Solução_Calculo_Stewart_6e

F.
116 ¤ CHAPTER 13 VECTORSANDTHEGEOMETRYOFSPACE ET TX.10 CHAPTER 12
13. (a) Since b × c is a vector, the dot product a · (b × c) is meaningful and is a scalar.
(b) b · c is a scalar, so a × (b · c) is meaningless, as the cross product is defined only for two vectors.
(c) Since b × c is a vector, the cross product a × (b × c) is meaningful and results in another vector.
(d) a · b is a scalar, so the cross product (a · b) × c is meaningless.
(e) Since (a · b) and (c · d) are both scalars, the cross product (a · b) × (c · d) is meaningless.
(f ) a × b and c × d are both vectors, so the dot product (a × b) · (c × d) is meaningful and is a scalar.
15. If we sketch u and v starting from the same initial point, we see that
the angle between them is 30 ◦ .UsingTheorem6,wehave
|u × v| = |u||v| sin 30 ◦ =(6)(8) 1
2
=24.
By the right-hand rule, u × v is directed into the page.
i j k
2 1
17. a × b =
1 2 1
=
0 1 3
1 3 i − 1 1
0 3 j + 1 2
k =(6− 1) i − (3 − 0) j +(1− 0) k =5i − 3 j + k
0 1 i j k
1 3
b × a =
0 1 3
=
1 2 1
2 1 i − 0 3
1 1 j + 0 1
k =(1− 6) i − (0 − 3) j +(0− 1) k = −5 i +3j − k
1 2 Notice a × b = −b × a here, as we know is always true by Theorem 8.
19. We know that the cross product of two vectors is orthogonal to both. So we calculate
i j k
−1 1
h1, −1, 1i×h0, 4, 4i =
1 −1 1
=
0 4 4
4 4 i − 1 1
0 4 j + 1 −1
k = −8 i − 4 j +4k.
0 4 h−8, −4, 4i
So two unit vectors orthogonal to both are ± √ = ± 64 + 16 + 16
and
√
2
6
,
√
1
6
, − √ 1
6
.
21. Let a = ha 1 ,a 2 ,a 3 i.Then
i j k 0 0
0 × a =
0 0 0 =
i −
a 2 a 3
a 1 a 2 a 3
0 0 j +
a 1 a 3
0 0 k = 0,
a 1 a 2
i j k
a 2 a 3
a × 0 =
a 1 a 2 a 3
=
0 0 0
0 0 i − a 1 a 3
0 0 j + a 1 a 2
0 0 k = 0.
23. a × b = ha 2b 3 − a 3b 2,a 3b 1 − a 1b 3,a 1b 2 − a 2b 1i
= h(−1)(b 2a 3 − b 3a 2) , (−1)(b 3a 1 − b 1a 3) , (−1)(b 1a 2 − b 2a 1)i
= − hb 2 a 3 − b 3 a 2 ,b 3 a 1 − b 1 a 3 ,b 1 a 2 − b 2 a 1 i = −b × a
h−8, −4, 4i
4 √ ,thatis, − √ 2
6
6
, − √ 1
6
,
√
1
6