 3) t(3, 1),
d1 (3, 1) l2: r (1, 4) u(6, 2), d2 (6, 2) 2(3, 1). Since d2 2d1, l1 and l2 are
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; d1 (2, 1).
l2: x u,
1 y 2u; d2 (1, 2). 3
Since d1 · d2 0, the two lines are perpendicular. x1 y4 c. l1: , z 1; d1 (2, 1, 0). 2
1
l2: x 4t, y 1 2t, z 6; d2 (4, 2, 0) 2(2, 1, 0). Since d 2d , l and l are parallel. 2
1
1
2
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),
d1 (1, 1, 1). l2: (x, y, z) (3, 0, 1) u(2, 2, 2), d2 (2, 2, 2) 2(1, 1, 1). Since d1 kd2, d1 · d2 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,
1 11 t , and the point is (3, , 0). 2 2 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
y3 x2 and the scalar equation is 5 1 5x y 13 0.
4 2t m, m 8; 3t n, n 6.
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). 3 1 3 4 2 4 (4, 3). A point on the line is (2, 2) and a vector equation is r (2, 2) t(4, 3).
c. The line y x has slope and a direction
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.
c. The line r (7, 0, 0) t(4, 1, 0) has direction
d (4, 1, 0), d 17 ; direction cosines 4 1 cos , cos , cos δ 0; 17 17 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 l2: x 4 4u, y 1 u, z 2u. Equating components and rearranging 4t 4u 4 ➀ 3t u 1 ➁ 4t 2u 2 ➂ 2 ➁ ➂: 10t 0, t 0, u 1. Substitute in ➀: 4(0) 4(1) 4 which verifies. 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.
z
4
1 5 6 2u, u 2 1 1 6 3 6u, u ; therefore the two lines 2 2 do not intersect.
(0, 8, 4) 8
y 14. a. P(2, 1, 3), Q(0, 4, 7). The distance between these
where QP (2, 5, 4) and points is QP 4 QP 25 16
6
45 . The shortest distance between P and Q is 35.
x
x 3 5
y6 2
z1 1
12. a. The line has direction
d (5, 2, 1), d 30 ; direction cosines
b. The distance from A(3, 7) to 2x 3y 7 0 is
Ax1 By1 C given by d A2 B2 6 21 7 13
1 5 2 cos , cos , cos δ ; 30 30 30 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 8 1 4 cosines cos , cos , cos δ ; 9 9 9 and direction angles 27, 96, δ 116.
106 Chapter 7: Lines in a Plane
22 d . 13 The shortest distance from the point to the line is
2213 . 13
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). (2, 2, 0), PA d (2, 2, 2) Now PA
d PA 23 line is 2 . d 6
x9
y2
3
1
c. Symmetric equations are . d. The scalar equation is x 3y 15 0.
d. From the point A(1, 3, 2) to the line
x1 y3 z7 . A direction of the line is 1 1 2 d (1, 1, 2). A point on the line is P(1, 3, 7). (0, 0, 5), AP d (5, 5, 0), Now AP d 52 AP , d 6 . The perpendicular distance from A to the line is d PA 52 53 given by .
15.
(6, 2) 2(3, 1). AB b. Parametric equations are x 9 3t, y 2 t.
The perpendicular distance from the point A to the
6
1. A line through A(9, 2), B(3, 4) has direction a. A vector equation is r (9, 2) t(3, 1).
d 23 PA , d 6 .
d
Chapter 7 Test
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 x y1 equation of l is and the scalar equation is 2 3 3x 2y 2 0. x2
y4
z2
6
3
3
3. The line has direction
3
d (6, 3, 3) 3(2, 1, 1). Parametric equations are x 2 2t, y 4 t, z 2 t. For an
Q (3, 2, 4)
intersection with the xy-plane, z 0, t 2, and
→
d=(5, 3, 4)
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
A
B(0, 3, 1). The intersection with the xz-plane is C(6, 0, 2). l
z
C
Let the foot of the perpendicular be A(6 5t, 7 3t, 3 4t) (9 5t, 9 3t, 7 4t). QA is perpendicular to the line, QA · d 0 Since QA
A
therefore 45 25t 27 9t 28 16t 0 50t 100 t 2. The coordinates of the foot of the perpendicular are (4, 1, 5).
y B
x
Chapter 7: Lines in a Plane 107
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, 5). AP d (1, 2, 3) and AP (3, 6, 3). AP d 9 36 9 31 4 1 36 .
d 3.
From ➂, 3(1) 3(4) 3 12 15 8 therefore 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.
The perpendicular distance from P to the line is
l1
P1 (10, 1, 2)
d AP 36 32. 3 d 5. A line through (0, 0, 0) has direction angles 120,
→
d2 = (2, 1, 3)
45.
l2
P2
Since cos cos cos 1, 2
2
2
c. Let the coordinates of P2 be (1 2s, 1 s, 3s).
cos2 cos2 120 cos2 45 1;
Now P P (11 2s, 2 s, 2 3s). 1 2 P · d 0, Since P P is perpendicular to l P
1 cos 120 , 2
1 2
1 2 cos 2 2
1 2
2
14s 14 s 1.
