Chapt_07

Prévia do material em texto

Part II Vectors, Matrices, and Vector Calculus
77 Vectors
EXERCISES 7.1
Vectors in 2-Space
1. (a) 6i + 12j (b) i + 8j (c) 3i (d)
√
65 (e) 3
2. (a) 〈3, 3〉 (b) 〈3, 4〉 (c) 〈−1,−2〉 (d) 5 (e) √5
3. (a) 〈12, 0〉 (b) 〈4,−5〉 (c) 〈4, 5〉 (d) √41 (e) √41
4. (a) 12 i− 12 j (b) 23 i + 23 j (c) − 13 i− j (d) 2
√
2/3 (e)
√
10/3
5. (a) −9i + 6j (b) −3i + 9j (c) −3i− 5j (d) 3√10 (e) √34
6. (a) 〈3, 9〉 (b) 〈−4,−12〉 (c) 〈6, 18〉 (d) 4√10 (e) 6√10
7. (a) −6i + 27j (b) 0 (c) −4i + 18j (d) 0 (e) 2√85
8. (a) 〈21, 30〉 (b) 〈8, 12〉 (c) 〈6, 8〉 (d) 4√13 (e) 10
9. (a) 〈4,−12〉 − 〈−2, 2〉 = 〈6,−14〉 (b) 〈−3, 9〉 − 〈−5, 5〉 = 〈2, 4〉
10. (a) (4i + 4j)− (6i− 4j) = −2i + 8j (b) (−3i− 3j)− (15i− 10j) = −18i + 7j
11. (a) (4i− 4j)− (−6i + 8j) = 10i− 12j (b) (−3i + 3j)− (−15i + 20j) = 12i− 17j
12. (a) 〈8, 0〉 − 〈0,−6〉 = 〈8, 6〉 (b) 〈−6, 0〉 − 〈0,−15〉 = 〈−6, 15〉
13. (a) 〈16, 40〉 − 〈−4,−12〉 = 〈20, 52〉 (b) 〈−12,−30〉 − 〈−10,−30〉 = 〈−2, 0〉
14. (a) 〈8, 12〉 − 〈10, 6〉 = 〈−2, 6〉 (b) 〈−6,−9〉 − 〈25, 15〉 = 〈−31,−24〉
339
7.1 Vectors in 2-Space
15.
−−−→
P1P2 = 〈2, 5〉
16.
−−−→
P1P2 = 〈6,−4〉
17.
−−−→
P1P2 = 〈2, 2〉
18.
−−−→
P1P2 = 〈2,−3〉
19. Since −−−→P1P2 = −−→OP2 − −−→OP1, −−→OP2 = −−−→P1P2 + −−→OP1 = (4i + 8j) + (−3i + 10j) = i + 18j, and the terminal point is
(1, 18).
20. Since −−−→P1P2 = −−→OP2 −−−→OP1, −−→OP1 = −−→OP2 −−−−→P1P2 = 〈4, 7〉 − 〈−5,−1〉 = 〈9, 8〉, and the initial point is (9, 8).
21. a(= −a), b(= − 14a), c(= 52a), e(= 2a), and f(= − 12a) are parallel to a.
22. We want −3b = a, so c = −3(9) = −27.
23. 〈6, 15〉
24. 〈5, 2〉
25. ‖a‖ = √4 + 4 = 2√2 ; (a) u = 1
2
√
2
〈2, 2〉 = 〈 1√
2
, 1√
2
〉; (b) −u = 〈− 1√
2
,− 1√
2
〉
26. ‖a‖ = √9 + 16 = 5; (a) u = 15 〈−3, 4〉 = 〈− 35 , 45 〉; (b) −u = 〈 35 ,− 45 〉
27. ‖a‖ = 5; (a) u = 15 〈0,−5〉 = 〈0,−1〉; (b) −u = 〈0, 1〉
28. ‖a‖ = √1 + 3 = 2; (a) u = 12 〈1,−
√
3 〉 = 〈 12 ,−
√
3
2 〉; (b) −u = 〈− 12 ,
√
3
2 〉
29. ‖a + b‖ = ‖〈5, 12〉‖ = √25 + 144 = 13; u = 113 〈5, 12〉 = 〈 513 , 1213 〉
30. ‖2a− 3b‖ = ‖〈−5, 4〉‖ = √25 + 16 = √41 ; u = 1√
41
〈−5, 4〉 = 〈− 5√
41
, 4√
41
〉
31. ‖a‖ = √9 + 49 = √58 ; b = 2( 1√
58
)(3i + 7j) = 6√
58
i + 14√
58
j
32. ‖a‖ =
√
1
4 +
1
4 =
1√
2
; b = 3( 1
1/
√
2
)( 12 i− 12 j) = 3
√
2
2 i− 3
√
2
2 j
33. − 34a = 〈−3,−15/2〉 34. 5(a + b) = 5〈0, 1〉 = 〈0, 5〉
35. 36.
37. x = −(a + b) = −a− b 38. x = 2(a− b) = 2a− 2b
340
7.1 Vectors in 2-Space
39.
b = (−c)− a; (b + c) + a = 0; a + b + c = 0
40.
From Problem 39, e + c + d = 0. But b = e− a
and e = a + b, so (a + b) + c + d = 0.
41. From 2i + 3j = k1b + k2c = k1(i + j) + k2(i − j) = (k1 + k2)i + (k1 − k2)j we obtain the system of equations
k1 + k2 = 2, k1 − k2 = 3. Solving, we find k1 = 52 and k2 = − 12 . Then a = 52b− 12c.
42. From 2i + 3j = k1b + k2c = k1(−2i + 4j) + k2(5i + 7j) = (−2k1 + 5k2)i + (4k1 + 7k2)j we obtain the system of
equations −2k1 + 5k2 = 2, 4k1 + 7k2 = 3. Solving, we find k1 = 134 and k2 = 717 .
43. From y′ = 12x we see that the slope of the tangent line at (2, 2) is 1. A vector with slope 1 is i+ j. A unit vector
is (i + j)/‖i + j‖ = (i + j)/√2 = 1√
2
i + 1√
2
j. Another unit vector tangent to the curve is − 1√
2
i− 1√
2
j.
44. From y′ = −2x + 3 we see that the slope of the tangent line at (0, 0) is 3. A vector with slope 3 is i + 3j. A
unit vector is (i + 3j)/‖i + 3j‖ = (i + 3j)/√10 = 1√
10
i + 1√
10
j. Another unit vector is − 1√
10
i− 1√
10
j.
45. (a) Since Ff = −Fg, ‖Fg‖ = ‖Ff‖ = µ‖Fn‖ and tan θ = ‖Fg‖/‖Fn‖ = µ‖Fn‖/‖Fn‖ = µ.
(b) θ = tan−1 0.6 ≈ 31◦
46. Since w + F1 + F2 = 0,
−200j + ‖F1‖ cos 20◦i + ‖F1‖ sin 20◦j− ‖F2‖ cos 15◦i + ‖F2‖ sin 15◦j = 0
or (‖F1‖ cos 20◦ − ‖F2‖ cos 15◦)i + (‖F1‖ sin 20◦ + ‖F2‖ sin 15◦ − 200)j = 0.
Thus, ‖F1‖ cos 20◦ − ‖F2‖ cos 15◦ = 0; ‖F1‖ sin 20◦ + ‖F2‖ sin 15◦ − 200 = 0. Solving this system for ‖F1‖ and
‖F2‖, we obtain
‖F1‖ = 200 cos 15
◦
sin 15◦ cos 20◦ + cos 15◦ sin 20◦
=
200 cos 15◦
sin(15◦ + 20◦)
=
200 cos 15◦
sin 35◦
≈ 336.8 lb
and
‖F2‖ = 200 cos 20
◦
sin 15◦ cos 20◦ + cos 15◦ sin 20◦
=
200 cos 20◦
sin 35◦
≈ 327.7 lb.
47. Since y/2a(L2 + y2)3/2 is an odd function on [−a, a], Fy = 0. Now, using the fact that L/(L2 + y2)3/2 is an
even function, we have∫ a
−a
Ldy
2a(L2 + y2)3/2
=
L
a
∫ a
0
dy
(L2 + y2)3/2
y = L tan θ, dy = L sec2 θ dθ
=
L
a
∫ tan−1 a/L
0
L sec2 θ dθ
L3(1 + tan2 θ)3/2
=
1
La
∫ tan−1 a/L
0
sec2 θ dθ
sec3 θ
=
1
La
∫ tan−1 a/L
0
cos θ dθ =
1
La
sin θ
∣∣∣∣tan−1 a/L
0
=
1
La
a√
L2 + a2
=
1
L
√
L2 + a2
.
Then Fx = qQ/4π�0L
√
L2 + a2 and F = (qQ/4π�0L
√
L2 + a2 )i.
48. Place one corner of the parallelogram at the origin and let two adja-
cent sides be −−→OP1 and −−→OP2. Let M be the midpoint of the diagonal
connecting P1 and P2 and N be the midpoint of the other diagonal.
Then −−→OM = 12 (
−−→
OP1 +
−−→
OP2). Since
−−→
OP1 +
−−→
OP2 is the main diagonal of the parallelogram and N is its midpoint,−−→
ON = 12 (
−−→
OP1 +
−−→
OP2). Thus,
−−→
OM = −−→ON and the diagonals bisect each other.
341
7.1 Vectors in 2-Space
49. By Problem 39, −−→AB +−−→BC +−→CA = 0 and −−→AD+−−→DE +−−→EC +−→CA = 0. From the first equation,
−−→
AB + −−→BC = −−→CA. Since D and E are midpoints, −−→AD = 12
−−→
AB and −−→EC = 12
−−→
BC. Then,
1
2
−−→
AB +−−→DE + 12
−−→
BC +−→CA = 0 and
−−→
DE = −−→CA− 1
2
(−−→AB +−−→BC) = −−→CA− 1
2
(−−→CA) = −1
2
−→
CA.
Thus, the line segment joining the midpoints D and E is parallel to the side AC and half its length.
50. We have −→OA = 150 cos 20◦i+150 sin 20◦j, −−→AB = 200 cos 113◦i+200 sin 113◦j, −−→BC = 240 cos 190◦i+240 sin 190◦j.
Then
r = (150 cos 20◦ + 200 cos 113◦ + 240 cos 190◦)i + (150 sin 20◦ + 200 sin 113◦ + 240 sin 190◦)j
≈ −173.55i + 193.73j
and ‖r‖ ≈ 260.09 miles.
EXERCISES 7.2
Vectors in 3-Space
1. – 6.
7. A plane perpendicular to the z-axis, 5 units above the xy-plane
8. A plane perpendicular to the x-axis, 1 unit in front of the yz-plane
9. A line perpendicular to the xy-plane at (2, 3, 0)
10. A single point located at (4,−1, 7)
11. (2, 0, 0), (2, 5, 0), (2, 0, 8), (2, 5, 8), (0, 5, 0), (0, 5, 8), (0, 0, 8), (0, 0, 0)
12.
13. (a) xy-plane: (−2, 5, 0), xz-plane: (−2, 0, 4), yz-plane: (0, 5, 4); (b) (−2, 5,−2)
342
7.2 Vectors in 3-Space
(c) Since the shortest distance between a point and a plane is a perpendicular line, the point in the plane x = 3
is (3, 5, 4).
14. We find planes that are parallel to coordinate planes: (a) z = −5; (b) x = 1 and y = −1; (c) z = 2
15. The union of the planes x = 0, y = 0, and z = 0
16. The origin (0, 0, 0)
17. The point (−1, 2,−3)
18. The union of the planes x = 2 and z = 8
19. The union of the planes z = 5 and z = −5
20. The line through the points (1, 1, 1), (−1,−1,−1), and the origin
21. d =
√
(3− 6)2 + (−1− 4)2 + (2− 8)2 = √70
22. d =
√
(−1− 0)2 + (−3− 4)2 + (5− 3)2 = 3√6
23. (a) 7; (b) d =
√
(−3)2 + (−4)2 = 5
24. (a) 2; (b) d =
√
(−6)2 + 22 + (−3)2 = 7
25. d(P1, P2) =
√
32 + 62 + (−6)2 = 9; d(P1, P3) =
√
22 + 12 + 22 = 3
d(P2, P3) =
√
(2− 3)2 + (1− 6)2 + (2− (−6))2 = √90 ; The triangle is a right triangle.
26. d(P1, P2) =
√
12 + 22 + 42 =
√
21 ; d(P1, P3) =
√
32 + 22 + (2
√
2)2 =
√
21
d(P2, P3) =
√
(3− 1)2 + (2− 2)2 + (2√2− 4)2 =
√
28− 16√2
The triangle is an isosceles triangle.
27. d(P1, P2) =
√
(4− 1)2 + (1− 2)2 + (3− 3)2 = √10
d(P1, P3) =
√
(4− 1)2 + (6− 2)2 + (4− 3)2 = √26
d(P2, P3) =
√
(4− 4)2 + (6− 1)2 + (4− 3)2 = √26 ; The triangle is an isosceles triangle.
28. d(P1, P2) =
√
(1− 1)2 + (1− 1)2 + (1− (−1))2 = 2
d(P1, P3) =
√
(0− 1)2 + (−1− 1)2 + (1− (−1))2 = 3
d(P2, P3) =
√
(0− 1)2 + (−1− 1)2 + (1− 1)2 = √5 ; The triangleis a right triangle.
29. d(P1, P2) =
√
(−2− 1)2 + (−2− 2)2 + (−3− 0)2 = √34
d(P1, P3) =
√
(7− 1)2 + (10− 2)2 + (6− 0)2 = 2√34
d(P2, P3) =
√
(7− (−2))2 + (10− (−2))2 + (6− (−3))2 = 3√34
Since d(P1, P2) + d(P1, P3) = d(P2, P3), the points P1, P2, and P3 are collinear.
30. d(P1, P2) =
√
(1− 2)2 + (4− 3)2 + (4− 2)2 = √6
d(P1, P3) =
√
(5− 2)2 + (0− 3)2 + (−4− 2)2 = 3√6
d(P2, P3) =
√
(5− 1)2 + (0− 4)2 + (−4− 4)2 = 4√6
Since d(P1, P2) + d(P1, P3) = d(P2, P3), the points P1, P2, and P3 are collinear.
31.
√
(2− x)2 + (1− 2)2 + (1− 3)2 =
√
21 =⇒ x2 − 4x + 9 = 21 =⇒ x2 − 4x + 4 = 16
=⇒ (x− 2)2 = 16 =⇒ x = 2± 4 or x = 6, −2
343
7.2 Vectors in 3-Space
32.
√
(0− x)2 + (3− x)2 + (5− 1)2 = 5 =⇒ 2x2 − 6x + 25 = 25 =⇒ x2 − 3x = 0 =⇒ x = 0, 3
33.
(
1 + 7
2
,
3 + (−2)
2
,
1/2 + 5/2
2
)
= (4, 1/2, 3/2)
34.
(
0 + 4
2
,
5 + 1
2
,
−8 + (−6)
2
)
= (2, 3,−7)
35. (x1 + 2)/2 = −1, x1 = −4; (y1 + 3)/2 = −4, y1 = −11; (z1 + 6)/2 = 8, z1 = 10
The coordinates of P1 are (−4,−11, 10).
36. (−3 + (−5))/2 = x3 = −4; (4 + 8)/2 = y3 = 6; (1 + 3)/2 = z3 = 2.
The coordinates of P3 are (−4, 6, 2).
