Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 13.3 THE DOT PRODUCT ET SECTION 12.3 ¤ 113
29. Since |h3, 4, 5i| = √ 9+16+25= √ 50 = 5 √ 2, using Equations 8 and 9 we have cos α = 3
5 √ , cos β = 4
2 5 √ ,and 2
cos γ = 5
5 √ = √ 1
2 2
. The direction angles are given by α =cos −1 3
5 √ ≈ 65 ◦ , β =cos −1 4
2
5 √ ≈ 56 ◦ ,and
2
γ =cos −1 √
1
2
=45 ◦ .
31. Since |2 i +3j − 6 k| = √ 4+9+36= √ 49 = 7, Equations 8 and 9 give cos α = 2 7 , cos β = 3 7
α =cos −1
2
7 ≈ 73 ◦ , β =cos −1
3
7 ≈ 65 ◦ ,andγ =cos −1
− 6 7 ≈ 149 ◦ .
,andcos γ =
−6
7 ,while
33. |hc, c, ci| = √ c 2 + c 2 + c 2 = √ 3c [since c>0], so cos α =cosβ =cosγ = c √
3c
=
1 √
3
and
α = β = γ =cos −1 √
1
3
≈ 55 ◦ .
35. |a| = 3 2 +(−4) 2 =5. The scalar projection of b onto a is comp a b = a · b
|a|
a · b a
projection of b onto a is proj a b =
|a| |a| =3· 1 h3, −4i = 9
, − 12
5 5 5 .
37. |a| = √ 9+36+4=7so the scalar projection of b onto a is comp ab = a · b
|a|
= 3 · 5+(−4) · 0
5
projection of b onto a is proj ab = 9 a
7 |a| = 9 · 1 h3, 6, −2i = 9 h3, 6, −2i = 27
, 54 , − 18
7 7 49 49 49 49 .
39. |a| = √ 4+1+16= √ 21 so the scalar projection of b onto a is comp a b = a · b
|a|
=3and the vector
= 1 (3 + 12 − 6) = 9 7 7
. The vector
= 0 − 1+2 √
21
= 1 √
21
while the vector
projection of b onto a is proj a b = 1 √
21
a
|a| = 1 √
21
· 2 i − j +4k √
21
= 1 21 (2 i − j +4k) = 2 21 i − 1 21 j + 4 21 k.
41. (orth a b) · a =(b − proj a b) · a = b · a − (proj a b) · a = b · a − a · b
|a| 2
So they are orthogonal by (7).
43. comp a b = a · b
|a|
a · a = b · a − a · b
|a| 2 |a|2 = b · a − a · b =0.
=2 ⇔ a · b =2|a| =2 √ 10. Ifb = hb 1 ,b 2 ,b 3 i,thenweneed3b 1 +0b 2 − 1b 3 =2 √ 10.
One possible solution is obtained by taking b 1 =0, b 2 =0, b 3 = −2 √ 10. In general, b = s, t, 3s − 2 √ 10 , s, t ∈ R.
45. The displacement vector is D =(6− 0) i +(12− 10) j +(20− 8) k =6i +2j +12k so by Equation 12 the work done is
W = F · D =(8i − 6 j +9k) · (6 i +2j +12k) =48− 12 + 108 = 144 joules.
47. Here |D| =80ft, |F| =30lb, and θ =40 ◦ . Thus
W = F · D = |F||D| cos θ = (30)(80) cos 40 ◦ = 2400 cos 40 ◦ ≈ 1839 ft-lb.
49. First note that n = ha, bi is perpendicular to the line, because if Q 1 =(a 1 ,b 1 ) and Q 2 =(a 2 ,b 2 ) lie on the line, then
n · −−−→
Q 1 Q 2 = aa 2 − aa 1 + bb 2 − bb 1 =0,sinceaa 2 + bb 2 = −c = aa 1 + bb 1 from the equation of the line.
Let P 2 =(x 2 ,y 2 ) lie on the line. Then the distance from P 1 to the line is the absolute value of the scalar projection