1 1 cos2 1 4 2
The coordinates of P2 are (1, 2, 3).
1 1 1 cos2 , cos or . 2 4 2 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 ➀ t 2s 8 ➁ 3t 3s 3 ➂ ➁ ➂ 3: 3s 9, s 3, t 2 from 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,
d1 (1, 2, 3) and
l2: x 1 2s, y 1 s, z 3s, d2 (2, 1, 3). Since the direction d kd , the lines are not parallel. 1
2
therefore 22 4s 2 s 6 9s 0
2
Equating components gives: t 2s 9 ➀ 2t s 2 ➁ 3t 3s 8 ➂ 2 ➁ ➀: 5t 5, t 1, s 4. 108 Chapter 7: Lines in a Plane
c. l1: r (1, 7) t(3, 4); l2: r (2, 0) s(4,
3). The direction vectors are d1 (3, 4) and d2 (4, 3). Since d · d 0 and d ≠ kd , the
Exercise 7.1
1
2
1
2
lines are neither perpendicular nor parallel. 8. b. The vector equation of a line passing through
3 2 2 P 2, with direction d , 6 (1, 9) 4 3 3
3 is r 2, t(1, 9). 4 1 3
(3, 7). 11. The direction of r (1, 8) t(3, 7) is d 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).
2 3
9. a. x 2t, y 3 t.
12. a. x 6, y 1 7t
2 2 The direction vector d (2, ) (3, 1). 3 3 A direction vector with integer components is (3, 1). There is no point on this line with integer coordinates.
3 2 3 4 1 1 1 d , 4, 3. A direction vector with 3 4 12
Since x 6, this line is parallel to the y-axis therefore does not intersect the y-axis. The line intersects the x-axis at the point (6, 0). y
4
1 1 1 1 b. r , t , . The direction vector
integer components is (4, 3). A point on the line with integer coordinates is (1, 1) when t 2.
2 1
1 2
c. r , 3 t , 5 . The direction vector
1 1 d , 5 (1, 10). A direction vector 2 2 with integer components is (1, 10). A point on the line with integer coordinates is (1, 18) when t 3.
–3
4
x
6
–2
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 line intersects the y-axis when x 0, t 5, and y 35, i.e., the point (0, 35). y
10. a. l1: x 1 3t , y 7 4t; l2: x 2 4s,
y 3s. The direction vectors are d1 (3, 4) and d (4, 3). Since d · d 0, the lines are 2
1
35
2
perpendicular. b. l1: r (1, 7), t(3, 4); l2: r (2, 0) s(3, 4),
the direction vectors are d1 (3, 4) and d2 (3, 4). Since d d , the lines are parallel. 1
Chapter 7: Lines in a Plane
2
–35
7 14
35
x
–21
91
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 line intersects
2 11 the y-axis when x 0, t , and y , i.e., 3 3 11 the point 0, . 3
14. a. i) The line r (2, 6) t(3, 4) has direction
d (3, 4). The direction of the positive x-axis is ˆi (1, 0). The angle the line makes with the x-axis is α. (a positive rotation about A). To use (3, 4). the dot product we use direction AB · ˆi AB ˆi cos α Therefore AB 3 5 cos α 3 cos α 5
y
α 127. The line makes an angle of 127 with the x-axis. 5 y B
x
–2
5
α
10
–2
A
x
→
d
P 13. l1: r (3, 9) t(2, 5) x 3 2t, y 9 5t when t 1; x 1, y 4 and (1, 4) lies on l1. 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 d1 (2, 5) and d2 (3, 1). The angle between the direction vectors is the angle of intersection of these lines. Now d · d d d cos 1
2
1
2
6 5 29 10 cos
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 · ˆi dˆi cos α 5 26 cos α 5 cos α 26 α 11. The line makes an angle of 11 with the x-axis.
1 cos 29 10
y 5
87. →
The acute angle between these lines is 87.
) d
P(6,1 α
–3
5 –2
92
Chapter 7: Lines in a Plane
x
b. (i) x 5 8t, y 3 5t.
b. Consider a line l that intersects the x-axis at
A(a, 0). Choose a point B(x1, y1), y1 0 on l. The angle of inclination of the line is ∠BAX α.
5 y3 A symmetric equation is x . 8 5
y
y
(ii) r (0, 4) t(4, 1). A symmetric equation
α
α
x
x A(a, 0)
A(a, 0)
(12, 2) 2(6, 1). B(5, 4) is AB A symmetric equation of this line is
Translate the line to the left a units
x7 y2 . 6 1
therefore A(a, 0) → A'(0, 0) B(x1, y1) → B'(x1 a, y1).