(a)
(−3 + (−4)
2
,
4 + 6
2
,
1 + 2
2
)
= (−7/2, 5, 3/2)
(b)
(−4 + (−5)
2
,
6 + 8
2
,
2 + 3
2
)
= (−9/2, 7, 5/2)
37. −−−→P1P2 = 〈−3,−6, 1〉 38. −−−→P1P2 = 〈8,−5/2, 8〉
39. −−−→P1P2 = 〈2, 1, 1〉 40. −−−→P1P2〈−3,−3, 7〉
41. a + (b + c) = 〈2, 4, 12〉
42. 2a− (b− c) = 〈2,−6, 4〉 − 〈−3,−5,−8〉 = 〈5,−1, 12〉
43. b + 2(a− 3c) = 〈−1, 1, 1〉+ 2〈−5,−21,−25〉 = 〈−11,−41,−49〉
44. 4(a + 2c)− 6b = 4〈5, 9, 20〉 − 〈−6, 6, 6〉 = 〈26, 30, 74〉
45. ‖a + c‖ = ‖〈3, 3, 11〉‖ = √9 + 9 + 121 = √139
46. ‖c‖‖2b‖ = (√4 + 36 + 81 )(2)(√1 + 1 + 1 ) = 22√3
47.
∥∥∥∥ a‖a‖
∥∥∥∥ + 5∥∥∥∥ b|b‖
∥∥∥∥ = 1‖a‖‖a‖+ 5 1‖b‖‖b‖ = 1 + 5 = 6
48. ‖b‖a + ‖a‖b = √1 + 1 + 1 〈1,−3, 2〉+√1 + 9 + 4 〈−1, 1, 1〉 = 〈
√
3 ,−3
√
3 , 2
√
3 〉+ 〈−
√
14 ,
√
14 ,
√
14 〉
= 〈
√
3−
√
14 ,−3
√
3 +
√
14 , 2
√
3 +
√
14 〉
49. ‖a‖ = √100 + 25 + 100 = 15; u = − 1
15
〈10,−5, 10〉 = 〈−2/3, 1/3,−2/3〉
50. ‖a‖ = √1 + 9 + 4 = √14 ; u = 1√
14
(i− 3j + 2k) = 1√
14
i− 3√
14
j +
2√
14
k
51. b = 4a = 4i− 4j + 4k
52. ‖a‖ = √36 + 9 + 4 = 7; b = −1
2
(
1
7
)
〈−6, 3,−2〉 = 〈3
7
,− 3
14
,
1
7
〉
53.
344
7.3 Dot Product
EXERCISES 7.3
Dot Product
1. a · b = 10(5) cos(π/4) = 25√2 2. a · b = 6(12) cos(π/6) = 36√3
3. a · b = 2(−1) + (−3)2 + 4(5) = 12 4. b · c = (−1)3 + 2(6) + 5(−1) = 4
5. a · c = 2(3) + (−3)6 + 4(−1) = −16 6. a · (b + c) = 2(2) + (−3)8 + 4(4) = −4
7. a · (4b) = 2(−4) + (−3)8 + 4(20) = 48 8. b · (a− c) = (−1)(−1) + 2(−9) + 5(5) = 8
9. a · a = 22 + (−3)2 + 42 = 29 10. (2b) · (3c) = (−2)9 + 4(18) + 10(−3) = 24
11. a · (a + b + c) = 2(4) + (−3)5 + 4(8) = 25
12. (2a) · (a− 2b) = 4(4) + (−6)(−7) + 8(−6) = 10
13.
(
a · b
b · b
)
b =
[
2(−1) + (−3)2 + 4(5)
(−1)2 + 22 + 52
]
〈−1, 2, 5〉 = 12
30
〈−1, 2, 5〉 = 〈−2/5, 4/5, 2〉
14. (c · b)a = [3(−1) + 6(2) + (−1)5]〈2,−3, 4〉 = 4〈2,−3, 4〉 = 〈8,−12, 16〉
15. a and f, b and e, c and d
16. (a) a · b = 2 · 3 + (−c)2 + 3(4) = 0 =⇒ c = 9
(b) a · b = c(−3) + 12 (4) + c2 = c2 − 3c + 2 = (c− 2)(c− 1) = 0 =⇒ c = 1, 2
17. Solving the system of equations 3x1 + y1 − 1 = 0, −3x1 + 2y1 + 2 = 0 gives x1 = 4/9 and y1 = −1/3. Thus,
v = 〈4/9,−1/3, 1〉.
18. If a and b represent adjacent sides of the rhombus, then ‖a‖ = ‖b‖, the diagonals of the rhombus are a + b
and a− b, and
(a + b) · (a− b) = a · a− a · b + b · a− b · b = a · a− b · b = ‖a‖2 − ‖b‖2 = 0.
Thus, the diagonals are perpendicular.
19. Since
c · a =
(
b− a · b‖a‖2 a
)
· a = b · a− a · b‖a‖2 (a · a) = b · a−
a · b
‖a‖2 ‖a‖
2 = b · a− a · b = 0,
the vectors c and a are orthogonal.
20. a · b = 1(1) + c(1) = c + 1; ‖a‖ = √1 + c2 , ‖b‖ =
√
2
cos 45◦ =
1√
2
=
c + 1√
1 + c2
√
2
=⇒
√
1 + c2 = c + 1 =⇒ 1 + c2 = c2 + 2c + 1 =⇒ c = 0
21. a · b = 3(2) + (−1)2 = 4; ‖a‖ = √10 , ‖b‖ = 2√2
cos θ =
4
(
√
10)(2
√
2)
=
1√
5
=⇒ θ = cos−1 1√
5
≈ 1.11 rad ≈ 63.43◦
22. a · b = 2(−3) + 1(−4) = −10; ‖a‖ = √5 , ‖b‖ = 5
345
7.3 Dot Product
cos θ =
−10
(
√
5 )5
= − 2√
5
=⇒ θ = cos−1(−2/
√
5 ) ≈ 2.68 rad ≈ 153.43◦
23. a · b = 2(−1) + 4(−1) + 0(4) = −6; ‖a‖ = 2√5 , ‖b‖ = 3√2
cos θ =
−6
(2
√
5)(3
√
2)
= − 1√
10
=⇒ θ = cos−1(−1/
√
10 ) ≈ 1.89 rad ≈ 108.43◦
24. a · b = 12 (2) + 12 (−4) + 32 (6) = 8; ‖a‖ =
√
11/2, ‖b‖ = 2√14
cos θ =
8
(
√
11/2)(2
√
14 )
=
8√
154
=⇒ θ = cos−1(8/
√
154 ) ≈ 0.87 rad ≈ 49.86◦
25. ‖a‖ = √14 ; cosα = 1/√14 , α ≈ 74.50◦; cosβ = 2/√14 , β ≈ 57.69◦; cos γ = 3/√14 , γ ≈ 36.70◦
26. ‖a‖ = 9; cosα = 2/3, α ≈ 48.19◦; cosβ = 2/3, β ≈ 48.19◦; cos γ = −1/3, γ ≈ 109.47◦
27. ‖a‖ = 2; cosα = 1/2, α = 60◦; cosβ = 0, β = 90◦; cos γ = −√3/2, γ = 150◦
28. ‖a‖ = √78 ; cosα = 5/√78 , α ≈ 55.52◦; cosβ = 7/√78 , β ≈ 37.57◦; cos γ = 2/√78 , γ ≈ 76.91◦
29. Let θ be the angle between −−→AD and −−→AB and a be the length of an edge of the cube. Then −−→AD = ai + aj + ak,
−−→
AB = ai and
cos θ =
−−→
AD · −−→AB
‖−−→AD‖ ‖−−→AB‖ =
a2√
3a2
√
a2
=
1√
3
so θ ≈ 0.955317 radian or 54.7356◦. Letting φ be the angle between −−→AD and −→AC and noting that −→AC = ai + aj
we have
cosφ =
−−→
AD · −→AC
‖−−→AD‖ ‖−→AC‖ =
a2 + a2√
3a2
√
2a2
=
√
2
3
so φ ≈ 0.61548 radian or 35.2644◦.
30. If a and b are orthogonal, then a · b = 0 and
cosα1 cosα2 + cosβ1 cosβ2 + cos γ1 cos γ2 =
a1
‖a‖
b1
‖b‖ +
a2
‖a‖
b2
‖b‖ +
a3
‖a‖
b3
‖b‖
=
1
‖a‖ ‖b‖ (a1b1 + a2b2 + a3b3) =
1
‖a‖ ‖b‖ (a · b) = 0.
31. a = 〈5, 7, 4〉; ‖a‖ = 3√10 ; cosα = 5/3√10 , α ≈ 58.19◦; cosβ = 7/3√10 , β ≈ 42.45◦; cos γ = 4/3√10 ,
γ ≈ 65.06◦
32. We want cosα = cosβ = cos γ or a1 = a2 = a3. Letting a1 = a2 = a3 = 1 we obtain the vector i+ j+k. A unit
vector in the same direction is 1√
3
i + 1√
3
j + 1√
3
k.
33. compba = a · b/‖b‖ = 〈1,−1, 3〉 · 〈2, 6, 3〉/7 = 5/7
34. compab = b · a/‖a‖ = 〈2, 6, 3〉 · 〈1,−1, 3〉/
√
11 = 5/
√
11
35. b− a = 〈1, 7, 0〉; compa(b− a) = (b− a) · a/‖a‖ = 〈1, 7, 0〉 · 〈1,−1, 3〉/
√
11 = −6/√11
36. a + b = 〈3, 5, 6〉; 2b = 〈4, 12, 6〉; comp2b(a + b) · 2b/|2b| = 〈3, 5, 6〉 · 〈4, 12, 6〉/14 = 54/7
37. −−→OP = 3i + 10j; ‖−−→OP‖ = √109 ; comp−→
OP
a = a · −−→OP/‖−−→OP‖ = (4i + 6j) · (3i + 10j)/√109 = 72/√109
38. −−→OP = 〈1,−1, 1〉; ‖−−→OP‖ = √3 ; comp−→
OP
a = a · −−→OP/‖−−→OP‖ = 〈2, 1,−1〉 · 〈1,−1, 1〉/√3 = 0
39. compba = a · b/‖b‖ = (−5i + 5j) · (−3i + 4j)/5 = 7
projba = (compba)b/‖b‖ = 7(−3i + 4j)/5 = − 215 i + 285 j
40. compba = a · b/‖b‖ = (4i + 2j) · (−3i + j)/
√
10 = −√10
projba = (compba)b/‖b‖ = −
√
10(−3i + j)/√10 = 3i− j
346
7.3 Dot Product
41. compba = a · b/‖b‖ = (−i− 2j + 7k) · (6i− 3j− 2k)/7 = −2
projba = (compba)b/‖b‖ = −2(6i− 3j− 2k)/7 = − 127 i + 67 j + 47k
42. compba = a · b/‖b‖ = 〈1, 1, 1〉 · 〈−2, 2,−1〉/3 = −1/3
projba = (compba)b/‖b‖ = − 13 〈−2, 2,−1〉/3 = 〈2/9,−2/9, 1/9〉
43. a + b = 3i + 4j; ‖a + b‖ = 5; comp(a+b)a = a · (a + b)/‖a + b‖ = (4i + 3j) · (3i + 4j)/5 = 24/5
proj(a+b)a = (comp(a+b)a)(a + b)/‖a + b‖ = 245 (3i + 4j)/5 = 7225 i + 9625 j
44. a− b = 5i + 2j; ‖a− b‖ = √29 ; comp(a−b)b = b · (a− b)/‖a− b‖ = (−i + j) · (5i + 2j)/
√
29 = −3/√29
proj(a−b)b = (comp(a−b)b)(a− b)/‖a− b‖ = − 3√29 (5i + 2j)/
√
29 = − 1529 i− 629 j
45. We identify ‖F‖ = 20, θ = 60◦ and ‖d‖ = 100. Then W = ‖F‖ ‖d‖ cos θ = 20(100)( 12 ) = 1000 ft-lb.
46. We identify d = −i + 3j + 8k. Then W = F · d = 〈4, 3, 5〉 · 〈−1, 3, 8〉 = 45 N-m.
47. (a) Since w and d are orthogonal, W = w · d = 0.
(b) We identify θ = 0◦. Then W = ‖F‖ ‖d‖ cos θ = 30(√42 + 32 ) = 150 N-m.
48. Using d = 6i + 2j and F = 3( 35 i +
4
5 j), W = F · d = 〈 95 , 125 〉 · 〈6, 2〉 = 785 ft-lb.
49. Let a and b be vectors from the center of the carbon atom to the centersof two distinct hydrogen atoms. The
distance between two hydrogen atoms is then
‖b− a‖ =
√
(b− a) · (b− a) = √b · b− 2a · b + a · a
=
√
‖b‖2 + ‖a‖2 − 2‖a‖ ‖b‖ cos θ =
√
(1.1)2 + (1.1)2 − 2(1.1)(1.1) cos 109.5◦
=
√
1.21 + 1.21− 2.42(−0.333807) ≈ 1.80 angstroms.
50. Using the fact that | cos θ| ≤ 1, we have |a · b| = ‖a‖ ‖b‖| cos θ| = ‖a‖ ‖b‖| cos θ| ≤ ‖a‖ ‖b‖.
51. ‖a + b‖2 = (a + b) · (a + b) = a · a + 2a · b + b · b = ‖a‖2 + 2a · b + ‖b‖2
≤ ‖a‖2 + 2|a · b|+ ‖b‖2 since x ≤ |x|
≤ ‖a‖2 + 2‖a‖ ‖b‖+ ‖b‖2 = (‖a‖+ ‖b‖)2 by Problem 50
Thus, since ‖a + b‖ and ‖a‖+ ‖b‖ are positive, ‖a + b‖ ≤ ‖a‖+ ‖b‖.
52. Let P1(x1, y1) and P2(x2, y2) be distinct points on the line ax + by = −c. Then
n · −−−→P1P2 = 〈a, b〉 · 〈x2 − x1, y2 − y1〉 = ax2 − ax1 + by2 − by1
= (ax2 + by2)− (ax1 + by1) = −c− (−c) = 0,
and the vectors are perpendicular. Thus, n is perpendicular to the line.
53. Let θ be the angle between n and −−−→P2P1. Then
d = ‖−−−→P1P2‖ | cos θ| = |n ·
−−−→
P2P1|
‖n‖ =
|〈a, b〉 · 〈x1 − x2, y1 − y2〉|√
a2 + b2
=
|ax1 − ax2 + by1 − by2|√
a2 + b2
=
|ax1 + by1 − (ax2 + by2)|√
a2 + b2
=
|ax1 + by1 − (−c)|√
a2 + b2
=
|ax1 + by1 + c|√
a2 + b2
.
