Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 12.3 THE DOT PRODUCT 0 121
21. Let a = a 1 i + a 2 j + a 3 k be a vector orthogonal to both i + j and i + k, Then a · (i + j ) = 0 a1 + a2 = 0 and
a . (i + k ) = 0 a 1 + a 3 = 0, so a 1 = - a2 ~ -a3. Furthermore a is to be a unit vector, so 1 = ai + a~+ a§ = 3ai
. . 1 h l . 1. l k d 1 '+ 1 ' + 1 k tw . h .
1mphes a 1 = ± 73
. T us a= 73
1- 7aJ- 73 .an a = - 73 1 7aJ 73 are o sue umt vectors.
29. The line 2x - y = 3 y ·= 2x - 3 has slope 2, so a vector parallel to the line is a = (1, 2}. The line 3x + y = 7
y = - 3x + 7 has slope - 3, so a vector parallel to the line is b = (1, - 3). The angle between the lines is the same as the
angle 8 between the vectors. Here we have a · b = (1)(1) + (2)( - 3) = - 5, lal = v'P + 2 2 = v'5, and
a · b -5 - 5 1 v'2
[hi = J1 2 + ( -3)2 = v'lQ, so cos 8 = I alibi = y'5. v'1Q = 5
J2 =- J2 or ""'2· Thus 8 = 135°, and the
acute angle between the lines is 180° - 135° = 45°.
31. The curves y = x 2 andy= x 3 meet when x 2 = x 3 x 3 - x 2 = 0 x 2 (x - 1) = 0 x = 0, x = 1. We have
i:-x 2 = 2x and .!£x 3 = 3x 2 , so the tangent lines of both curv~s have slope 0 at x = 0. Thus the angle between the curves is
dx
dx
0° at th~ point (0, 0). For x = 1, ! x 2 1:c=l = 2 and d~ xt:=l = 3 so the tangent lines at the point (1, 1) have slopes 2 and
3. Vectors parallel to the tangent lines are (1, 2} and (1, 3}, and the angle (} between them is given by
cos(} = (1, 2} . (1, 3} = 1 + 6 = _ 7_
1(1, 2} 1 1(1,3}1 v'5v'W 5v'2
~3 . Since 1(2, 1, 2)1 = v'4 + 1 + 4 = V9 = 3, using Equations 8 and 9 we have cos a=~ . cos/3 =~. and cos-y = ~· The
direction angles are given by a = cos- 1 ( i) ~ 48°, /3 = cos- 1 (~) ~ 71°, and 1 = cos- 1 (~) = 48°.
35. Since I i- 2j-3kl = v'1 + 4 + 9 = v'l4, Equations 8 and 9 give cos a = 7IT• cos/3 =*' and cos-y = 7&, while
a = cos - 1 (~ ) ~74° , {3 = cos - 1 (-~) ~ 122 °, and /=cos - 1 (- vih) ~ 143°.
37. l(c, c, c) I = v'c 2 + c 2 + c 2 = v'3c [since c > 0], so .cos a = cos/3 = cos:-y = ~ = ~ and
v3c v3
r;an a · b -5 · 4 + 12 · 6
39. Ia I = J( -5)2 + 12 2 = v w:.r = 13. The scalar projection of b onto a is comp 11
b = -
1
-
1
= = 4 and the
a 13
vector projection ofb onto a is, proj 11 b = ( j~~)
: = 4 · f3 (-5, 12} = (-~, - ~ ).
1 1
41 . la l = v'9 + 36 + 4 = 7 so the scalar projection ofb onto a is compab = j~~ = ~ (3 + 12..:. 6) = l The vector
. . f b . . . b 9 a 9 l (3 6 2} o (3 6 2} ( 27 54 18 )
proJeCtiOn o onto a IS, prOJa = 7j;f = 7 · 7 , , - = 4o , , - = 49 • 49• -49 ·
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