17. a. An equation of the line passing through A(7, 3)
y
1
α
α
with direction vector d (2, 5) is r (7, 3) t(2, 5). If t 1, r (5, 8) then
'
'
x
x y4 is . 4 1 c. A direction of the line through A(7, 2),
y1 0 y1 The slope of line l is m . x1 a x1 a
y
x5 y3 Solving for t gives t and t. 8 5
x
P(5, 8) is on the line. If t 5, r2 (17, 22), then Q(17, 22) is on the line. b. The line segment PQ is defined by the parametric
y1 By definition tan α x1 a y1 but m (the slope of l) x1 a therefore tan α m. Note: if x1 a, the slope is undefined and α 90.
equation x 7 2t, y 3 5t, for 1 t 5. OP OQ 18. a. A direction vector of the line is PQ p q. The vector equation of the line, passing through P , is r p t(p q) with direction PQ r (1 t)p tq.
15. Point A(24, 96) with velocity vector v (85, 65)
y
(units in km and km/h). a. Parametric equations of the highway line are
x 24 85t, y 96 65t.
Q
b. The horizontal velocity component is 85 km/h. The
102 time taken to travel 102 km at 85 km/h is 85 hours (1h 12 min). 102 c. When t , x 126, y 18. The coordinates 85 of P at that time will be (126, 18).
P O
x
16. a. Parametric equations of a line are x x0 ab1,
y y0 bt.
xx yy Solving for t gives 0 t, 0 t. a b xx y y0 Therefore 0 . a b
Chapter 7: Lines in a Plane
93
· d m d cos m 1 1
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.
5 cos m1 2m2 m
c. When t 2, r p 2q 2q, SR p OS OPSR is a parallelogram PQ QR R divides PQ in the ratio 2:1.
m1 2m2 cos . 5m Equating cos gives S
11m1 2m2
Q
m1 2m2
55m 5m
R
11m1 2m2 5m1 10m2
P
16m1 8m2 1 m1 m2· 2
O
1 d. If t , R will be the midpoint of PQ. Therefore 2 1 for t > , R will be closer to Q than to P. 2 19. a. b.
l1 →
m →
d2
→
The vector equation of l3 is r (5, 2) s(1, 2). To determine the bisector of the other angle we use directions d1 and d2 which gives equation
11m1 2m2 5m1 10m2 6m1 12m2
A(5, 2)
d 2
m1 2m2
55m 5m
→
l4
3
11m1 2m2
l3
d1
Choose m2 2. Therefore m1 1 and the (1, 2). direction of l is m
m1 2m2. Choose m2 1. Therefore m1 2 and the
l2
direction of this line is (2, 1). The vector
l1: r1 (5, 2) t(3, 6), d1 (1, 2) l : r (5, 2) u(11, 2), d (11, 2). 2
2
2
Lines intersect at A(5, 2).
equation of the second angle bisector is l : r (5, 2) v(2, 1). 4
c. The direction of the two lines are (1, 2) and
(2, 1). Since (1, 2) · (2, 1) 0, the two lines are perpendicular.
Let l3 represent the line bisecting the angle between l1 and l2 and let the angle between l1 and l3, and l2
Exercise 7.2
and l3, be . Let the direction vector of l3 be (m , m ). m
9. a. 5x 3y 15 0.
1
2
· d m d cos m 2 2 m 125 11m1 2m cos 2 11m1 2m2 cos . 55 m
94
Chapter 7: Lines in a Plane
5 Solve for y: y 5. 3 Let x 3t, y 5t 5 A vector equation is r (0, 5) t(3, 5). Scalar equations are x 3t y 5 5t. x y5 A symmetric equation is . 3 5
x3 2
b. 4x 6y 9 0.
y4 7
b. therefore 7x 21 2y 8
2 3 Solve for y: y x . 3 2 3 Let x 3t, y 2t . 2
7x 2y 13 0 21 4 13 and d 49 4
3 A vector equation is r 0, t(3, 2). 2 Parametric equations are x 3t
12 d . 53
1 1 c. r (3, 7) t , . A direction vector
3 y 2t . 2
5 6 (6, 5) therefore a normal n (5, 6). A point m on the line is P(3, 7).
3
y 2 x A symmetric equation is . 3 2 10. Let P be a point not on the line l and D be on l, the foot
of the perpendicular from P. Choose a point A on l other than D. Now PAD is a right triangle where PA is the hypotenuse. Therefore PA is the longest side and PA PD. Therefore the shortest distance from a point to a line is the perpendicular distance from the point to the line. P
· n PQ Now d n (6, 5) · (5 6) n 0 therefore the point is on the line. d. The distance from Q to x 5 is 8.
l
x = –5
A
2 2
D x = –5
–5
11. The distance d from Q(3, 2) to each of the following
3
lines:
Q (3, –2)
a. 3x 2y 6 0
Ax1 By1 C d 2 A B2 946 13 7 d . 13
12. Equation of the line is 6x 3y 10 0. a. Point (4, 7)
24 21 10 d 36 9
5 35 · 35 5 75 d . 3
Chapter 7: Lines in a Plane
95
b. Point (4, 8)
b. If two lines in a plane are perpendicular, their
normals are perpendicular. Let two lines l1 and l2 and m . Let m (a, b). Since have direction m
24 24 10
d 35
1
2
1
· m 0, therefore m (kb, ka). l1 ⊥ l2, m 1 2 2 A normal to l1 is (b, a) and to l2 is (ka, kb). Now
10 35
(b, a) · (ka, kb) kab kab 0, therefore the normals are perpendicular.