347
7.3 Dot Product
EXERCISES 7.4
Cross Product
7.4 Cross Product
1. a× b =
∣∣∣∣∣∣∣
i j k
1 −1 0
0 3 5
∣∣∣∣∣∣∣ =
∣∣∣∣−1 03 5
∣∣∣∣ i− ∣∣∣∣ 1 00 5
∣∣∣∣ j + ∣∣∣∣ 1 −10 3
∣∣∣∣k = −5i− 5j + 3k
2. a× b =
∣∣∣∣∣∣∣
i j k
2 1 0
4 0 −1
∣∣∣∣∣∣∣ =
∣∣∣∣ 1 00 −1
∣∣∣∣ i− ∣∣∣∣ 2 04 −1
∣∣∣∣ j + ∣∣∣∣ 2 14 0
∣∣∣∣k = −i + 2j− 4k
3. a× b =
∣∣∣∣∣∣∣
i j k
1 −3 1
2 0 4
∣∣∣∣∣∣∣ =
∣∣∣∣−3 10 4
∣∣∣∣ i− ∣∣∣∣ 1 12 4
∣∣∣∣ j + ∣∣∣∣ 1 −32 0
∣∣∣∣k = 〈−12,−2, 6〉
4. a× b =
∣∣∣∣∣∣∣
i j k
1 1 1
−5 2 3
∣∣∣∣∣∣∣ =
∣∣∣∣ 1 12 3
∣∣∣∣ i− ∣∣∣∣ 1 1−5 3
∣∣∣∣ j + ∣∣∣∣ 1 1−5 2
∣∣∣∣k = 〈1,−8, 7〉
5. a× b =
∣∣∣∣∣∣∣
i j k
2 −1 2
−1 3 −1
∣∣∣∣∣∣∣ =
∣∣∣∣−1 23 −1
∣∣∣∣ i− ∣∣∣∣ 2 2−1 −1
∣∣∣∣ j + ∣∣∣∣ 2 −1−1 3
∣∣∣∣k = −5i + 5k
6. a× b =
∣∣∣∣∣∣∣
i j k
4 1 −5
2 3 −1
∣∣∣∣∣∣∣ =
∣∣∣∣ 1 −53 −1
∣∣∣∣ i− ∣∣∣∣ 4 −52 −1
∣∣∣∣ j + ∣∣∣∣ 4 12 3
∣∣∣∣k = 14i− 6j + 10k
7. a× b =
∣∣∣∣∣∣∣
i j k
1/2 0 1/2
4 6 0
∣∣∣∣∣∣∣ =
∣∣∣∣ 0 1/26 0
∣∣∣∣ i− ∣∣∣∣ 1/2 1/24 0
∣∣∣∣ j + ∣∣∣∣ 1/2 04 6
∣∣∣∣k = 〈−3, 2, 3〉
8. a× b =
∣∣∣∣∣∣∣
i j k
0 5 0
2 −3 4
∣∣∣∣∣∣∣ =
∣∣∣∣ 5 0−3 4
∣∣∣∣ i− ∣∣∣∣ 0 02 4
∣∣∣∣ j + ∣∣∣∣ 0 52 −3
∣∣∣∣k = 〈20, 0,−10〉
9. a× b =
∣∣∣∣∣∣∣
i j k
2 2 −4
−3 −3 6
∣∣∣∣∣∣∣ =
∣∣∣∣ 2 −4−3 6
∣∣∣∣ i− ∣∣∣∣ 2 −4−3 6
∣∣∣∣ j + ∣∣∣∣ 2 2−3 −3
∣∣∣∣k = 〈0, 0, 0〉
10. a× b =
∣∣∣∣∣∣∣
i j k
8 1 −6
1 −2 10
∣∣∣∣∣∣∣ =
∣∣∣∣ 1 −6−2 10
∣∣∣∣ i− ∣∣∣∣ 8 −61 10
∣∣∣∣ j + ∣∣∣∣ 8 11 −2
∣∣∣∣k = 〈−2,−86,−17〉
11. −−−→P1P2 = (−2, 2,−4); −−−→P1P3 = (−3, 1, 1)
−−−→
P1P2 ×−−−→P1P3 =
∣∣∣∣∣∣∣
i j k
−2 2 −4
−3 1 1
∣∣∣∣∣∣∣ =
∣∣∣∣ 2 −41 1
∣∣∣∣ i− ∣∣∣∣−2 −4−3 1
∣∣∣∣ j + ∣∣∣∣−2 2−3 1
∣∣∣∣k = 6i + 14j + 4k
348
7.4 Cross Product
12. −−−→P1P2 = (0, 1, 1); −−−→P1P3 = (1, 2, 2); −−−→P1P2 ×−−−→P1P3 =
∣∣∣∣∣∣∣
i j k
0 1 1
1 2 2
∣∣∣∣∣∣∣ =
∣∣∣∣ 1 12 2
∣∣∣∣ i− ∣∣∣∣ 0 11 2
∣∣∣∣ j + ∣∣∣∣ 0 11 2
∣∣∣∣k = j− k
13. a× b =
∣∣∣∣∣∣∣
i j k
2 7 −4
1 1 −1
∣∣∣∣∣∣∣ =
∣∣∣∣ 7 −41 −1
∣∣∣∣ i− ∣∣∣∣ 2 −41 −1
∣∣∣∣ j + ∣∣∣∣ 2 71 1
∣∣∣∣k = −3i− 2j− 5k
is perpendicular to both a and b.
14. a× b =
∣∣∣∣∣∣∣
i j k
−1 −2 4
4 −1 0
∣∣∣∣∣∣∣ =
∣∣∣∣−2 4−1 0
∣∣∣∣ i− ∣∣∣∣−1 44 0
∣∣∣∣ j + ∣∣∣∣−1 −24 −1
∣∣∣∣k = 〈4, 16, 9〉
is perpendicular to both a and b.
15. a× b =
∣∣∣∣∣∣∣
i j k
5 −2 1
2 0 −7
∣∣∣∣∣∣∣ =
∣∣∣∣−2 10 −7
∣∣∣∣ i− ∣∣∣∣ 5 12 −7
∣∣∣∣ j + ∣∣∣∣ 5 −22 0
∣∣∣∣k = 〈14, 37, 4〉
a · (a× b) = 〈5,−2,−1〉 · 〈14, 37, 4〉 = 70− 74 + 4 = 0; b · (a× b) = 〈2, 0,−7〉 · 〈14, 37, 4〉 = 28 + 0− 28 = 0
16. a× b =
∣∣∣∣∣∣∣
i j k
1/2 −1/4 0
2 −2 6
∣∣∣∣∣∣∣ =
∣∣∣∣−1/4 0−2 6
∣∣∣∣ i− ∣∣∣∣ 1/2 02 6
∣∣∣∣ j + ∣∣∣∣ 1/2 −1/42 −2
∣∣∣∣k = − 32 i− 3j− 12k
a · (a× b) = (12 i− 14 j) · (− 32 i− 3j− 12k) = − 34 + 34 + 0 = 0
b · (a× b) = (2i− 2j + 6k) · (− 32 i− 3j− 12k) = −3 + 6− 3 = 0
17. (a) b× c =
∣∣∣∣∣∣∣
i j k
2 1 1
3 1 1
∣∣∣∣∣∣∣ =
∣∣∣∣ 1 11 1
∣∣∣∣ i− ∣∣∣∣ 2 13 1
∣∣∣∣ j + ∣∣∣∣ 2 13 1
∣∣∣∣k = j− k
a× (b× c) =
∣∣∣∣∣∣∣
i j k
1 −1 2
0 1 −1
∣∣∣∣∣∣∣ =
∣∣∣∣−1 21 −1
∣∣∣∣ i− ∣∣∣∣ 1 20 −1
∣∣∣∣ j + ∣∣∣∣ 1 −10 1
∣∣∣∣k = −i + j + k
(b) a · c = (i− j + 2k) · (3i + j + k) = 4; (a · c)b = 4(2i + j + k) = 8i + 4j + 4k
a · b = (i− j + 2k) · (2i + j + k) = 3; (a · b)c = 3(3i + j + k) = 9i + 3j + 3k
a× (b× c) = (a · c)b− (a · b)c = (8i + 4j + 4k)− (9i + 3j + 3k) = −i + j + k
18. (a) b× c =
∣∣∣∣∣∣∣
i j k
1 2 −1
−1 5 8
∣∣∣∣∣∣∣ =
∣∣∣∣ 2 −15 8
∣∣∣∣ i− ∣∣∣∣ 1 −1−1 8
∣∣∣∣ j + ∣∣∣∣ 1 2−1 5
∣∣∣∣k = 21i− 7j + 7k
a× (b× c) =
∣∣∣∣∣∣∣
i j k
3 0 −4
21 −7 7
∣∣∣∣∣∣∣ =
∣∣∣∣ 0 −4−7 7
∣∣∣∣ i− ∣∣∣∣ 3 −421 7
∣∣∣∣ j + ∣∣∣∣ 3 021 −7
∣∣∣∣k = −28i− 105j− 21k
(b) a · c = (3i− 4k) · (−i + 5j + 8k) = −35; (a · c)b = −35(i + 2j− k) = −35i− 70j + 35k
a · b = (3i− 4k) · (i + 2j− k) = 7; (a · b)c = 7(−i + 5j + 8k) = −7i + 35j + 56k
a× (b× c) = (a · c)b− (a · b)c = (−35i− 70j + 35k)− (−7i + 35j + 56k) = −28i− 105j− 21k
19. (2i)× j = 2(i× j) = 2k
20. i× (−3k) = −3(i× k) = −3(−j) = 3j
349
7.4 Cross Product
21. k× (2i− j) = k× (2i) + k× (−j) = 2(k× i)− (k× j) = 2j− (−i) = i + 2j
22. i× (j× k) = i× i = 0
23. [(2k)× (3j)]× (4j) = [2 · 3(k× j)× (4j)] = 6(−i)× 4j = (−6)(4)(i× j) = −24k
24. (2i− j + 5k)× i = (2i× i) + (−j× i) + (5k× i) = 2(i× i) + (i× j) + 5(k× i) = 5j + k
25. (i + j)× (i + 5k) = [(i + j)× i] + [(i + j)× 5k] = (i× i) + (j× i) + (i× 5k) + (j× 5k)
= −k + 5(−j) + 5i = 5i− 5j− k
26. i× k− 2(j× i) = −j− 2(−k) = −j + 2k
27. k · (j× k) = k · i = 0
28. i · [j× (−k)] = i · [−(j× k)] = i · (−i) = −(i · i) = −1
29. ‖4j− 5(i× j)‖ = ‖4j− 5k‖ = √41
30. (i× j) · (3j× i) = k · (−3k) = −3(k · k) = −3
31. i× (i× j) = i× k = −j 32. (i× j)× i = k× i = j
33. (i× i)× j = 0× j = 0 34. (i · i)(i× j) = 1(k) = k
35. 2j · [i× (j− 3k)] = 2j · [(i× j) + (i× (−3k)] = 2j · [k + 3(k× i)] = 2j · (k + 3j) = 2j · k + 2j · 3j
= 2(j · k) + 6(j · j) = 2(0) + 6(1) = 6
36. (i× k)× (j× i) = (−j)× (−k) = (−1)(−1)(j× k) = j× k = i
37. a× (3b) = 3(a× b) = 3(4i− 3j + 6k) = 12i− 9j + 18k
38. b× a = −a× b = −(a× b) = −4i + 3j− 6k
39. (−a)× b = −(a× b) = −4i + 3j− 6k
40. |a× b| = √42 + (−3)2 + 62 = √61
41. (a× b)× c =
∣∣∣∣∣∣∣
i j k
4 −3 6
2 4 −1
∣∣∣∣∣∣∣ =
∣∣∣∣−3 64 −1
∣∣∣∣ i− ∣∣∣∣ 4 62 −1
∣∣∣∣ j + ∣∣∣∣ 4 −32 −4
∣∣∣∣k = −21i + 16j + 22k
42. (a× b) · c = 4(2) + (−3)4 + 6(−1) = −10
43. a · (b× c) = (a× b) · c = 4(2) + (−3)4 + 6(−1) = −10
44. (4a) · (b× c) = (4a× b) · c = 4(a× b) · c = 16(2) + (−12)4 + 24(−1) = −40
45. (a) Let A = (1, 3, 0), B = (2, 0, 0), C = (0, 0, 4), and D = (1,−3, 4). Then −−→AB = i − 3j, −→AC = −i − 3j + 4k,
−−→
CD = i−3j, and −−→BD = −i−3j+4k. Since −−→AB = −−→CD and −→AC = −−→BD, the quadrilateral is a parallelogram.
(b) Computing
−−→
AB ×−→AC =
∣∣∣∣∣∣∣
i j k
1 −3 0
−1 −3 4
∣∣∣∣∣∣∣ = −12i− 4j− 6k
we find that the area is ‖ − 12i− 4j− 6k‖ = √144 + 16 + 36 = 14.
46. (a) Let A = (3, 4, 1), B = (−1, 4, 2), C = (2, 0, 2) and D = (−2, 0, 3). Then −−→AB = −4i + k, −→AC = −i− 4j + k,
−−→
CD = −4i+k, and −−→BD = −i−4j+k. Since −−→AB = −−→CD and −→AC = −−→BD, the quadrilateral is a parallelogram.
(b) Computing
−−→
AB ×−→AC =
∣∣∣∣∣∣∣
i j k
−4 0 1
−1 −4 1
∣∣∣∣∣∣∣ = 4i + 3j + 16k
350
7.4 Cross Product
we find that the area is ‖4i + 3j + 16k‖ = √16 + 9 + 256 = √281 ≈ 16.76.
47. −−−→P1P2 = j; −−−→P2P3 = −j + k
−−−→
P1P2 ×−−−→P2P3 =
∣∣∣∣∣∣∣
i j k
0 1 0
0 −1 1
∣∣∣∣∣∣∣ =
∣∣∣∣ 1 0−1 1
∣∣∣∣ i− ∣∣∣∣ 0 00 1
∣∣∣∣ j + ∣∣∣∣ 0 10 −1
∣∣∣∣k = i; A = 12‖i‖ = 12 sq. unit
48. −−−→P1P2 = j + 2k; −−−→P2P3 = 2i + j− 2k
−−−→
P1P2 ×−−−→P2P3 =
∣∣∣∣∣∣∣
i j k
0 1 2
2 1 −2
∣∣∣∣∣∣∣ =
∣∣∣∣ 1 21 −2
∣∣∣∣ i− ∣∣∣∣ 0 22 −2
∣∣∣∣ j + ∣∣∣∣ 0 12 1
∣∣∣∣k = −4i + 4j− 2k
A = 12‖ − 4i + 4j− 2k‖ =3 sq. units
49. −−−→P1P2 = −3j− k; −−−→P2P3 = −2i− k
−−−→
P1P2 ×−−−→P2P3 =
∣∣∣∣∣∣∣
i j k
0 −3 −1
−2 0 −1
∣∣∣∣∣∣∣ =
∣∣∣∣−3 −10 −1
∣∣∣∣ i− ∣∣∣∣ 0 −1−2 −1
∣∣∣∣ j + ∣∣∣∣ 0 −3−2 0
∣∣∣∣k = 3i + 2j− 6k
A = 12‖3i + 2j− 6k‖ = 72 sq. units
50. −−−→P1P2 = −i + 3k; −−−→P2P3 = 2i + 4j− k
−−−→
P1P2 ×−−−→P2P3 =
∣∣∣∣∣∣∣
i j k
−1 0 3
2 4 −1
∣∣∣∣∣∣∣ =
∣∣∣∣ 0 34 −1
∣∣∣∣ i− ∣∣∣∣−1 32 −1
∣∣∣∣ j + ∣∣∣∣−1 02 4
∣∣∣∣k = −12i + 5j− 4k
A = 12‖ − 12i + 5j− 4k‖ =
√
185
2 sq. units
51. b× c =
∣∣∣∣∣∣∣
i j k
−1 4 0
2 2 2
∣∣∣∣∣∣∣ =
∣∣∣∣ 4 02 2
∣∣∣∣ i− ∣∣∣∣−1 02 2
∣∣∣∣ j + ∣∣∣∣−1 42 2
∣∣∣∣k = 8i + 2j− 10k
v = |a · (b× c)| = |(i + j) · (8i + 2j− 10k)| = |8 + 2 + 0| = 10 cu. units
52. b× c =
∣∣∣∣∣∣∣
i j k
1 4 1
1 1 5
∣∣∣∣∣∣∣ =
∣∣∣∣ 4 11 5
∣∣∣∣ i− ∣∣∣∣ 1 11 5
∣∣∣∣ j + ∣∣∣∣ 1 41 1
∣∣∣∣k = 19i− 4j− 3k
v = |a · (b× c)| = |(3i + j + k) · (19i− 4j− 3k)| = |57− 4− 3| = 50 cu. units
53. b× c =
∣∣∣∣∣∣∣
i j k
−2 6 −6
5/2 3 1/2
∣∣∣∣∣∣∣ =
∣∣∣∣ 6 −63 2/2
∣∣∣∣ i− ∣∣∣∣ −2 −65/2 1/2
∣∣∣∣ j + ∣∣∣∣ −2 65/2 3
∣∣∣∣k = 21i− 14j− 21k
a · (b× c) = (4i + 6j) · (21i− 14j− 21k) = 84− 84 + 0 = 0. The vectors are coplanar.
