Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 13.2 VECTORS ET SECTION 12.2 ¤ 109
15. h0, 1, 2i + h0, 0, −3i = h0+0, 1+0, 2+(−3)i
= h0, 1, −1i
17. a + b = h5+(−3) , −12 + (−6)i = h2, −18i
2a +3b = h10, −24i + h−9, −18i = h1, −42i
|a| = 5 2 +(−12) 2 = √ 169 = 13
|a − b| = |h5 − (−3), −12 − (−6)i| = |h8, −6i| = 8 2 +(−6) 2 = √ 100 = 10
19. a + b =(i +2j − 3 k)+(−2 i − j +5k) =− i + j +2k
2a +3b =2(i +2j − 3 k)+3(−2 i − j +5k) =2i +4j − 6 k − 6 i − 3 j +15k = − 4 i + j +9k
|a| = 1 2 +2 2 +(−3) 2 = √ 14
|a − b| = |(i +2j − 3 k) − (−2 i − j +5k)| = |3 i +3j − 8 k| = 3 2 +3 2 +(−8) 2 = √ 82
21. |−3 i +7j| = (−3) 2 +7 2 = √ 58,sou = 1 √
58
(−3 i +7j) =− 3 √
58
i + 7 √
58
j.
23. The vector 8 i − j +4k has length |8 i − j +4k| = 8 2 +(−1) 2 +4 2 = √ 81 = 9, so by Equation 4 the unit vector with
the same direction is 1 9 (8 i − j +4k) = 8 9 i − 1 9 j + 4 9 k.
25. From the figure, we see that the x-component of v is
v 1 = |v| cos(π/3) = 4 · 1 =2and the y-component is
2
v 2 = |v| sin(π/3) = 4 · √3
v = hv 1 ,v 2 i = 2, 2 √ 3 .
2 =2√ 3. Thus
27. The velocity vector v makes an angle of 40 ◦ with the horizontal and
has magnitude equal to the speed at which the football was thrown.
From the figure, we see that the horizontal component of v is
|v| cos 40 ◦ =60cos40 ◦ ≈ 45.96 ft/s and the vertical component is
|v| sin 40 ◦ =60sin40 ◦ ≈ 38.57 ft/s.
29. The given force vectors can be expressed in terms of their horizontal and vertical components as −300 i and
200 cos 60 ◦ i +200sin60 ◦ j = 200 √3
1
2 i +200 j = 100 i + 100 √ 2
3 j. TheresultantforceF is the sum of these two
vectors: F =(−300 + 100) i + 0 + 100 √ 3 j = −200 i + 100 √ 3 j. Thenwehave