25 d .
If the normals of two lines are perpendicular, then
3
the two lines are perpendicular. Let two lines l1 and l2 have normals n1 (A, B) and n . Since n ⊥ n , n · n 0 and
c. Point (0, 5)
15 10 d 35
2
1
2
1
2
n2 (kB, kA). The normal to l1 is n1 (A, B), (B, A). Similarly a therefore a direction is m 1
(kA · kB). Now direction of l2 will be m 2 · m 0, therefore m ⊥ m and the two lines m 1 2 1 2
5
35
are perpendicular.
5 d 3.
Therefore two lines in a plane are perpendicular if 20 3
d. Point 5,
and only if their normals are perpendicular.
y
14. a.
l
30 20 10
d 35 d 0.
B →
m
13. a. i) Given two lines l1 and l2. Let the direction
A
α
x E
vector of l1 be d1 (a, b). Since l1l2, the direction vector d of l is a multiple of d . Let 2
2
1
d2 (ka, kb). Now the normal of l1 is n1 (b, a) and of l is n – (kb, ka). But n 2
2
2
(kb, ka) k(b, a) kn1 therefore n1n2. ii) Given two lines l1 and l2 having normals n1 and
angle of inclination of l is ∠BAX α. Let the m (m , m ). direction vector of l be AB 1
2
n2. Let the normal of l1 be n1 (A, B) since n n , n kn , and n (kA, kB). The
cos α In ∆ABC, AE m1 m sin α. and EB m m
(B, A) and direction of l1 and l2 will be m 1 m (kB, kA). Since m (kB, kA)
Therefore a direction of the line is (cos α, sin α)
. m m and the lines l and l k(B, A) km 1 2 1 1 2
equation of a line is Ax By C 0 where
are parallel, therefore two lines in a plane are
(A, B) is a normal therefore the equation is
parallel if and only if their normals are parallel.
x sin α y cos α C 0.
1
2
2
1
2
2
96
Given a line l intersecting the x-axis at A. The
Chapter 7: Lines in a Plane
2
2
and a normal is (sin α, cos α). The scalar
b. 2x 4y 9 0 has normal (1, 2) therefore a
16. a.
direction is (2, 1).
→
n
m2 tan m1 P(x, y)
1 2
O
153. The angle of inclination is 153.
l
c. The equation will be x sin 120 y cos 120 D 0.
3 1 sin 120 2, cos 120
n is a normal to the line l. P(x, y) is a point on the is the position vector of P. Rotate the line and OP
therefore we have 3x y 2D 0.
line about P until it passes through the origin. Now are coincident. the line l and the position vector OP
2
Now (6, 4) is on the line. 63 4 2D 0, 2D 4 63 and the equation is 3x y 4 63 0.
Since n ⊥ l, it will now be perpendicular to OP 0. and n · OP is a b. If the line goes through the origin then OP
15. a.
direction vector of the line. n is normal to the line 0. therefore n · OP
B(8, 10)
y l
5
0 then n is perpendicular to OP . But n If n · OP N(x, y)
is a normal to the line hence is perpendicular to l. and the line, Since n is perpendicular to both OP
A(2, 2) 5
–5
x
and the line are parallel. But P is a point on the OP line, hence they are coincident and the line passes
–5
(x 2, y 2), BN (x 8, y 10) AN · BN 0 AN (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) and radius 5.
through the origin.
Exercise 7.3 7. Given the points A(2, 3, 2) and B(4, 1, 5). The
3 midpoint of AB is M 3, 1, . The line passes through 2 C(0, 1, 1) and M, therefore a direction is
1 CM 3, 2, . 2 Using direction vector (6, 4, 1) and point C(0, 1, 1), the parametric equations are x 6t, y 1 4t, z 1 t.
b. The midpoint of AB is (5, 6), which is the centre of
the circle. Since both A and B are on the circle, AB is a diameter of the circle.