54. The four points will be coplanar if the three vectors −−−→P1P2 = 〈3,−1,−1〉, −−−→P2P3 = 〈−3,−5, 13〉, and −−−→P3P4 =
〈−8, 7,−6〉 are coplanar.
−−−→
P2P3 ×−−−→P3P4 =
∣∣∣∣∣∣∣
i j k
−3 −5 13
−8 7 −6
∣∣∣∣∣∣∣ =
∣∣∣∣−5 137 −6
∣∣∣∣ i− ∣∣∣∣−3 13−8 −6
∣∣∣∣ j + ∣∣∣∣−3 −5−8 7
∣∣∣∣k = 〈−61,−122,−61〉
−−−→
P1P2 · (−−−→P2P3 ×−−−→P3P4) = 〈3,−1,−1〉 · 〈−61,−122,−61〉 = −183 + 122 + 61 = 0
The four points are coplanar.
55. (a) Since θ = 90◦, ‖a× b‖ = ‖a‖ ‖b‖ | sin 90◦| = 6.4(5) = 32.
351
7.4 Cross Product
(b) The direction of a× b is into the fourth quadrant of the xy-plane or to the left of the plane determined by
a and b as shown in Figure 7.54 in the text. It makes an angle of 30◦ with the positive x-axis.
(c) We identify n = (
√
3 i− j)/2. Then a× b = 32n = 16√3 i− 16j.
56. Using Definition 7.4, a × b = √27 (8) sin 120◦n = 24√3 (√3/2)n = 36n. By the right-hand rule, n = j or
n = −j. Thus, a× b = 36j or −36j.
57. (a) We note first that a × b = k, b × c = 12 (i − k), c × a = 12 (j − k), a · (b × c) = 12 , b · (c × a) = 12 , and
c · (a× b) = 12 . Then
A =
1
2 (i− k)
1
2
= i− k, B =
1
2 (j− k)
1
2
= j− k, and C = k1
2
= 2k.
(b) We need to compute A · (B×C). Using formula (10) in the text we have
B×C = (c× a)× (a× b)
[b · (c× a)][c · (a× b)] =
[(c× a) · b]a− [(c× a) · a]b
[b · (c× a)][c · (a× b)]
=
a
c · (a× b) since (c× a) · a = 0.
Then
A · (B×C) = b× c
a · (b× c) ·
a
c · (a× b) =
1
c · (a× b)
and the volume of the unit cell of the reciprocal latrice is the reciprocal of the volume of the unit cell of
the original lattice.
58. a× (b + c) =
∣∣∣∣ a2 a3b2 + c2 b3 + c3
∣∣∣∣ i− ∣∣∣∣ a1 a3b1 + c1 b3 + c3
∣∣∣∣ j + ∣∣∣∣ a1 a2b1 + c1 b2 + c2
∣∣∣∣k
= (a2b3 − a3b2)i + (a2c3 − a3c2)i− [(a1b3 − a3b1)j + (a1c3 − a3c1)j] + (a1b2 − a2b1)k + (a1c2 − a2c1)k
= (a2b3 − a3b2)i− (a1b3 − a3b1)j + (a1b2 − a2b1)k + (a2c3 − a3c2)i− (a1c3 − a3c1)j + (a1c2 − a2c1)k
= a× b + a× c
59. b× c = (b2c3 − b3c2)i− (b1c3 − b3c1)j + (b1c2 − b2c1)k
a× (b× c) = [a2(b1c2 − b2c1) + a3(b1c3 − b3c1)]i− [a1(b1c2 − b2c1)− a3(b2c3 − b3c2)]j
+ [−a1(b1c3 − b3c1)− a2(b2c3 − b3c2)]k
= (a2b1c2 − a2b2c1 + a3b1c3 − a3b3c1)i− (a1b1c2 − a1b2c1 − a3b2c3 + a3b3c2)j
− (a1b1c3 − a1b3c1 + a2b2c3 − a2b3c2)k
(a · c)b− (a · b)c = (a1c1 + a2c2 + a3c3)(b1i + b2j + b3k)− (a1b1 + a2b2 + a3b3)(c1i + c2j + c3k)
= (a2b1c2 − a2b2c1 + a3b1c3 − a3b3c1)i− (a1b1c2 − a1b2c1 − a3b2c3 + a3b3c2)j
− (a1b1c3 − a1b3c1 + a2b2c3 − a2b3c2)k
60. The statement is false since i× (i× j) = i× k = −j and (i× i)× j = 0× j = 0.
61. Using equation 9 in the text,
a · (b× c) =
∣∣∣∣∣∣∣
a1 a2 a3
b1 b2 b3
c1 c2 c3
∣∣∣∣∣∣∣ and (a× b) · c = c · (a× b) =
∣∣∣∣∣∣∣
c1 c2 c3
a1 a2 a3
b1 b2 b3
∣∣∣∣∣∣∣ .
Expanding these determinants out we obtain a · (b× c) = a1b2c3 + a2b3c1 + a3b1c2 − a3b2c1 − a1b3c2 − a2b1c3
and c · (a×b) = a2b3c1 + a3b1c2 + a1b2c3− a2b1c3− a3b2c1− a1b3c2. These are equal so a · (b× c) = (a×b) · c.
352
7.5 Lines and Planes in 3-Space
62. a× (b× c) + b× (c× a) + c× (a× b)
= (a · c)b− (a · b)c + (b · a)c− (b · c)a + (c · b)a− (c · a)b
= [(a · c)b− (c · a)b] + [(b · a)c− (a · b)c] + [(c · b)a− (b · c)a] = 0
63. Since
‖a× b‖2 = (a2b3 − a3b2)2 + (a1b3 − a3b1)2 + (a1b2 − a2b1)2
= a22b
2
3 − 2a2b3a3b2 + a23b22 + a21b23 − 2a1b3a3b1 + a23b21 + a21b22 − 2a1b2a2b1 + a22b21
and
‖a‖2‖b‖2 − (a · b)2 = (a21 + a22 + a23)(b21 + b22 + b23)− (a1b1 + a2b2 + a3b3)2
= a21a
2
2 + a
2
1b
2
2 + a
2
1b
2
3 + a
2
2b
2
1 + a
2
2b
2
2 + a
2
2b
2
3 + a
2
3b
2
1 + a
2
3b
2
2 + a
2
3b
2
3
− a21b21 − a22b22 − a23b23 − 2a1b1a2b2 − 2a1b1a3b3 − 2a2b2a3b3
= a21b
2
2 + a
2
1b
2
3 + a
2
2b
2
1 + a
2
2b
2
3 + a
2
3b
2
1 + a
2
3b
2
2 − 2a1a2b1b2 − 2a1a3b1b3 − 2a2a3b2b3
we see that ‖a× b‖2 = ‖a‖2‖b‖2 − (a · b)2.
64. No. For example i× (i + j) = i× j by the distributive law (iii) in the text, and the fact that i× i = 0. But i + j
does not equal j.
65. By the distributive law (iii) in the text:
(a + b)× (a− b) = (a + b)× a− (a + b)× b = a× a + b× a− a× b− b× b = 2b× a
since a× a = 0, b× b = 0, and −a× b = b× a.
EXERCISES 7.5
Lines and Planes in 3-Space
The equation of a line through P1 and P2 in 3-space with r1 =
−−→
OP1 and r2 =
−−→
OP2 can be expressed as r = r1 + t(ka)
or r = r2 + t(ka) where a = r2 − r1 and k is any non-zero scalar. Thus, the form of the equation of a line is not
unique. (See the alternate solution to Problem 1.)
1. a = 〈1− 3, 2− 5, 1− (−2)〉 = 〈−2,−3, 3〉; 〈x, y, z〉 = 〈1, 2, 1〉+ t〈−2,−3, 3〉
Alternate Solution: a = 〈3− 1, 5− 2,−2− 1〉 = 〈2, 3,−3〉; 〈x, y, z〉 = 〈3, 5,−2〉+ t〈2, 3,−3〉
2. a = 〈0− (−2), 4− 6, 5− 3〉 = 〈2,−2, 2〉; 〈x, y, z〉 = 〈0, 4, 5〉+ t〈2,−2, 2〉
3. a = 〈1/2− (−3/2),−1/2− 5/2, 1− (−1/2)〉 = 〈2,−3, 3/2〉; 〈x, y, z〉 = 〈1/2,−1/2, 1〉+ t〈2,−3, 3/2〉
4. a = 〈10− 5, 2− (−3),−10− 5〉 = 〈5, 5,−15〉; 〈x, y, z〉 = 〈10, 2,−10〉+ t〈5, 5,−15〉
5. a = 〈1− (−4), 1− 1,−1− (−1)〉 = 〈5, 0, 0〉; 〈x, y, z〉 = 〈1, 1,−1〉+ t〈5, 0, 0〉
6. a = 〈3− 5/2, 2− 1, 1− (−2)〉 = 〈1/2, 1, 3〉; 〈x, y, z〉 = 〈3, 2, 1〉+ t〈1/2, 1, 3〉
7. a = 〈2− 6, 3− (−1), 5− 8〉 = 〈−4, 4,−3〉; x = 2− 4t, y = 3 + 4t, z = 5− 3t
8. a = 〈2− 0, 0− 4, 0− 9〉 = 〈2,−4,−9〉; x = 2 + 2t, y = −4t, z = −9t
9. a = 〈1− 3, 0− (−2), 0− (−7)〉 = 〈−2, 2, 7〉; x = 1− 2t, y = 2t, z = 7t
10. a = 〈0− (−2), 0− 4, 5− 0〉 = 〈2,−4, 5〉; x = 2t, y = −4t, z = 5 + 5t
353
7.5 Lines and Planes in 3-Space
11. a = 〈4− (−6), 1/2− (−1/4), 1/3− 1/6〉 = 〈10, 3/4, 1/6〉; x = 4 + 10t, y = 1
2
+
3
4
t, z =
1
3
+
1
6
t
12. a = 〈−3− 4, 7− (−8), 9− (−1)〉 = 〈−7, 15, 10〉; x = −3− 7t, y = 7 + 15t, z = 9 + 10t
13. a1 = 10− 1 = 9, a2 = 14− 4 = 10, a3 = −2− (−9) = 7; x− 109 =
y − 14
10
=
z + 2
7
14. a1 = 1− 2/3 = 1/3, a2 = 3− 0 = 3, a3 = 1/4− (−1/4) = 1/2; x− 11/3 =
y − 3
3
=
z − 1/4
1/2
15. a1 = −7− 4 = −11, a2 = 2− 2 = 0, a3 = 5− 1 = 4; x + 7−11 =
z − 5
4
, y = 2
16. a1 = 1− (−5) = 6, a2 = 1− (−2) = 3, a3 = 2− (−4) = 6; x− 16 =
y − 1
3
=
z − 2
6
17. a1 = 5− 5 = 0, a2 = 10− 1 = 9, a3 = −2− (−14) = 12; x = 5, y − 109 =
z + 2
12
18. a1 = 5/6− 1/3 = 1/2; a2 = −1/4− 3/8 = −5/8; a3 = 1/5− 1/10 = 1/10
x− 5/6
1/2
=
y + 1/4
−5/8 =
z − 1/5
1/10
19. parametric: x = 4 + 3t, y = 6 + t/2, z = −7− 3t/2; symmetric: x− 4
3
=
y − 6
1/2
=
z + 7
−3/2
20. parametric: x = 1− 7t, y = 8− 8t, z = −2; symmetric: x− 1−7 =
y − 8
−8 , z = −2
21. parametric: x = 5t, y = 9t, z = 4t; symmetric:
x
5
=
y
9
=
z
4
22. parametric: x = 12t, y = −3− 5t, z = 10− 6t; symmetric: x
12
=
y + 3
−5 =
z − 10
−6
23. Writing the given line in the form x/2 = (y − 1)/(−3) = (z − 5)/6, we seethat a direction vector is 〈2,−3, 6〉.
Parametric equations for the line are x = 6 + 2t, y = 4− 3t, z = −2 + 6t.
24. A direction vector is 〈5, 1/3,−2〉. Symmetric equations for the line are (x−4)/5 = (y+11)/(1/3) = (z+7)/(−2).
25. A direction vector parallel to both the xz- and xy-planes is i = 〈1, 0, 0〉. Parametric equations for the line are
x = 2 + t, y = −2, z = 15.
26. (a) Since the unit vector j = 〈0, 1, 0〉 lies along the y-axis, we have x = 1, y = 2 + t, z = 8.
(b) since the unit vector k = 〈0, 0, 1〉 is perpendicular to the xy-plane, we have x = 1, y = 2, z = 8 + t.
27. Both lines go through the points (0, 0, 0) and (6, 6, 6). Since two points determine a line, the lines are the same.
28. a and f are parallel since 〈9,−12, 6〉 = −3〈−3, 4,−2〉. c and d are orthogonal since 〈2,−3, 4〉 · 〈1, 4, 5/2〉 = 0.
29. In the xy-plane, z = 9 + 3t = 0 and t = −3. Then x = 4− 2(−3) = 10 and y = 1 + 2(−3) = −5. The point is
(10,−5, 0). In the xz-plane, y = 1+2t = 0 and t = −1/2. Then x = 4−2(−1/2) = 5 and z = 9+3(−1/2) = 15/2.
The point is (5, 0, 15/2). In the yz-plane, x = 4−2t = 0 and t = 2. Then y = 1+2(2) = 5 and z = 9+3(2) = 15.
The point is (0, 5, 15).
30. The parametric equations for the line are x = 1 + 2t, y = −2 + 3t, z = 4 + 2t. In the xy-plane, z = 4 + 2t = 0
and t = −2. Then x = 1 + 2(−2) = −3 and y = −2 + 3(−2) = −8. The point is (−3,−8, 0). In the xz-plane,
y = −2+3t = 0 and t = 2/3. Then x = 1+2(2/3) = 7/3 and z = 4+2(2/3) = 16/3. The point is (7/3, 0, 16/3).
In the yz-plane, x = 1 + 2t = 0 and t = −1/2. Then y = −2 + 3(−1/2) = −7/2 and z = 4 + 2(−1/2) = 3. The
point is (0,−7/2, 3).
354
7.5 Lines and Planes in 3-Space
31. Solving the system 4 + t = 6 + 2s, 5 + t = 11 + 4s, −1 + 2t = −3 + s, or t − 2s = 2, t − 4s = 6, 2t − s = −2
yields s = −2 and t = −2 in all three equations. Thus, the lines intersect at the point x = 4 + (−2) = 2,
y = 5 + (−2) = 3, z = −1 + 2(−2) = −5, or (2, 3,−5).
32. Solving the system 1 + t = 2− s, 2− t = 1 + s, 3t = 6s, or t + s = 1, t + s = 1, t− 2s = 0 yields s = 1/3 and
t = 2/3 in all three equations. Thus, the lines intersect at the point x = 1 + 2/3 = 5/3, y = 2 − 2/3 = 4/3,
z = 3(2/3) = 2, or (5/3, 4/3, 2).
