Solução_Calculo_Stewart_6e

F.
112 ¤ CHAPTER 13 VECTORSANDTHEGEOMETRYOFSPACE ET TX.10 CHAPTER 12
17. |a| = 3 2 +(−1) 2 +5 2 = √ 35, |b| = (−2) 2 +4 2 +3 2 = √ 29,anda · b =(3)(−2) + (−1)(4) + (5)(3) = 5. Then
cos θ = a · b
|a||b| = 5
√ √ = 5
√ and the angle between a and b is θ =cos −1 √
5
35 · 29 1015
1015
≈ 81 ◦ .
19. |a| = √ 0 2 +1 2 +1 2 = √ 2, |b| = 1 2 +2 2 +(−3) 2 = √ 14, anda · b = (0)(1) + (1)(2) + (1)(−3) = −1.
Then cos θ = a · b
|a||b| = −1
√ √ =
−1
2 · 14 2 √ 7 and θ =cos−1 − 1
2 √ ≈ 101 ◦ .
7
21. Let a, b,andc be the angles at vertices A, B,andC respectively.
Then a is the angle between vectors −→
AB and −→
AC, b is the angle
between vectors −→
BA and −→
BC,andc is the angle between vectors
−→
CA and −→
CB.
−→
AB · −→
AC
Thus cos a =
−→
AB
−→
h2, 6i·h−2, 4i
= √ AC
22 +6 2 (−2) 2 +4 = 1
√ √ (−4+24)= 20
√
2
√ =
2 40 20 800 2 and
√ 2
a =cos −1 =45 ◦ . Similarly, cos b =
2
−→ −→
BA · BC
−→
BA
−→ = BC
h−2, −6i·h−4, −2i
√ 4+36
√ 16 + 4
=
1
√ √ (8 + 12) = 20
√
2
√ =
40 20 800 2
√ 2
so b =cos −1 =45 ◦ and c =180 ◦ − (45 ◦ + 45 ◦ )=90 ◦ .
2
−→
2 −→
2 −→
2
BC − AB − AC
Alternate solution: Apply the Law of Cosines three times as follows: cos a =
2 −→
AB
−→
,
AC
−→
AC 2 − −→
AB 2 − −→
2
BC −→
AB 2 − −→
AC 2 − −→
2
BC
cos b =
2 −→
AB
−→ ,andcos c =
BC 2 −→
AC
−→ . BC
23. (a) a · b =(−5)(6) + (3)(−8) + (7)(2) = −40 6= 0,soa and b are not orthogonal. Also, since a is not a scalar multiple
25.
of b, a and b are not parallel.
(b) a · b =(4)(−3) + (6)(2) = 0,soa and b are orthogonal (and not parallel).
(c) a · b =(−1)(3) + (2)(4) + (5)(−1) = 0,soa and b are orthogonal (and not parallel).
(d) Because a = − 2 b, a and b are parallel.
3
−→
QP = h−1, −3, 2i, −→
QR = h4, −2, −1i,and −→ −→
−→ −→
QP · QR = −4+6− 2=0. Thus QP and QR are orthogonal, so the angle of
thetriangleatvertexQ is a right angle.
27. Let a = a 1 i + a 2 j + a 3 k be a vector orthogonal to both i + j and i + k. Thena · (i + j) =0 ⇔ a 1 + a 2 =0and
a · (i + k) =0 ⇔ a 1 + a 3 =0,so a 1 = −a 2 = −a 3 .Furthermorea istobeaunitvector,so1=a 2 1 + a 2 2 + a 2 3 =3a 2 1
implies a 1 = ± 1 √
3
.Thusa = 1 √
3
i − 1 √
3
j − 1 √
3
k and a = − 1 √
3
i + 1 √
3
j + 1 √
3
k are two such unit vectors.