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
10. a. x t, y 2, z 1: perpendicular to the yz-plane
passing through (0, 2, 1),
(6, 7, 2) and symmetric equation
z
x y z . 6 7 2 9. a. line l1: r (1, 0, 3) t(3, 6, 3) with direction
d1 3(1, 2, 1) and l2: r (2, 2, 5) t(2, 4, 2) with direction d 2(1, 2, 1) since 2
y
x x = t, y = 2, z = –1
3 d1 d2, the lines are parallel. The symmetric 2 equation of l1 is y x1 z3 . 1 1 2
b. x 0, y 1 t, z 1 t: a line in the yz-plane
having y-intercept 2 and z-intercept 2. z
(2, 2, 5) is on l2. Check to see if it is on l1. 2
21 2 53 . 1 2 1 Therefore the lines are parallel and distinct.
2
y
3(1, 0, 2). b. l1: r (2, 1, 4) s(3, 0, 6); d 1
l2: r (3, 0, 1) t(2, 0, 2); d2 2(1, 0, 1). Since d kd , the lines are not parallel nor the 1
x
2
same line. 2(3, 1, 0). c. l1: r (1, 1, 1) s(6, 2, 0); d 1 l2: r (5, 3, 1) t(9, 3, 0); d 3 (3, 1, 0).
c. x 5, y 2 t, z 2 t represents a line in
the plane x 5, a plane parallel to the yz-plane. In this yz-plane the line has equation y z, a line passing through (5, s, s) for all s R.
2
z
2 Since d1 d2, the lines are parallel. 3 x1 y1 Symmetric equation of l1 is ; z 1. 3 1
(5, s, s) 5
Check to see if (5, 3, 1) is on l1.
l
5 1 3 1 2; z 1. 3 1
y
Since (5, 3, 1) lies on l1 and l1 is parallel to l2, l1 and l2 are the same line.
98
Chapter 7: Lines in a Plane
x
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, the
line will be perpendicular to one of the coordinate planes; 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.
b. The line segment AB is the set of points on the line
x 3t, y t, z 2 6t, for 3 t 2. x 11 3
y8 1
z4 1
14. A line l has equation .
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 (7 3t, 13 t, 1 t). Since l and of line m is AT 0 m are perpendicular, d · AT 21 9t 13 t 1 t 0 11t 33. l
T
12. A line l1 passes through the point A(6, 4, 2) and is
perpendicular to both y 10 z2 x l1: and 6 3 4 x5 y5 z5 l2: . 3 2 4 A direction of l1 is d1 (4, 6, 3) and of l2 is d2 (3, 2, 4). The direction of l1d1 is perpendicular to both l1 and l2. Therefore d d d (30, 25, 10) 5(6, 5, 2). The 1
2
symmetric equation of a line through A(6, 4, 3) x6 y4 having direction (6, 5, 2) is 6 5 z3 . 2
→
d = (3, 1, 1) A(4, 5, 5)
m
(2, 10, 4). Line m passes t 3. Now AT 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. →
d P
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 by equating components 3t 9, t 3, 2 6t 16 t 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 14 t 2 t2 Since t 2 generates the point B, B(6, 2, 14) is on the line l.
l
F
Q
l is a line passing through P and having direction vector d. Q is a point, not on the line and FQ is the ⊥ l. is the angle distance from Q to the line. FQ and d. between PQ PQ sin From ∆PFQ, FQ
d sin PQ d d sin PQ . d
Chapter 7: Lines in a Plane
99
d PQ d sin But PQ
Exercise 7.4
d PQ . therefore FQ d
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 5u yt y 8 2u z 3 3t z 6 3u Equating 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 ➂ ➀ ➂ 3: 4t 8, t 2, u 3 From 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. The distance from Q(1, 2, 3) to the line
r (3, 1, 0) t(1, 1, 2). A point on the line is (2, 3, 3) and P(3, 1, 0), therefore PQ d (3, 1, 1) d (1, 1, 2). PQ d PQ 91 1 11 ,
d 11 4 6. Therefore the distance from Q
66 11 to the line is . 6 6
b. line 1
c.
l1 A(2, 2, 1) l2 F
B(2, 1, 2)
Two lines l1: r (2, 2, 1) t(7, 3, 4), l : r (2, 1, 2) u(u, 3, 4). 2
The distance between two parallel lines is the perpendicular distance from a point on one of the lines to the other line. A(2, 2, 1) is a point on l1. The direction of l2 is d (7, 3, 4) and a point on l is B(2, 1, 2). 2
2
(4, 3, 3), BA d (21, 5, 33). Now BA d BA Therefore AF d
1555 . 74 The distance between the two parallel lines is
1555 . 74
line 2 x1t x 3 2u y 1 2t y 5 4u z 1 3t z 5 6u Equating components and rearranging terms: 1 t 3 2u t 2u 2 ➀ 1 2t 5 4u 2t 4u 4 ➁ 1 3t 5 6u 3t 6u 6 ➂ Each of equations ➀, ➁, and ➂ are equivalent. Note that d (1, 2, 3), d (2, 4, 6) 1
2
2(1, 2, 3) 2d1 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: x2t x 1 2u y 1 2t y1u z 3t z2u Equating components and rearranging gives: t 2u 3 ➀ 2t u 2 ➁ 3t u 2 ➂ ➁ ➂: 5t 0, t 0 and u 2.