33. The system of equations 2− t = 4 + s, 3 + t = 1 + s, 1 + t = 1− s, or t + s = −2, t− s = −2, t + s = 0 has no
solution since −2 	= 0. Thus, the lines do not intersect.
34. Solving the system 3− t = 2 + 2s, 2 + t = −2 + 3s, 8 + 2t = −2 + 8s, or t + 2s = 1, t− 3s = −4, 2t− 8s = −10
yields s = 1 and t = −1 in all three equations. Thus, the lines intersect at the point x = 3 − (−1) = 4,
y = 2 + (−1) = 1, z = 8 + 2(−1) = 6, or (4, 1, 6).
35. a = 〈−1, 2,−2〉, b = 〈2, 3,−6〉, a · b = 16, ‖a‖ = 3, ‖b‖ = 7; cos θ = a · b‖a‖ ‖b‖ =
16
3 · 7 ;
θ = cos−1
16
21
≈ 40.37◦
36. a = 〈2, 7,−1〉, b = 〈−2, 1, 4〉, a · b = −1, ‖a‖ = 3√6 , ‖b‖ = √21 ;
cos θ =
a · b
‖a‖ ‖b‖ =
−1
(3
√
6 )(
√
21 )
= − 1
9
√
14
; θ = cos−1(− 1
9
√
14
) ≈ 91.70◦
37. A direction vector perpendicular to the given lines will be 〈1, 1, 1〉 × 〈−2, 1,−5〉 = 〈−6, 3, 3〉. Equations of the
line are x = 4− 6t, y = 1 + 3t, z = 6 + 3t.
38. The direction vectors of the given lines are 〈3, 2, 4〉 and 〈6, 4, 8〉 = 2〈3, 2, 4〉. These are parallel, so we need a
third vector parallel to the plane containing the lines which is not parallel to them. The point (1,−1, 0) is on
the first line and (−4, 6, 10) is on the second line. A third vector is then 〈1,−1, 0〉 − 〈−4, 6, 10〉 = 〈5,−7,−10〉.
Now a direction vector perpendicular to the plane is 〈3, 2, 4〉× 〈5,−7,−10〉 = 〈8, 50,−31〉. Equations of the line
through (1,−1, 0) and perpendicular to the plane are x = 1 + 8t, y = −1 + 50t, z = −31t.
39. 2(x− 5)− 3(y − 1) + 4(z − 3) = 0; 2x− 3y + 4z = 19
40. 4(x− 1)− 2(y − 2) + 0(z − 5) = 0; 4x− 2y = 0
41. −5(x− 6) + 0(y − 10) + 3(z + 7) = 0; −5x + 3z = −51
42. 6x− y + 3z = 0
43. 6(x− 1/2) + 8(y − 3/4)− 4(z + 1/2) = 0; 6x + 8y − 4z = 11
44. −(x + 1) + (y − 1)− (z − 0) = 0; −x + y − z = 2
45. From the points (3, 5, 2) and (2, 3, 1) we obtain the vector u = i + 2j + k. From the points (2, 3, 1) and
(−1,−1, 4) we obtain the vector v = 3i+4j− 3k. From the points (−1,−1, 4) and (x, y, z) we obtain the vector
w = (x + 1)i + (y + 1)j + (z − 4)k. Then, a normal vector is
u× v =
∣∣∣∣∣∣∣
i j k
1 2 1
3 4 −3
∣∣∣∣∣∣∣ = −10i + 6j− 2k.
A vector equation of the plane is −10(x + 1) + 6(y + 1)− 2(z − 4) = 0 or 5x− 3y + z = 2.
46. From the points (0, 1, 0) and (0, 1, 1) we obtain the vector u = k. From the points (0, 1, 1) and (1, 3,−1) we obtain
the vector v = i + 2j − 2k. From the points (1, 3,−1) and (x, y, z) we obtain the vector
355
7.5 Lines and Planes in 3-Space
w = (x− 1)i + (y − 3)j + (z + 1)k. Then, a normal vector is
u× v =
∣∣∣∣∣∣∣
i j k
0 0 1
1 2 −2
∣∣∣∣∣∣∣ = −2i + j.
A vector equation of the plane is −2(x− 1) + (y − 3) + 0(z + 1) = 0 or −2x + y = 1.
47. From the points (0, 0, 0) and (1, 1, 1) we obtain the vector u = i + j + k. From the points (1, 1, 1) and
(3, 2,−1) we obtain the vector v = 2i + j − 2k. From the points (3, 2,−1) and (x, y, z) we obtain the vector
w = (x− 3)i + (y − 2)j + (z + 1)k. Then, a normal vector is
u× v =
∣∣∣∣∣∣∣
i j k
1 1 1
2 1 −2
∣∣∣∣∣∣∣ = −3i + 4j− k.
A vector equation of the plane is −3(x− 3) + 4(y − 2)− (z + 1) = 0 or −3x + 4y − z = 0.
48. The three points are not colinear and all satisfy x = 0, which is the equation of the plane.
49. From the points (1, 2,−1) and (4, 3, 1) we obtain the vector u = 3i + j + 2k. From the points (4, 3, 1) and
(7, 4, 3) we obtain the vector v = 3i + j + 2k. From the points (7, 4, 3) and (x, y, z) we obtain the vector
w = (x− 7)i + (y − 4)j + (z − 3)k. Since u× v = 0, the points are colinear.
50. From the points (2, 1, 2) and (4, 1, 0) we obtain the vector u = 2i − 2k. From the points (4, 1, 0) and
(5, 0,−5) we obtain the vector v = i − j − 5k. From the points (5, 0,−5) and (x, y, z) we obtain the vector
w = (x− 5)i + yj + (z + 5)k. Then, a normal vector is
u× v =
∣∣∣∣∣∣∣
i j k
2 0 −2
1 −1 −5
∣∣∣∣∣∣∣ = −2i + 8j− 2k.
A vector equation of the plane is −2(x− 5) + 8y − 2(z + 5) = 0 or x− 4y + z = 0.
51. A normal vector to x + y − 4z = 1 is 〈1, 1,−4〉. The equation of the parallel plane is
(x− 2) + (y − 3)− 4(z + 5) = 0 or x + y − 4z = 25.
52. A normal vector to 5x−y+z = 6 is 〈5,−1, 1, 〉. The equation of the parallel plane is 5(x−0)−(y−0)+(z−0) = 0
or 5x− y + z = 0.
53. A normal vector to the xy-plane is 〈0, 0, 1〉. The equation of the parallel plane is z − 12 = 0 or z = 12.
54. A normal vector is 〈0, 1, 0〉. The equation of the plane is y + 5 = 0 or y = −5.
55. Direction vectors of the lines are 〈3,−1, 1〉. and 〈4, 2, 1〉. A normal vector to the plane is 〈3,−1, 1〉 × 〈4, 2, 1〉 =
〈−3, 1, 10〉. A point on the first line, and thus in the plane, is 〈1, 1, 2〉. The equation of the plane is
−3(x− 1) + (y − 1) + 10(z − 2) = 0 or −3x + y + 10z = 18.
56. Direction vectors of the lines are 〈2,−1, 6〉 and 〈1, 1,−3〉. A normal vector to the plane is 〈2,−1, 6〉×〈1, 1,−3〉 =
〈−3, 12, 3〉. A point on the first line, and thus in the plane, is (1,−1, 5). The equation of the plane is
−3(x− 1) + 12(y + 1) + 3(z − 5) = 0 or −x + 4y + z = 0.
57. A direction vector for the two lines is 〈1, 2, 1〉. Points on the lines are (1, 1, 3) and (3, 0,−2). Thus, another
vector parallel to the plane is 〈1−3, 1−0, 3+2〉 = 〈−2, 1, 5〉. A normal vector to the plane is 〈1, 2, 1〉×〈−2, 1, 5〉 =
〈9,−7, 5〉. Using the point (3, 0,−2) in the plane, the equation of the plane is 9(x− 3)− 7(y− 0) + 5(z + 2) = 0
or 9x− 7y + 5z = 17.
356
7.5 Lines and Planes in 3-Space
58. A direction vector for the line is 〈3, 2,−2〉. Letting t = 0, we see that the origin is on the line and hence in the
plane. Thus, another vector parallel to the plane is 〈4 − 0, 0 − 0,−6 − 0〉 = 〈4, 0,−6〉. A normal vector to the
plane is 〈3, 2,−2〉×〈4, 0,−6〉 = 〈−12, 10,−8〉. The equation of the plane is −12(x−0)+10(y−0)−8(z−0) = 0
or6x− 5y + 4z = 0.
59. A direction vector for the line, and hence a normal vector to the plane, is 〈−3, 1,−1/2〉. The equation of the
plane is −3(x− 2) + (y − 4)− 12 (z − 8) = 0 or −3x + y − 12z = −6.
60. A normal vector to the plane is 〈2− 1, 6− 0,−3 + 2〉 = 〈1, 6,−1〉. The equation of the plane is
(x− 1) + 6(y − 1)− (z − 1) = 0 or x + 6y − z = 6.
61. Normal vectors to the planes are (a) 〈2,−1, 3〉, (b) 〈1, 2, 2〉, (c) 〈1, 1,−3/2〉, (d) 〈−5, 2, 4〉,
(e) 〈−8,−8, 12〉, (f) 〈−2, 1,−3〉. Parallel planes are (c) and (e), and (a) and (f). Perpendicular planes
are (a) and (d), (b) and (c), (b) and (e), and (d) and (f).
62. A normal vector to the plane is 〈−7, 2, 3〉. This is a direction vector for the line and the equations of the line
are x = −4− 7t, y = 1 + 2t, z = 7 + 3t.
63. A direction vector of the line is 〈−6, 9, 3〉, and the normal vectors of the planes are (a) 〈4, 1, 2〉, (b) 〈2,−3, 1〉,
(c) 〈10,−15,−5〉, (d) 〈−4, 6, 2〉. Vectors (c) and (d) are multiples of the direction vector and hence the
corresponding planes are perpendicular to the line.
64. A direction vector of the line is 〈−2, 4, 1〉, and normal vectors to the planes are (a) 〈1,−1, 3〉,
(b) 〈6,−3, 0〉, (c) 〈1,−2, 5〉, (d) 〈−2, 1,−2〉. Since the dot product of each normal vector with the direc-
tion vector is non-zero, none of the planes are parallel to the line.
65. Letting z = t in both equations and solving 5x− 4y = 8 + 9t, x + 4y = 4− 3t, we obtain x = 2 + t, y = 12 − t,
z = t.
66. Letting y = t in both equations and solving x − z = 2 − 2t, 3x + 2z = 1 + t, we obtain x = 1 − 35 t, y = t,
z = −1 + 75 t or, letting t = 5s, x = 1− 3s, y = 5s, z = −1 + 7s.
67. Letting z = t in both equations and solving 4x− 2y = 1 + t, x + y = 1− 2t, we obtain x = 12 − 12 t, y = 12 − 32 t,
z = t.
68. Letting z = t and using y = 0 in the first equation, we obtain x = − 12 t, y = 0, z = t.
69. Substituting the parametric equations into the equation of the plane, we obtain 2(1+2t)−3(2−t)+2(−3t) = −7
or t = −3. Letting t = −3 in the equation of the line, we obtain the point of intersection (−5, 5, 9).
70. Substituting the parametric equations into the equation of the plane, we obtain (3−2t)+(1+6t)+4(2− 12 t) = 12
or 2t = 0. Letting t = 0 in the equation of the line, we obtain the point of intersection (3, 1, 2).
71. Substituting the parametric equations into the equation of the plane, we obtain 1 + 2− (1 + t) = 8 or t = −6.
Letting t = −6 in the equation of the line, we obtain the point of intersection (1, 2,−5).
72. Substituting the parametric equations into the equation of the plane, we obtain 4 + t− 3(2 + t) + 2(1 + 5t) = 0
or t = 0. Letting t = 0 in the equation of the line, we obtain the point of intersection (4, 2, 1).
357
7.5 Lines and Planes in 3-Space
In Problems 73 and 74, the cross product of the normal vectors to the two planes will be a vector parallel to both
planes, and hence a direction vector for a line parallel to the two planes.
73. Normal vectors are 〈1, 1,−4〉 and 〈2,−1, 1〉. A direction vector is
〈1, 1,−4〉 × 〈2,−1, 1〉 = 〈−3,−9,−3〉 = −3〈1, 3, 1〉.
Equations of the line are x = 5 + t, y = 6 + 3t, z = −12 + t.
74. Normal vectors are 〈2, 0, 1〉 and 〈−1, 3, 1〉. A direction vector is
〈2, 0, 1〉 × 〈−1, 3, 1〉 = 〈−3,−3, 6〉 = −3〈1, 1,−2〉.
Equations of the line are x = −3 + t, y = 5 + t, z = −1− 2t.
In Problems 75 and 76, the cross product of the direction vector of the line with the normal vector of the given plane
will be a normal vector to the desired plane.
75. A direction vector of the line is 〈3,−1, 5〉 and a normal vector to the given plane is 〈1, 1, 1〉. A normal vector
to the desired plane is 〈3,−1, 5〉 × 〈1, 1, 1〉 = 〈−6, 2, 4〉. A point on the line, and hence in the plane, is (4, 0, 1).
The equation of the plane is −6(x− 4) + 2(y − 0) + 4(z − 1) = 0 or 3x− y − 2z = 10.
76. A direction vector of the line is 〈3, 5, 2〉 and a normal vector to the given plane is 〈2,−4,−1〉. A normal vector to
the desired plane is 〈−3, 5, 2〉 × 〈2,−4,−1〉 = 〈3, 1, 2〉. A point on the line, and hence in the plane, is (2,−2, 8).
The equation of the plane is 3(x− 2) + (y + 2) + 2(z − 8) = 0 or 3x + y + 2z = 20.
77. 78. 79.
80. 81. 82.
358
7.6 Vector Spaces
EXERCISES 7.6
Vector Spaces
1. Not a vector space. Axiom (vi) is not satisfied. 2. Not a vector space. Axiom (i) is not satisfied.
3. Not a vector space. Axiom (x) is not satisfied. 4. A vector space
5. A vector space 6. A vector space
7. Not a vector space. Axiom (ii) is not satisfied. 8. A vector space
9. A vector space 10. Not a vector space. Axiom (i) is not satisfied.
11. A subspace 12. Not a subspace. Axiom (i) is not satisfied.
13. Not a subspace. Axiom (ii) is not satisfied. 14. A subspace
15. A subspace 16. A subspace
17. A subspace 18. A subspace
19. Not a subspace. Neither axioms (i) nor (ii) are satisfied.
20. A subspace
21. Let (x1, y1, z1) and (x2, y2, z2) be in S. Then
(x1, y1, z1) + (x2, y2, z2) = (at1, bt1, ct1) + (at2, bt2, ct2) = (a(t1 + t2), b(t1 + t2), c(t1 + t2))
is in S. Also, for (x, y, z) in S then k(x, y, z) = (kx, ky, kz) = (a(kt), b(kt), c(kt)) is also in S.
22. Let (x1, y1, z1) and (x2, y2, z2) be in S. Then ax1 + by1 + cz1 = 0 and ax2 + by2 + cz2 = 0. Adding gives
a(x1 +x2)+ b(y1 + y2)+ c(z1 + z2) = 0 and so (x1, y1, z1)+ (x2, y2, z2) = (x1 +x2, y1 + y2, z1 + z2) is in S. Also,
for (x, y, z) then ax + by + cz = 0 implies k(ax + by + cz) = k · 0 = 0 and a(kx) + b(ky) + c(kz) = 0. this means
k(x, y, z) = (kx,ky, kz) is in S.
