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

122 0 CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
43. lal = J4 + 1 + 16 = J2I so the scalar projection ofb onto a is compa b = a
1a 21 J2I
'
· bl = 0 -;; 2 = - 1 - while the vector
. . fb . . b 1 a 1 2i-j + 4 k 1 ( • • ) 2 1 4
prOJeCtion 0 Onto a IS prOJa = J2T jaj = J2T · J2T = 2l 2 1 - J + 4 k = 21 i - 2T j + 2T k.
a · b a·b
45. ( ortha b ) · a = (b - proja b ) · a = b · a - (proja b } · a = b · a - --
2
2 a· a = b · a - - - 2
Ia I = b · a - a· b = 0.
Ja l
lal
So they are orthogonal by (7).
a·b
47. compa b = j;f = 2 ~ a · b = 2lal = 2 v'IO. lfb = (b1, b:l, b3) , then we need 3bl + Ob2- 1b3 = 2 v'IO.
One possible solution is obtained by taking b1 = 0, b2 = 0, bs = - 2 v'IO. In general, b = ( s, t, 3s - 2 v'10 ), s, t E llt
49. The displacement vector is D = (6 - 0) i + (12 - 10} j + (20- 8) k = 6 i + 2j + 12 k so, by Equation 12, the work done is
W = F · D = (8 i - 6j + 9 k ) · (6i + 2j + 12k} = 48 - 12 + 108 = 144joules.
51. Here ID I = 80 ft, IF I = 30 lb, and B = 40°. Thus
w =F . D = IF IIDI cos B = (30)(80) cos 40° = 2400 cos 40° ~ 1839 ft-lb.
53. First note that n = (a, b) is perpendicular to the line, because if Q1 = (a1, bt) and Q2 = (a2 , b2) lie on the line, then
-----+
n · Q1 Q2 = aa2 -
aa1 + ~ - bb1 = 0, since aa2 + bb:l = - c = aa1 + bb1 from the equation of the line.
I
Let P2 = (x2, y2) lie on the line. Then the distance from P1 to the line is tJ:te absolute value of the scalar projection
o fp ----+p.
(P-----t
1 2 onto n . compn 1 2 = I I = r:?'7l:? = r::?7'l:?
P. ) In · (x2- x1, Y2 - Yl )l lax2- ax1 + by2 -lnJ1I lax1 + by1 + cl
n v a2 + b2 v a2 + b2
since ax2 + by2 =-c. The required distance is 1( 3 )(- 2 ) + (- 4 )( 3 ) + 5 1 = 1 3 .
. )32 + (- 4)2 5
,I
55. For convenience, consider the unit cube positioned so that its back left corner is at the origin, and its edges lie along the
coordinate axes. The diagonal of the cube that begins at the origin and ends at (1, 1, 1} has vector representation (1, 1, 1).
The angle B between this vector and the vector of the edge which also begins at the ori~i n and runs along the x-liXis [that is,
. . b B (1, 1, 1) · (1, 0, 0) 1
( )]
1,0,0 1sg1ven ycos = I( 1 , 1 , 1 }I I( 1 , 0, 0)I = J3
57. Consider the H- C-H combination consisting of the sole carbon atom and the two hydrogen atoms that are at (1, 0, 0) and
(0, 1, 0) (or any H-C-H combination, for that matter). Vector representations of the line segments emanating from the
carbon atom and extending to these two hydrogen atoms are ( 1 - t, 0 - t, 0 - t ) = ( t, - t, -t) and
(0- t. 1-t. 0- t) ~ (- t, t, - t ). The bond angle, B, is therefore given by
® 2012 Cengage Lcoming. All Righrs Reser\'cd. Me~y no1 be scanned, copied. or duplicated, or posted to lt publicly accessible wcbshc, in whole or in p:.trt.