From ➀ t 2u 0 4 4 3, the lines do not intersect. Since d (1, 2, 3), d 1
Therefore the lines are skew.
100 Chapter 7: Lines in a Plane
2
(2, 1, 1) and d1 ≠ kd2, the lines are not parallel.
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 2 t 2u 2 t 2u 1 there is no solution to this system of equation hence the lines do not intersect. d (1, 1, 1), 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 ➁ 4t 3u 5 ➂ 1 ➀ ➁ ÷ 2: 8t 4, t , u 1 which also 2 satisfies ➂ therefore the two lines intersect at
d2 (2, 2, 2) 2(1, 1, 1). Since d 2d , the lines are parallel. Therefore
(2, 3, 1). The directions are d1 (1, 2, 1) and d (1, 2, 3).
the lines are parallel and distinct.
d1 · d2 1 4 3 0. Therefore the two lines
1
x3 4
2
y2 1
e. l1: 0 z 2;
x2 y1 z 2 l2: 3 2 1 The parametric equations are: x 3 4t x 2 3u y2t y 1 2u z2t z 2u Equating components and rearranging gives: 4t 3u 1 ➀ t 2u 3 ➁ tu0 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 ➁ t 3u 1 ➂ ➀ ➁ 2: 2t 2, t 1, u 0. t 1, u 0 also satisfies ➂. Therefore the two lines intersect at A(2, 3, 0).
2
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 true for t 4. Therefore the x-intercept is 24 7(4) 4. For the y-intercept, both x 0 and z 0 for the same t, 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 a direction 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 ➁ ➀ 5 ➁: 29y 29 21 y 1, x . 2 21 The point of intersection is , 1 . 2
9. Three lines l1, l2, and l3 are in the same plane. Possible
intersections are:
a. All intersect in a common point.
l1
(3, 2, 1) and b. The direction of the two lines are d 1 d2 (1, 2, 3). d d (4, 8, 4). 1
2
l2
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).
l3
Chapter 7: Lines in a Plane 101
b. Pairs of lines intersect, but there is no common
b. Two parallel lines and the third intersecting one of
intersection.
the two. l1 l1
l2 l2
A
l3 l3
l1l2, l3 intersects l2 at A.
c. One line intersects two distinct parallel lines. 11.
l1
l
l2
T →
l3
(1,
2,
) 2
d=
B (7, 13, 8)
A (5, 4, 2) m
d. One line intersects two coincident lines.
l1
l 2 and l 3 e. The three lines are coincident.
l1, l2 , and l3 10. Three lines l1, l2, l3 are in space. The possible
intersections, in addition to those of question 9, are:
a. A line intersecting two skew lines. l1
Given line l: r (7, 13, 8) t(1, 2, 2), point A(5, 4, 2). The line m, through A, intersects l at right angles. Let T be the point of intersection of l and m. Since T is on l, we represent its coordinates by T(7 t, 13 2t, 8 2t). (12 t, 9 2t, 6 2t) is a direction of m. AT Since l and m are perpendicular, their directions are 0. perpendicular, hence d · AT (1, 2, 2) · (12 t, 9 2t, 6 2t) 0 12 t 18 4t 12 4t 0 9t 18 t 2. (14, 5, 2) and the equation of m is Now AT 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
l2
equations x t, y 5 3t, z 3 2t, intersects the sphere x2 y2 z2 6. (The centre of the sphere is
B
C(0, 0, 0) and its radius is 6.) Substituting for x, y, A
l3
l1 and l2 are skew. l3 intersects l1 and l2 at A and B. 102 Chapter 7: Lines in a Plane
and z into the equation of the sphere: t2 (5 3t)2 (3 2t)2 6 2 t 25 30t 9t2 9 12t 4t2 6 14t2 42t 28 0 t2 3t 2 0 (t 2)(t 1) 0 t 2 or 1
therefore the line intersects the sphere at A(1, 2, 1)
3 1 and B(2, 1, 1). The midpoint of AB is , , 0 2 2
A direction for l3 is (17, 15, 20) and the equation of l is r s(17, 15, 20). 3
NOTE:
which is not the centre of the sphere, therefore AB is not a diameter of the sphere. 13.
B l1 l2 z
To determine the equation of l3 we require a 121 direction. Once t is established, a 16 direction is evident and further substitutions to determine u are not required. Upon substitution, one would find that u 41.
14. a.
y
m
A l
y
O
N
x l3 x
O
Two lines: l1: r (2, 16, 19) t(1, 1, 4); l : r (14, 19, 2) u(2, 1, 2). A line l passes 2
3
through the origin and intersects l2 at A and l1 at B.
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
Since B is on l1, we represent its coordinates as B(2 t, 16 t, 19 4t).
direction of the line m along ON. The equation of m is r t(A, B); x At, y Bt.