23. (a) c1u1 + c2u2 + c3u3 = 0 if and only if c1 + c2 + c3 = 0, c2 + c3 = 0, c3 = 0. The only solution of this system
is c1 = 0, c2 = 0, c3 = 0.
(b) Solving the system c1 + c2 + c3 = 3, c2 + c3 = −4, c3 = 8 gives c1 = 7, c2 = −12, c3 = 8. Thus
a = 7u1 − 12u2 + 8u3.
24. (a) The assumption c1p1 + c2p2 = 0 is equivalent to (c1 + c2)x + (c1 − c2) = 0. Thus c1 + c2 = 0, c1 − c2 = 0.
The only solution of this system is c1 = 0, c2 = 0.
(b) Solving the system c1 + c2 = 5, c1 − c2 = 2 gives c1 = 72 , c2 = 32 . Thus p(x) = 72p1(x) + 32p2(x)
25. Linearly dependent since 〈−6, 12〉 = − 32 〈4,−8〉
26. Linearly dependent since 2〈1, 1〉+ 3〈0, 1〉+ (−1)〈2, 5〉 = 〈0, 0〉
359
7.6 Vector Spaces
27. Linearly independent
28. Linearly dependent since for all x (1) · 1 + (−2)(x + 1) + (1)(x + 1)2 + (−1)x2 = 0.
29. f is discontinuous at x = −1 and at x = −3.
30. (x, sinx) =
∫ 2π
0
x sinx dx = (−x cosx + sinx)
∣∣∣2π
0
= −2π
31. ‖x‖2 =
∫ 2π
0
x2 dx =
1
3
x3
∣∣∣∣2π
0
=
8
3
π3 and so ‖x‖ = 2
√
2π3
3
. Now
‖ sinx‖2 =
∫ 2π
0
sin2 x dx =
1
2
∫ 2π
0
(1− cos 2x) dx = 1
2
(
x− 1
2
sin 2x
) ∣∣∣2π
0
= π
and so ‖ sinx‖ = √π .
32. A basis could be 1, x, ex cos 3x, ex sin 3x.
33. We need to show that Span{x1, x2, . . . , xn} is closed under vector addition and scalar multiplication. Suppose
u and v are in Span{x1, x2, . . . , xn}. Then u = a1x1 + a2x2 + · · ·+ anxn and v = b1x1 + b2x2 + · · ·+ bnxn, so
that
u + v = (a1 + b1)x1 + (a2 + b2)x2 + · · ·+ (an + bn)xn,
which is in Span{x1, x2, . . . , xn}. Also, for any real number k,
ku = k(a1x1 + a2x2 + · · ·+ anxn) = ka1x1 + ka2x2 + · · ·+ kanxn,
which is in Span{x1, x2, . . . , xn}. Thus, Span{x1, x2, . . . , xn} is a subspace of V.
34. R2 is not a subspace of either R3 or R4 and R3 is not a subspace of R4. The vectors in R2 are ordered pairs,
while the vectors in R3 are ordered triples.
35. Since a basis for M22 is
B =
{[
1 0
0 0
]
,
[
0 1
0 0
]
,
[
0 0
1 0
]
,
[
0 0
0 1
]}
,
the dimension of M22 is 4.
36. To show that the set of nonzero orthogonal vectors is linearly independent we set c1v1 + c2v2 + · · ·+ cnvn = 0.
For 0 ≤ i ≤ n,
(c1v1 + c2v2 + · · ·+ civi · · ·+ cnvn) · vi = c1v1 · vi + c2v2 · vi + · · ·+ civi · vi · · ·+ cnvn · vi = ci||vi||2,
so ci||vi||2 = 0 because
(c1v1 + c2v2 + · · ·+ civi · · ·+ cnvn) · vi = 0 · vi = 0.
Since vi is a nonzerovector, ci = 0. Thus, the assumption that c1v1 + c2v2 + · · · + cnvn = 0 leads to
c1 = c2 = · · · = cn = 0, and the set is linearly independent.
37. We verify the four properties:
(i) (u,v) = u1v1 + 4u2v2 = v1u1 + 4v2u2 = (v,u)
(ii) (ku,v) = (ku1)v1 + 4(ku2)v2 = k(u1v1 + 4u2v2) = k(u,v)
(iii) (u,u) = u21 + 4ku
2
2 > 0 for u 	= 0. Furthermore, u21 + 4ku22 = 0 if and only if u1 = 0 and u2 = 0, or
equivalently, u = 0.
(iv) (u,v + w) = u1(v1 + w1) + 4u2(v2 + w2) = (u1v1 + 4u2v2) + (u1w1 + 4u2w2) = (u,v) + (u,w)
360
7.7 Gram-Schmidt Orthogonalization Process
38. (a) Let u = 〈2, 1〉 and v = 〈2,−1〉 be nonzero vectors in R2. With respect to the standard inner or dot product
on R2,
u · v = 〈2, 1〉 · 〈2,−1〉 = 2 · 2 + 1 · (−1) = 3.
We see that u and v are not orthogonal with respect to that inner product. But using the inner product
in Problem 37, we have
(u,v) = 2 · 2 + 4(1) · (−1) = 0,
and so u and v are orthogonal with respect to that inner product.
(b) Consider f(x) = sinx and g(x) = cosx in C[0, 2π]. Since∫ 2π
0
sinx cosx dx =
1
2
∫ 2π
0
sin 2x dx = − 1
4
cos 2x
∣∣∣2π
0
= − 1
4
(1− 1) = 0,
these functions are orthogonal in C[0, 2π].
EXERCISES 7.7
Gram-Schmidt Orthogonalization Process
1. Letting w1 = 〈 1213 , 513 〉 and w2 = 〈 513 ,− 1213 〉, we have
w1 ·w2 =
(
12
13
) (
5
13
)
+
(
5
13
) (
−12
13
)
= 0,
so the vectors are orthogonal. Also,
||w1|| =
√(
12
13
)2
+
(
5
13
)2
= 1 and ||w2|| =
√(
5
13
)2
+
(
−12
13
)2
= 1,
so the basis is orthonormal. To express u = 〈4, 2〉 in terms of w1 and w2 we compute
u ·w1 = 〈4, 2〉 ·
〈
12
13
,
5
13
〉
= (4)
(
12
13
)
+ (2)
(
5
13
)
=
58
13
u ·w2 = 〈4, 2〉 ·
〈
5
13
,−12
13
〉
= (4)
(
5
13
)
+ (2)
(
−12
13
)
= − 4
13
,
so
u =
58
13
w1 − 413 w2.
2. Letting w1 = 〈1/
√
3, 1/
√
3,−1/√3〉, w2 = 〈0,−1/
√
2,−1/√2〉, and w3 = 〈−2/
√
6, 1/
√
6,−1/√6〉, we have
w1 ·w2 =
(
1√
3
)
(0) +
(
1√
3
) (
− 1√
2
)
+
(
− 1√
3
) (
− 1√
2
)
= 0
w1 ·w3 =
(
1√
3
) (
− 2√
6
)
+
(
1√
3
) (
1√
6
)
+
(
− 1√
3
) (
− 1√
6
)
= 0
w2 ·w3 = (0)
(
− 2√
6
)
+
(
− 1√
2
) (
1√
6
)
+
(
− 1√
2
) (
− 1√
6
)
= 0,
361
7.7 Gram-Schmidt Orthogonalization Process
so the vectors are orthogonal. Also,
||w1|| =
√(
1√
3
)2
+
(
1√
3
)2
+
(
− 1√
3
)2
= 1, ||w2|| =
√
02 +
(
− 1√
2
)2
+
(
− 1√
2
)2
= 1,
and ||w3|| =
√(
− 2√
6
)2
+
(
1√
6
)2
+
(
− 1√
6
)2
= 1,
so the basis is orthonormal. To express u = 〈5,−1, 6〉 in terms of w1, w2, and w3 we compute
u ·w1 = 〈5,−1, 6〉 ·
〈
1√
3
,
1√
3
,− 1√
3
〉
= (5)
(
1√
3
)
+ (−1)
(
1√
3
)
+ (6)
(
− 1√
3
)
= − 2√
3
u ·w2 = 〈5,−1, 6〉 ·
〈
0,− 1√
2
,− 1√
2
〉
= (5)(0) + (−1)
(
− 1√
2
)
+ (6)
(
− 1√
2
)
= − 5√
2
u ·w3 = 〈5,−1, 6〉 ·
〈
− 2√
6
,
1√
6
,− 1√
6
〉
= (5)
(
− 2√
6
)
+ (−1)
(
1√
6
)
+ (6)
(
− 1√
6
)
= − 17√
6
so
u = − 2√
3
w1 − 5√
2
w2 − 17√
6
w3.
Since the basis vectors in Problems 3 and 4 are orthogonal but not orthonormal, the result of Theorem 7.5 must be
slightly modified to read
u =
u ·w1
||w1||2 w1 +
u ·w2
||w2||2 w2 + · · ·+
u ·wn
||wn||2 wn.
The proof is very similar to that given in the text for Theorem 7.5.
3. Letting w1 = 〈1, 0, 1〉,w2 = 〈0, 1, 0〉, and w3 = 〈−1, 0, 1〉 we have
w1 ·w2 = (1)(0) + (0)(1) + (1)(0) = 0
w1 ·w3 = (1)(−1) + (0)(0) + (1)(1) = 0
w2 ·w3 = (0)(−1) + (1)(0) + (0)(1) = 0
so the vectors are orthogonal. We also compute
||w1||2 = 12 + 02 + 12 = 2
||w2||2 = 02 + 12 + 02 = 1
||w3||2 = (−1)2 + 02 + 12 = 2
and, with u = 〈10, 7,−13〉,
u ·w1 = (10)(1) + (7)(0) + (−13)(1) = −3
u ·w2 = (10)(0) + (7)(1) + (−13)(0) = 7
u ·w3 = (10)(−1) + (7)(0) + (−13)(1) = −23.
Then, using the result given before the solution to this problem, we have
u = −3
2
w1 + 7w2 − 232 w3.
362
7.7 Gram-Schmidt Orthogonalization Process
4. Letting w1 = 〈2, 1,−2, 0〉,w2 = 〈1, 2, 2, 1〉, w3 = 〈3,−4, 1, 3〉, and w4 = 〈5,−2, 4,−9〉 we have
w1 ·w2 = (2)(1) + (1)(2) + (−2)(2) + (0)(1) = 0
w1 ·w3 = (2)(3) + (1)(−4) + (−2)(1) + (0)(3) = 0
w1 ·w4 = (2)(5) + (1)(−2) + (−2)(4) + (0)(−9) = 0
w2 ·w3 = (1)(3) + (2)(−4) + (2)(1) + (1)(3) = 0
w2 ·w4 = (1)(5) + (2)(−2) + (2)(4) + (1)(−9) = 0
w3 ·w4 = (3)(5) + (−4)(−2) + (1)(4) + (3)(−9) = 0
so the vectors are orthogonal. We also compute
||w1||2 = 22 + 12 + (−2)2 + 02 = 9
||w2||2 = 12 + 22 + 22 + 12 = 10
||w3||2 = 32 + (−4)2 + 12 + 32 = 35
||w4||2 = 52 + (−2)2 + 42 + (−9)2 = 126
and, with u = 〈1, 2, 4, 3〉,
u ·w1 = (1)(2) + (2)(1) + (4)(−2) + (3)(0) = −4
u ·w2 = (1)(1) + (2)(2) + (4)(2) + (3)(1) = 16
u ·w3 = (1)(3) + (2)(−4) + (4)(1) + (3)(3) = 8
u ·w4 = (1)(5) + (2)(−2) + (4)(4) + (3)(−9) = −10.
Then, using the result given before the solution to this problem, we have
u = −4
9
w1 +
8
5
w2 +
8
35
w3 − 563 w4.
5. (a) We have u1 = 〈−3, 2〉 and u2 = 〈−1,−1〉. Taking v1 = u1 = 〈−3, 2〉, and using u2 ·v1 = 1 and v1 ·v1 = 13
we obtain
v2 = u2 − u2 · v1v1 · v1 v1 = 〈−1,−1〉 −
1
13
〈−3, 2〉 =
〈
−10
13
,−15
13
〉
.
Thus, an orthogonal basis is {〈−3, 2〉, 〈− 1013 ,− 1513 〉} and an orthonormal basis is {w′1,w′2}, where
w′1 =
1
||〈−3, 2〉|| 〈−3, 2〉 =
1√
13
〈−3, 2〉 =
〈
− 3√
13
,
2√
13
〉
and
w′2 =
1
||〈−1013 ,− 1513 〉||
〈
−10
13
,−15
13
〉
=
1
5/
√
13
〈
−10
13
,−15
13
〉
=
〈
− 2√
13
,− 3√
13
〉
.
(b) We have u1 = 〈−3, 2〉 and u2 = 〈−1,−1〉. Taking v1 = u2 = 〈−1,−1〉, and using u1 ·v1 = 1 and v1 ·v1 = 2
we obtain
v2 = u1 − u1 · v1v1 · v1 v1 = 〈−3, 2〉 −
1
2
〈−1,−1〉 =
〈
−5
2
,
5
2
〉
.
Thus, an orthogonal basis is {〈−1,−1〉, 〈− 52 , 52 〉} and an orthonormal basis is {w′′3 ,w′′4}, where
w′′3 =
1
||〈−1,−1〉|| 〈−1,−1〉 =
1√
2
〈−1,−1〉 =
〈
− 1√
2
,− 1√
2
〉
and
w′′4 =
1
||〈−52 , 52 〉||
〈
−5
2
,
5
2
〉
=
1
5/
√
2
〈
−5
2
,
5
2
〉
=
〈
− 1√
2
,
1√
2
〉
.
363
-4 -2 2 4
-4
-2
2
4
-1 -0.5 0.5 1
-1
-0.5
0.5
1
-1 -0.5 0.5 1
-1
-0.5
0.5
1
-4 -2 2 4
-4
-2
2
4
-1 -0.5 0.5 1
-1
-0.5
0.5
1
-1 -0.5 0.5 1
-1
-0.5
0.5
1
7.7 Gram-Schmidt Orthogonalization Process
(c)
u
v
w1
w2
w4
w3
6. (a) We have u1 = 〈−3, 4〉 and u2 = 〈−1, 0〉. Taking v1 = u1 = 〈−3, 4〉, and using u2 · v1 = 3 and v1 · v1 = 25
we obtain
v2 = u2 − u2 · v1v1 · v1 v1 = 〈−1, 0〉 −
3
25
〈−3, 4〉 =
〈
−16
25
,−12
25
〉
.
Thus, an orthogonal basis is {〈−3, 4〉, 〈− 1625 ,− 1225 〉} and an orthonormal basis is {w′1,w′2}, where
w′1 =
1
||〈−3, 4〉|| 〈−3, 4〉 =
1
5
〈−3, 4〉 =
〈
−3
5
,
4
5
〉
and
w′2 =
1
||〈−1625 ,− 1225 〉||
〈
−16
25
,−12
25
〉
=
1
4/5
〈
−16
25
,−12
25
〉
=
〈
−4
5
,−3
5
〉
.
(b) We have u1 = 〈−3, 4〉 and u2 = 〈−1, 0〉. Taking v1 = u2 = 〈−1, 0〉, and using u1 · v1 = 3 and v1 · v1 = 1
we obtain
v2 = u1 − u1 · v1v1 · v1 v1 = 〈−3, 4〉 −
3
1
〈−1, 0〉 = 〈0, 4〉 .