Similarly the coordinates of A will be A(14 2u, 19 u, 2 2u). Since O, A, and B are collinear kOA OB (2 t, 16 t, 19 4t) k(14 2u, 19 u, 2 2u). Equating components: 2 t 14k 2ku ➀ 16 t 19k ku ➁ 19 4t 2k 2ku ➂ We solve by first eliminating ku: ➀ ➂: 21 3t 12k ➃ ➀ 2 ➁: 30 3t 52k ➄ add: 9 64k 9 k . 64 Substitute in ➃: 21 3t 12k 7 t 4k 9 t 7 4 · 64 121 t . 16 121 With t , we can find the coordinates of B hence 16 153 135 180 we have OB , , . 16 16 16
Substituting in l: A2t B2t C 0 C t . 2 A B2 BC AC The coordinates of N are , . 2 2 A B A2 B2
b.
BC AC ON , 2 2 A B A2 B2
ON
A2C2 B2C2 (A2 B2)2
C2(A2 B2) (A2 B2) C2 2 A B2 C
. ON A2 B2
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). (2 s 2t, 3 2s t, s 2t). BA
z 3 s. Two points P(6, 4, 0) and Q(0, 5, 3). d (0, 1, 1), d (2, 0, 1). 1
2
(6, 9, 3), Now PQ n d d (1, 2, 2). 1
2
· n PQ 6 18 6 18 6. 14 4 3 n The distance between the lines is 6.
l1
A
b. x 6, y 4 t, z t; x 2s, y 5,
Review Exercise
2. a. A line through A(3, 9), B(4, 2) has a direction AB
B l2
(7, 7). An equation is r (3, 9) t(1, 1). b. A line passes through A(5, 3) and is parallel to r
l
is perpendicular to l therefore BA · BA d1 0 1 2 s 2t 6 4s 2t s 2t 0 6s 2t 4, 3s t 2 ➀ is perpendicular to l , therefore BA · BA d2 0 2 4 2s 4t 3 2s t 2s 4t 0 2s 9t 7 ➁ 9 ➀ ➁: 25s 25 s 1, t 1. The coordinates of the points of intersection are A(1, 1, 1) and B(0, 1, 2). td , 16. The distance between two skew lines r OP 1 · n PQ
sd is given by where n d d . r OQ 2 1 2 n
r (0, 5, 0) s(1, 1, 2). The two points P(0, 2, 6), Q(0, 5, 0) and directions d (2, 1, 1), d (1, 1, 2).
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
2 . A direction is (3, 2) and parametric 3 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. (4, 2) AB 2(2, 1). The line has equation x 4 2t, y t.
2
(0, 3, 6), n d d (3, 3, 3) Now PQ 1 2
9 18 3 3 . 3 33 n · n PQ
The distance between the lines is 3 .
104 Chapter 7: Lines in a Plane
c. A line passes through A(0, 3) and is perpendicular
c. A line through A(4, 0), B(0, 2) has direction
a. r (0, 2, 6) t(2, 1, 1);
1
(4, 0) t(0, 5). A direction is (0, 1) and an equation is r (5, 3) s(0, 1).
4. a. The line passes through A(2, 0, 3), B(3, 2, 2).
(5, 2, 1) and a vector A direction is AB equation is r (2, 0, 3) t(5, 2, 1).
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 (7, 4, 0). A vector equation is a direction is AB r (7, 0, 0) t(7, 4, 0). x5 y2 z6 c. A line l, parallel to , 4 2 5 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
x1 y2 the line z 3. A direction of 3 2 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 parallel
to the y-axis. A direction is (0, 1, 0). Parametric equations 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 parallel
to 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 is
perpendicular to x 2 t, y 3 2t. A normal of 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 (5, 2, 3) and parametric equations x 6 BC 5t, y 4 2t, z 3t. b. Since (4, m, n) l, 6 5t 4, t 2;
8. a. l1: r (2, 3) t(3, 1),
d1 (3, 1) l2: r (1, 4) u(6, 2), d2 (6, 2) 2(3, 1). Since d2 2d1, l1 and l2 are
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; d1 (2, 1).
l2: x u,
1 y 2u; d2 (1, 2). 3
Since d1 · d2 0, the two lines are perpendicular. x1 y4 c. l1: , z 1; d1 (2, 1, 0). 2
1
l2: x 4t, y 1 2t, z 6; d2 (4, 2, 0) 2(2, 1, 0). Since d 2d , l and l are parallel. 2
1
1
2
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),
d1 (1, 1, 1). l2: (x, y, z) (3, 0, 1) u(2, 2, 2), d2 (2, 2, 2) 2(1, 1, 1). Since d1 kd2, d1 · d2 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,
1 11 t , and the point is (3, , 0). 2 2 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
y3 x2 and the scalar equation is 5 1 5x y 13 0.
4 2t m, m 8; 3t n, n 6.
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). 3 1 3 4 2 4 (4, 3). A point on the line is (2, 2) and a vector equation is r (2, 2) t(4, 3).
c. The line y x has slope and a direction
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.
c. The line r (7, 0, 0) t(4, 1, 0) has direction
d (4, 1, 0), d 17 ; direction cosines 4 1 cos , cos , cos δ 0; 17 17 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 l2: x 4 4u, y 1 u, z 2u. Equating components and rearranging 4t 4u 4 ➀ 3t u 1 ➁ 4t 2u 2 ➂ 2 ➁ ➂: 10t 0, t 0, u 1. Substitute in ➀: 4(0) 4(1) 4 which verifies. 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.
z
4
1 5 6 2u, u 2 1 1 6 3 6u, u ; therefore the two lines 2 2 do not intersect.
(0, 8, 4) 8
y 14. a. P(2, 1, 3), Q(0, 4, 7). The distance between these
where QP (2, 5, 4) and points is QP 4 QP 25 16
6
45 . The shortest distance between P and Q is 35.
x
x 3 5
y6 2
z1 1
12. a. The line has direction
d (5, 2, 1), d 30 ; direction cosines
b. The distance from A(3, 7) to 2x 3y 7 0 is
Ax1 By1 C given by d A2 B2 6 21 7 13
1 5 2 cos , cos , cos δ ; 30 30 30 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 8 1 4 cosines cos , cos , cos δ ; 9 9 9 and direction angles 27, 96, δ 116.
106 Chapter 7: Lines in a Plane
22 d . 13 The shortest distance from the point to the line is
2213 . 13
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). (2, 2, 0), PA d (2, 2, 2) Now PA
d PA 23 line is 2 . d 6
x9
y2
3
1
c. Symmetric equations are . d. The scalar equation is x 3y 15 0.
d. From the point A(1, 3, 2) to the line
x1 y3 z7 . A direction of the line is 1 1 2 d (1, 1, 2). A point on the line is P(1, 3, 7). (0, 0, 5), AP d (5, 5, 0), Now AP d 52 AP , d 6 . The perpendicular distance from A to the line is d PA 52 53 given by .
15.
(6, 2) 2(3, 1). AB b. Parametric equations are x 9 3t, y 2 t.
The perpendicular distance from the point A to the
6
1. A line through A(9, 2), B(3, 4) has direction a. A vector equation is r (9, 2) t(3, 1).
d 23 PA , d 6 .
d
Chapter 7 Test
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 x y1 equation of l is and the scalar equation is 2 3 3x 2y 2 0. x2
y4
z2
6
3
3
3. The line has direction
3
d (6, 3, 3) 3(2, 1, 1). Parametric equations are x 2 2t, y 4 t, z 2 t. For an
Q (3, 2, 4)
intersection with the xy-plane, z 0, t 2, and
→
d=(5, 3, 4)
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
A
B(0, 3, 1). The intersection with the xz-plane is C(6, 0, 2). l
z
C
Let the foot of the perpendicular be A(6 5t, 7 3t, 3 4t) (9 5t, 9 3t, 7 4t). QA is perpendicular to the line, QA · d 0 Since QA
A
therefore 45 25t 27 9t 28 16t 0 50t 100 t 2. The coordinates of the foot of the perpendicular are (4, 1, 5).
y B
x
Chapter 7: Lines in a Plane 107
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, 5). AP d (1, 2, 3) and AP (3, 6, 3). AP d 9 36 9 31 4 1 36 .
d 3.
From ➂, 3(1) 3(4) 3 12 15 8 therefore 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.
The perpendicular distance from P to the line is
l1
P1 (10, 1, 2)
d AP 36 32. 3 d 5. A line through (0, 0, 0) has direction angles 120,
→
d2 = (2, 1, 3)
45.
l2
P2
Since cos cos cos 1, 2
2
2
c. Let the coordinates of P2 be (1 2s, 1 s, 3s).
cos2 cos2 120 cos2 45 1;
Now P P (11 2s, 2 s, 2 3s). 1 2 P · d 0, Since P P is perpendicular to l P
1 cos 120 , 2
1 2
1 2 cos 2 2
1 2
2
14s 14 s 1.
1 1 cos2 1 4 2
The coordinates of P2 are (1, 2, 3).
1 1 1 cos2 , cos or . 2 4 2 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 ➀ t 2s 8 ➁ 3t 3s 3 ➂ ➁ ➂ 3: 3s 9, s 3, t 2 from 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,
d1 (1, 2, 3) and
l2: x 1 2s, y 1 s, z 3s, d2 (2, 1, 3). Since the direction d kd , the lines are not parallel. 1
2
therefore 22 4s 2 s 6 9s 0
2
Equating components gives: t 2s 9 ➀ 2t s 2 ➁ 3t 3s 8 ➂ 2 ➁ ➀: 5t 5, t 1, s 4. 108 Chapter 7: Lines in a Plane