Thus, an orthogonal basis is {〈−1, 0〉, 〈0, 4〉} and an orthonormal basis is {w′′3 ,w′′4}, where
w′′3 =
1
||〈−1, 0〉|| 〈−1, 0〉 =
1
1
〈−1, 0〉 = 〈−1, 0〉
and
w′′4 =
1
||〈0, 4〉|| 〈0, 4〉 =
1
4
〈0, 4〉 = 〈0, 1〉 .
(c) u
v
w1
w2
w4
w3
7. (a) We have u1 = 〈1, 1〉 and u2 = 〈1, 0〉. Taking v1 = u1 = 〈1, 1〉, and using u2 · v1 = 1 and v1 · v1 = 2 we
obtain
v2 = u2 − u2 · v1v1 · v1 v1 = 〈1, 0〉 −
1
2
〈1, 1〉 =
〈
1
2
,−1
2
〉
.
Thus, an orthogonal basis is {〈1, 1〉,〈 12 ,− 12 〉} and an orthonormal basis is {w′1,w′2}, where
w′1 =
1
||〈1, 1〉|| 〈1, 1〉 =
1√
2
〈1, 1〉 =
〈
1√
2
,
1√
2
〉
and
w′2 =
1
||〈 1√
2
,− 1√
2
〉||
〈
1√
2
,− 1√
2
〉
=
1
1
〈
1√
2
,− 1√
2
〉
=
〈
1√
2
,− 1√
2
〉
.
364
-2 -1 1 2
-2
-1.5
-1
-0.5
0.5
1
1.5
2
-1 -0.5 0.5 1
-1
-0.5
0.5
1
-1 -0.5 0.5 1
-1
-0.5
0.5
1
-7.5 -5 -2.5 2.5 5 7.5
-8
-6
-4
-2
2
4
6
8
-1 -0.5 0.5 1
-1
-0.5
0.5
1
-1 -0.5 0.5 1
-1
-0.5
0.5
1
7.7 Gram-Schmidt Orthogonalization Process
(b) We have u1 = 〈1, 1〉 and u2 = 〈1, 0〉. Taking v1 = u2 = 〈1, 0〉, and using u1 · v1 = 1 and v1 · v1 = 1 we
obtain
v2 = u1 − u1 · v1v1 · v1 v1 = 〈1, 1〉 −
1
1
〈1, 0〉 = 〈0, 1〉 .
Thus, an orthogonal basis is {〈1, 0〉, 〈0, 1〉}, which is also an orthonormal basis.
(c)
u
v
w1
w2
w4
w3
8. (a) We have u1 = 〈5, 7〉 and u2 = 〈1,−2〉. Taking v1 = u1 = 〈5, 7〉, and using u2 · v1 = −9 and v1 · v1 = 74
we obtain
v2 = u2 − u2 · v1v1 · v1 v1 = 〈1,−2〉 −
9
74
〈5, 7〉 =
〈
119
74
,−85
74
〉
.
Thus, an orthogonal basis is {〈5, 7〉, 〈 11974 ,− 8574 〉} and an orthonormal basis is {w′1,w′2}, where
w′1 =
1
||〈5, 7〉|| 〈5, 7〉 =
1√
74
〈5, 7〉 =
〈
5√
74
,
7√
74
〉
and
w′2 =
1
||〈 11974 ,− 8574 〉||
〈
119
74
,−85
74
〉
=
1
17/
√
74
〈
119
74
,−85
74
〉
=
〈
7√
74
,− 5√
74
〉
.
(b) We have u1 = 〈5, 7〉 and u2 = 〈1,−2〉. Taking v1 = u2 = 〈1,−2〉, and using u1 · v1 = −9 and v1 · v1 = 5
we obtain
v2 = u1 − u1 · v1v1 · v1 v1 = 〈5, 7〉 −
9
5
〈1,−2〉 =
〈
34
5
,
17
5
〉
.
Thus, an orthogonal basis is {〈1,−2〉, 〈 345 , 175 〉} and an orthonormal basis is {w′′3 ,w′′4}, where
w′′3 =
1
||〈1,−2〉|| 〈1,−2〉 =
1√
5
〈1,−2〉 =
〈
1√
5
,− 2√
5
〉
and
w′′4 =
1
||〈 345 , 175 〉||
〈
34
5
,
17
5
〉
=
1
17/
√
5
〈
34
5
,
17
5
〉
=
〈
2√
5
,
1√
5
〉
.
(c)
u
v
w1
w2
w4
w3
9. We have u1 = 〈1, 1, 0〉,u2 = 〈1, 2, 2〉, and u3 = 〈2, 2, 1〉. Taking v1 = u1 = 〈1, 1, 0〉 and using u2 · v1 = 3 and
v1 · v1 = 2 we obtain
v2 = u2 − u2 · v1v1 · v1 v1 = 〈1, 2, 2〉 −
3
2
〈1, 1, 0〉 =
〈
−1
2
,
1
2
, 2
〉
.
365
7.7 Gram-Schmidt Orthogonalization Process
Next, using u3 · v1 = 4,u3 · v2 = 2, and v2 · v2 = 92 , we obtain
v3 = u3 − u3 · v1v1 · v1 v1 −
u3 · v2
v2 · v2 v2 = 〈2, 2, 1〉 −
4
2
〈1, 1, 0〉 − 2
9/2
〈
−1
2
,
1
2
, 2
〉
=
〈
2
9
,−2
9
,
1
9
〉
.
Thus, an orthogonal basis is
B′ =
{
〈1, 1, 0〉 ,
〈
−1
2
,
1
2
, 2
〉
,
〈
2
9
,−2
9
,
1
9
〉}
,
and an orthonormal basis is
B′′ =
{〈
1√
2
,
1√
2
, 0
〉
,
〈
− 1
3
√
2
,
1
3
√
2
,
4
3
√
2
〉
,
〈
2
3
,−2
3
,
1
3
〉}
.
10. We have u1 = 〈−3, 1, 1〉,u2 = 〈1, 1, 0〉, and u3 = 〈−1, 4, 1〉. Taking v1 = u1 = 〈−3, 1, 1〉 and using u2 · v1 = −2
and v1 · v1 = 11 we obtain
v2 = u2 − u2 · v1v1 · v1 v1 = 〈1, 1, 0〉 −
−2
11
〈−3, 1, 1〉 =
〈
5
11
,
13
11
,
2
11
〉
.
Next, using u3 · v1 = 8,u3 · v2 = 4911 , and v2 · v2 = 1811 , we obtain
v3 = u3 − u3 · v1v1 · v1 v1 −
u3 · v2
v2 · v2 v2 = 〈−1, 4, 1〉 −
8
11
〈−3, 1, 1〉 − 49/11
18/11
〈
5
11
,
13
11
,
2
11
〉
=
〈
− 1
18
,
1
18
,−2
9
〉
.
Thus, an orthogonal basis is
B′ =
{
〈−3, 1, 1〉 ,
〈
5
11
,
13
11
,
2
11
〉
,
〈
− 1
18
,
1
18
,−2
9
〉}
,
and an orthonormal basis is
B′′ =
{〈
− 3√
11
,
1√
11
,
1√
11
〉
,
〈
5
3
√
22
,
13
3
√
22
,
2
3
√
22
〉
,
〈
− 1
3
√
2
,
1
3
√
2
,
4
3
√
2
〉}
.
11. We have u1 = 〈 12 , 12 , 1〉,u2 = 〈−1, 1,− 12 〉, and u3 = 〈−1, 12 , 1〉. Taking v1 = u1 = 〈 12 , 12 , 1〉 and using u2 ·v1 = − 12
and v1 · v1 = 32 we obtain
v2 = u2 − u2 · v1v1 · v1 v1 =
〈
−1, 1,−1
2
〉
− −1/2
3/2
〈
1
2
,
1
2
, 1
〉
=
〈
−5
6
,
7
6
,−1
6
〉
.
Next, using u3 · v1 = 34 ,u3 · v2 = 54 , and v2 · v2 = 2512 , we obtain
v3 = u3 − u3 · v1v1 · v1 v1 −
u3 · v2
v2 · v2 v2 =
〈
−1, 1
2
, 1
〉
− 3/4
3/2
〈
1
2
,
1
2
, 1
〉
− 5/4
25/12
〈
−5
6
,
7
6
,−1
6
〉
=
〈
−3
4
,− 9
20
,
3
5
〉
.
Thus, an orthogonal basis is
B′ =
{〈
1
2
,
1
2
, 1
〉
,
〈
−5
6
,
7
6
,−1
6
〉
,
〈
−3
4
,− 9
20
,
3
5
〉}
,
and an orthonormal basis is
B′′ =
{〈
1√
6
,
1√
6
,
2√
6
〉
,
〈
− 1√
3
,
7
5
√
3
,− 1
5
√
3
〉
,
〈
− 1√
2
,− 3
5
√
2
,
4
5
√
2
〉}
.
12. We have u1 = 〈1, 1, 1〉,u2 = 〈9,−1, 1〉, and u3 = 〈−1, 4,−2〉. Taking v1 = u1 = 〈1, 1, 1〉 and using u2 · v1 = 9
and v1 · v1 = 3 we obtain
v2 = u2 − u2 · v1v1 · v1 v1 = 〈9,−1, 1〉 −
9
3
〈1, 1, 1〉 = 〈6,−4,−2〉 .
366
7.7 Gram-Schmidt Orthogonalization Process
Next, using u3 · v1 = 1,u3 · v2 = −18, and v2 · v2 = 56, we obtain
v3 = u3 − u3 · v1v1 · v1 v1 −
u3 · v2
v2 · v2 v2 = 〈−1, 4,−2〉 −
1
3
〈1, 1, 1〉 − −18
56
〈6,−4,−2〉 =
〈
25
42
,
50
21
,−125
42
〉
.
Thus, an orthogonal basis is
B′ =
{
〈1, 1, 1〉 , 〈6,−4,−2〉 ,
〈
25
42
,
50
21
,−125
42
〉}
,
and an orthonormal basis is
B′′ =
{〈
1√
3
,
1√
3
,
1√
3
〉
,
〈
3√
14
,− 2√
14
,− 1√
14
〉
,
〈
1√
42
,
4√
42
,− 5√
42
〉}
.
13. We have u1 = 〈1, 5, 2〉, and u2 = 〈−2, 1, 1〉. Taking v1 = u1 = 〈1, 5, 2〉 and using u2 · v1 = 5 and v1 · v1 = 30
we obtain
v2 = u2 − u2 · v1v1 · v1 v1 = 〈−2, 1, 1〉 −
5
30
〈1, 5, 2〉 =
〈
−13
6
,
1
6
,
2
3
〉
.
Thus, an orthogonal basis is B′ =
{〈1, 5, 2〉 , 〈− 136 , 16 , 23〉} , and an orthonormal basis is
B′′ =
{〈
1√
30
,
5√
30
,
2√
30
〉
,
〈
− 13√
186
,
1√
186
,
4√
186
〉}
.
14. We have u1 = 〈1, 2, 3〉, and u2 = 〈3, 4, 1〉. Taking v1 = u1 = 〈1, 2, 3〉 and using u2 · v1 = 14 and v1 · v1 = 14 we
obtain
v2 = u2 − u2 · v1v1 · v1 v1 = 〈3, 4, 1〉 −
14
14
〈1, 2, 3〉 = 〈2, 2,−2〉 .
Thus, an orthogonal basis is B′ = {〈1, 2, 3〉 , 〈2, 2,−2〉} , and an orthonormal basis is
B′′ =
{〈
1√
14
,
2√
14
,
3√
14
〉
,
〈
1√
3
,
1√
3
,− 1√
3
〉}
.
15. We have u1 = 〈1,−1, 1,−1〉, and u2 = 〈1, 3, 0, 1〉. Taking v1 = u1 = 〈1,−1, 1,−1〉 and using u2 · v1 = −3 and
v1 · v1 = 4 we obtain
v2 = u2 − u2 · v1v1 · v1 v1 = 〈1, 3, 0, 1〉 −
−3
4
〈1,−1, 1,−1〉 =
〈
7
4
,
9
4
,
3
4
,
1
4
〉
.
Thus, an orthogonal basis is B′ =
{〈1,−1, 1,−1〉 , 〈 74 , 94 , 34 , 14〉} , and an orthonormal basis is
B′′ =
{〈
1
2
,−1
2
,
1
2
,−1
2
〉
,
〈
7
2
√
35
,
9
2
√
35
,
3
2
√
35
,
1
2
√
35
〉}
.
16. We have u1 = 〈4, 0, 2,−1〉,u2 = 〈2, 1,−1, 1〉, and u3 = 〈1, 1,−1, 0〉. Taking v1 = u1 = 〈4, 0, 2,−1〉 and using
u2 · v1 = 5 and v1 · v1 = 21 we obtain
v2 = u2 − u2 · v1v1 · v1 v1 = 〈2, 1,−1, 1〉 −
5
21
〈4, 0, 2,−1〉 =
〈
22
21
, 1,−31
21
,
26
21
〉
.
Next, using u3 · v1 = 2,u3 · v2 = 7421 , and v2 · v2 = 12221 , we obtain
v3 = u3 − u3 · v1v1 · v1 v1 −
u3 · v2
v2 · v2 v2
= 〈1, 1,−1, 0〉 − 2
21
〈4, 0, 2,−1〉 − 74/21
122/21
〈
22
21
, 1,−31
21
,
26
21
〉
=
〈
− 1
61
,
24
61
,−18
61
,−40
61
〉
.
Thus, an orthogonal basis is
B′ =
{
〈4, 0, 2,−1〉 ,
〈
22
21
, 1,−31
21
,
26
21
〉
,
〈
− 1
61
,
24
61
,−18
61
,−40
61
〉}
,
367
7.7 Gram-Schmidt Orthogonalization Process
and an orthonormal basis is
B′′ =
{〈
4√
21
, 0,
2√
21
,− 1√
21
〉
,
〈
22√
2562
,
21√
2562
,− 31√
2562
,
26√
2562
〉
,
〈
− 1√
2501
,
24√
2501
,− 18√
2501
,−40√
2501
〉 }
.
17. We have u1 = 1, u2 = x, and u3 = x2. Taking v1 = u1 = 1 and using
(u2, v1) =
∫ 1
−1
1 · x2 dx = 0 and (v1, v1) =
∫ 1
−1
x · x dx = 2
we obtain
v2 = u2 − (u2, v1)(v1, v1) v1 = x−
0
2
x = x.
Next, using
(u3, v1) =
∫ 1
−1
x2 · 1 dx = 2
3
, (u3, v2) =
∫ 1
−1
x2 · x dx = 0, and (v2, v2) =
∫ 1
−1
x · x dx = 2
3
,
we obtain
v3 = u3 − (u3, v1)(v1, v1) v1 −
(u3, v2)
(v2, v2)
v2 = x2 − 2/32 1−
0
2/3
x = x2 − 1
3
.
Thus, an orthogonal basis is B′ =
{
1, x, x2 − 13
}
.
18. We have u1 = x2 − x, u2 = x2 + 1, and u3 = 1− x2. Taking v1 = u1 = x2 − x and using
(u2, v1) =
∫ 1
−1
(x2 + 1)(x2 − x)dx = 16
15
and (v1, v1) =
∫ 1
−1
(x2 − x)(x2 − x)dx = 16
15
we obtain
v2 = u2 − (u2, v1)(v1, v1) v1 = x
2 + 1− 16/15
16/15
(x2 − x) = x + 1.
Next, using
(u3, v1) =
∫ 1
−1
(1− x2)(x2 − x)dx = 4
15
, (u3, v2) =
∫ 1
−1
(1− x2)(x + 1)dx = 4
3
,
and
(v2, v2) =
∫ 1
−1
(x + 1)(x + 1)dx =
8
3
,
we obtain
v3 = u3 − (u3, v1)(v1, v1) v1 −
(u3, v2)
(v2, v2)
v2 = 1− x2 − 4/1516/15(x
2 − x)− 4/3
8/3
(x + 1) = −5
4
x3 − 1
4
x +
1
2
.
Thus, an orthogonal basis is B′ =
{
x2 − x, x + 1,− 54 x3 − 14 x + 12
}
.
19. Using the solution of Problem 17 and computing
||v1||2 = (v1, v1) =
∫ 1
−1
1 · 1 dx = 2, ||v2||2 = (v2, v2) =
∫ 1
−1
x · x dx = 2
3
,
and
||v3||2 = (v3, v3) =
∫ 1
−1
(
x2 − 1
3
) (
x2 − 1
3
)
dx =
8
45
,
368
7.7 Gram-Schmidt Orthogonalization Process
we see that an orthonormal basis is
B′′ =
{
1√
2
,
x√
2/3
,
x2 − 1/3√
8/45
}
=
{
1√
2
,
3√
6
x,
15
2
√
10
(
x2 − 1
3
)}
.
20. Using the solution of Problem 18 and computing
||v1||2 = (v1, v1) =
∫ 1
−1
(x2 − x)(x2 − x)dx = 16
15
, ||v2||2 = (v2, v2) =
∫ 1
−1
(x + 1)(x + 1)dx =
8
3
,
and
||v3||2 = (v3, v3) =
∫ 1
−1
(
−5
4
x3 − 1
4
x +
1
2
) (
−5
4
x3 − 1
4
x +
1
2
)
dx =
1
3
,
we see that an orthonormal basis is
B′′ =
{√
15
4
(x2 − x), 3
2
√
6
(x + 1),
√
3
4
(−5x2 − x + 2)
}
.
21. Using w1 = 1/
√
2, w2 = 3x/
√
6, and w3 = (15/2
√
10)(x2 − 1/3), and computing
(p, w1) =
∫ 1
−1
(9x2 − 6x + 5) 1√
2
dx = 8
√
2,
(p, w2) =
∫ 1
−1
(9x2 − 6x + 5) 3√
6
x dx = −2
√
6
(p, w3) =
∫ 1
−1
(9x2 − 6x + 5)
[
15
2
√
10
(
x2 − 1
3
)]
dx =
12√
10
,
we find from Theorem 7.5
p(x) = 9x2 − 6x + 5 = (p, w1)w1 + (p, w2)w2 + (p, w3)w3 = 8
√
2w1 − 2
√
6w2 +
12√
10
w3.
22. Using w1 = (
√
15/4)(x2 − x), w2 = (3/2
√
6)(x + 1), and w3 = −(
√
3/4)(5x2 + x− 2), and computing
(p, w1) =
∫ 1
−1
(9x2 − 6x + 5)
[√
15
4
(
x2 − x)] dx = 41√
15
,
(p, w2) =
∫ 1
−1
(9x2 − 6x + 5)
[
3
2
√
6
(x + 1)
]
dx = 3
√
6
(p, w3) =
∫ 1
−1
(9x2 − 6x + 5)
[
−
√
3
4
(5x2 + x− 2)
]
dx =
1√
3
,
we find from Theorem 7.5
p(x) = 9x2 − 6x + 5 = (p, w1)w1 + (p, w2)w2 + (p, w3)w3 = 41√
15
w1 + 3
√
6w2 +
1√
3
w3.
23. Since u3 depends on u1 and u2 we would expect the Gram-Schmidt process to yield a pair of orthogonal vectors
v1 and v2, with a third vector v3 that is 0. This is because u3 lies in the subspace W2 of R3 spanned by u1
and u2, and hence the projection of u3 onto W2 is u3 itself. In other words,
u3 = projW3u3 =
u3 · v1
v1 · v1 v1 +
u3 · v2
v2 · v2 v2 so v3 = u3 −
u3 · v1
v1 · v1 v1 +
u3 · v2
v2 · v2 v2 = 0.
To carry out the orthogonalization process we take v1 = u1 = 〈1, 1, 3〉. Then, using u2 ·v1 = 8 and v1 ·v1 = 11
we obtain
v2 = u2 − u2 · v1v1 · v1 v1 = 〈1, 4, 1〉 −
8
11
〈1, 1, 3〉 =
〈
3
11
,
36
11
,−13
11
〉
.
369
7.7 Gram-Schmidt Orthogonalization Process
Next, using u3 · v1 = 2,u3 · v2 = 40211 , and v2 · v2 = 13411 , we obtain
v3 = u3 − u3 · v1v1 · v1 v1 −
u3 · v2
v2 · v2 v2 = 〈1, 10,−3〉 −
2
11
〈1, 1, 3〉 − 402/11
134/11
〈
3
11
,
36
11
,−13
11
〉
= 〈0, 0, 0〉 .
In this case {v1,v2} = {〈1, 1, 3}, 〈 311 , 3611 ,− 1311 〉} is an orthogonal subset of R3 containing the third vector
u3 = 〈1, 10,−3〉.
CHAPTER 7 REVIEW EXERCISES
1. True
2. False; the points must be non-collinear.
3. False; since a normal to the plane is 〈2, 3,−4〉 which is not a multiple of the direction vector 〈5,−2, 1〉 of the
line.
4. True 5. True 6. True 7. True 8. True 9. True
10. True; since a× b and c× d are both normal to the plane and hence parallel (unless a× b = 0 or c× d = 0.)
11. 9i + 2j + 2k 12. orthogonal
13. −5(k× j) = −5(−i) = 5i 14. i · (i× j) = i× k = 0
15.
√
(−12)2 + 42 + 62 = 14
16. (−1− 20)i− (−2− 0)j + (8− 0)k = −21i + 2j + 8k
17. −6i + j− 7k
18. The coordinates of (1,−2,−10) satisfy the given equation.
19. Writing the line in parametric form, we have x = 1+ t, y = −2+3t, z = −1+2t. Substituting into the equation
of the plane yields (1+ t)+2(−2+3t)− (−1+2t) = 13 or t = 3. Thus, the point of intersection is x = 1+3 = 4,
y = −2 + 3(3) = 7, z = −1 + 2(3) = 5, or (4, 7, 5).
20. |a| = √42 + 32 + (−5)2 = 5√2 ; u = − 1
5
√
2
(4i + 3j− 5k) = − 4
5
√
2
i− 3
5
√
2
j +
1√
2
k
21. x2 − 2 = 3, x2 = 5; y2 − 1 = 5, y2 = 6; z2 − 7 = −4, z2 = 3; P2 = (5, 6, 3)
22. (5, 1/2, 5/2)
23. (7.2)(10) cos 135◦ = −36√2
24. 2b = 〈−2, 4, 2〉; 4c = 〈0,−8, 8〉; a · (2b + 4c) = 〈3, 1, 0〉 · 〈−2,−4, 10〉 = −10
25. 12, −8, 6
26. cos θ =
a · b
|a||b| =
1√
2
√
2
=
1
2
; θ = 60◦
27. A =
1
2
|5i− 4j− 7k| = 3
√
10
2
370
CHAPTER 7 REVIEW EXERCISES
28. From 3(x− 3) + 0(y − 6) + (1)(z − (−2)) = 0 we obtain 3x + z = 7.
29. | − 5− (−3)| = 2
30. parallel: −2c = 5, c = −5/2; orthogonal: 1(−2) + 3(−6) + c(5) = 0, c = 4
31. a × b =
∣∣∣∣∣∣∣
i j k
1 1 0
1 −2 1
∣∣∣∣∣∣∣ =
∣∣∣∣ 1 0−2 1
∣∣∣∣ i − ∣∣∣∣ 1 01 1
∣∣∣∣ j + ∣∣∣∣ 1 11 −2
∣∣∣∣k = i − j − 3k A unit vector perpendicular to both a
and b is
a× b
‖a× b‖ =
1√
1 + 1 + 9
(i− j− 3k) = 1√
11
i− 1√
11
j− 3√
11
k.
32. ‖a‖ =
√
1/4 + 1/4 + 1/6 =
3
4
; cosα =
1/2
3/4
=
2
3
, α ≈ 48.19◦; cosβ = 1/2
3/4
=
2
3
, β ≈ 48.19◦;
cos γ =
−1/4
3/4
= −1
3
, γ ≈ 109.47◦
33. compba = a · b/‖b‖ = 〈1, 2,−2〉 · 〈4, 3, 0〉/5 = 2
34. compab = b · a/‖a‖ = 〈4, 3, 0〉 · 〈1, 2,−2〉/3 = 10/3
projab = (compab)a/‖a‖ = (10/3)〈1, 2,−2〉/3 = 〈10/9, 20/9,−20/9〉
35. a + b = 〈1, 2,−2〉+ 〈4, 3, 0〉 = 〈5, 5,−2〉
compa(a + b) = (a + b) · a/
√
1 + 4 + 4 = 13 (a · a + b · a) = 13 [(1 + 4 + 4) + (4 + 6 + 0)] = 193
proja(a + b) = [compa(a + b)](a/‖a‖) = 193 〈 13 , 23 ,− 23 〉 = 〈 199 , 389 ,− 389 〉
36. a− b = 〈1, 2,−2〉 − 〈4, 3, 0〉 = 〈−3,−1,−2〉
compb(a− b) = (a− b) · b/
√
16 + 9 = 15 (a · b− b · b) = 15 [(4 + 6 + 0)− (16 + 9)] = −3
projb(a− b) = [compb(a− b)](b/‖b‖) = −3〈 45 , 35 , 0〉 = 〈− 125 ,− 95 , 0〉
37. Let a = 〈a, b, c〉 and r = 〈x, y, z〉. Then
(a) (r− a) · r = 〈x− a, y − b, z − c〉 · 〈x, y, z〉 = x2 − ax + y2 − by + z2 − zc = 0 implies
(x− a
2
)2 + (y − b
2
)2 + (z − c
2
)2 =
a2 + b2 + c2
4
. The surface is a sphere.
(b) (r− a) · a = 〈x− a, y − b, z − c〉 · 〈a, b, c〉 = a(x− a) + b(y − b) + c(z − c) = 0
The surface is a plane.
38. 〈4, 2,−2〉 − 〈2, 4,−3〉 = 〈2,−2, 1〉; 〈2, 4,−3〉 − 〈6, 7,−5〉 = 〈−4,−3, 2〉; 〈2,−2, 1〉 · 〈−4,−3, 2〉 = 0
The points are the vertices of a right triangle.
39. A direction vector of the given line is 〈4,−2, 6〉. A parallel line containing (7, 3,−5) is (x−7)/4 = (y−3)/(−2) =
(z + 5)/6.
40. A normal to the plane is 〈8, 3,−4〉. The line with this direction vector and through (5,−9, 3) is x = 5 + 8t,
y = −9 + 3t, z = 3− 4t.
41. The direction vectors are 〈−2, 3, 1〉 and 〈2, 1, 1〉. Since 〈−2, 3, 1〉 · 〈2, 1, 1〉 = 0, the lines are orthogonal. Solving
1 − 2t = x = 1 + 2s, 3t = y = −4 + s, we obtain t = −1 and s = 1. The point (3,−3,0) obtained by letting
t = −1 and s = 1 is common to the two lines, so they do intersect.
42. Vectors in the plane are 〈2, 3, 1〉 and 〈1, 0, 2〉. A normal vector is 〈2, 3, 1〉×〈1, 0, 2〉 = 〈6,−3,−3〉 = 3〈2,−1,−1〉.
An equation of the plane is 2x− y − z = 0
43. The lines are parallel with direction vector 〈1, 4,−2〉. Since (0, 0, 0) is on the first line and (1, 1, 3) is on the second
line, the vector 〈1, 1, 3〉 is in the plane. A normal vector to the plane is thus 〈1, 4,−2〉 × 〈1, 1, 3〉 = 〈14,−5,−3〉.
An equation of the plane is 14x− 5y − 3z = 0.
371
CHAPTER 7 REVIEW EXERCISES
44. Letting z = t in the equations of the plane and solving −x + y = 4 + 8t, 3x − y = −2t, we obtain x = 2 + 3t,
y = 6 + 11t, z = t. Thus, a normal to the plane is 〈3, 11, 1〉 and an equation of the plane is
3(x− 1) + 11(y − 7) + (z + 1) = 0 or 3x + 11y + z = 79.
45. F = 10
a
‖a‖ =
10√
2
(i + j) = 5
√
2 i + 5
√
2 j; d = 〈7, 4, 0〉 − 〈4, 1, 0〉 = 3i + 3j
W = F · d = 15√2 + 15√2 = 30√2 N-m
46. F = 5
√
2 i + 5
√
2 j + 50i = (5
√
2 + 50)i + 5
√
2 j; d = 3i + 3j
W = 15
√
2 + 150 + 15
√
2 = 30
√
2 + 150 N-m ≈ 192.4 N-m
47. Since F2 = 200(i + j)/
√
2 = 100
√
2 i + 100
√
2 j, F3 = F2 − F1 = (100
√
2− 200)i + 100√2 j and
‖F3‖ =
√
(100
√
2− 200)2 + (100√2)2 = 200
√
2−√2 ≈ 153 lb.
48. Let ‖F1‖ = F1 and ‖F2‖ = F2. Then F1 = F1[(cos 45◦)i + (sin 45◦)j] and F2 = F2[(cos 120◦)i + (sin 120◦)j], or
F1 = F1( 1√2 i +
1√
2
j) and F2 = F2(− 12 i +
√
3
2 j). Since w + F1 + F2 = 0,
F1(
1√
2
i +
1√
2
j) + F2(−12 i +
√
3
2
j) = 50j, (
1√
2
F1 − 12F2)i + (
1√
2
F1 +
√
3
2
F2)j = 50j
and
1√
2
F1 − 12F2 = 0,
1√
2
F1 +
√
3
2
F2 = 50.
Solving, we obtain F1 = 25(
√
6−√2 ) ≈ 25.9 lb and F2 = 50(
√
3− 1) ≈ 36.6 lb.
49. Not a vector space. Axiom (viii) is not satisfied.
50. The vectors are linearly independent. The only solution of the system
c1 = 0, c1 + 2c2 + c3 = 0, 2c1 + 3c2 − c3 = 0
is c1 = 0, c2 = 0, c3 = 0.
51. Let p1 and p2 be in Pn such that
d2p1
dx2
= 0 and
d2p2
dx2
= 0. Since
0 =
d2p1
dx2
+
d2p2
dx2
=
d2
dx2
(p1 + p2) and 0 = k
d2p1
dx2
=
d2
dx2
(kp1)
we conclude that the set of polynomials with the given property is a subspace of Pn. A basis for the subspace
is 1, x.
52. The intersection W1 ∩W2 is a subspace of V . If x and y are in W1 ∩W2 then x and y are in each subspace
and so x + y is in each subspace. That is, x + y is in W1 ∩W2. Similarly, if x is in W1 ∩W2 then x is in each
subspace and so kx is in each subspace. That is, kx is in W1 ∩W2 for any scalar k.
The union W1∪W2 is generally not a subspace. For example, W1 = {〈x, y〉
∣∣ y = x} and W2 = {〈x, y〉 ∣∣ y = 2x}
are subspaces of R2. Now 〈1, 1〉 is in W1 and 〈1, 2〉 is in W2 but 〈1, 1〉+ 〈1, 2〉 = 〈2, 3〉 is not in W1 ∪W2.
372