Thomas Calculus 13th [Solutions]

Section 12.2 Vectors 885
F 1 sin 40 100sin 35 w . Solving the first equation for F 1 results in:
Substituting this result into the second equation gives us: w 126.093 N.
F 100cos35
1 cos 40
106.933 N.
48. F1 F1 cos , F1
sin 75cos , 75sin , F2 F2 cos , F 2 sin 75cos , 75sin , and
w = 0, 25 . Since F1 F2 0, 25 75cos 75cos , 75sin 75sin 0, 25 150sin 25
9.59 .
49. (a) The tree is located at the tip of the vector OP
5 5 3 5 5 3
5 cos 60 i 5sin 60 j i j P ,
2 2 2 2
(b) The telephone pole is located at the point Q, which is the tip of the vector OP
5 5 3 5 10 2 5 3 10 2 5 10 2 5 3 10 2
i j 10cos315 i 10sin 315 j i j Q ,
2 2 2 2 2 2 2 2
q
50. Let t p q
and q
s
p q
. Choose T on OP 1 so that TQ is
parallel to OP 2 , so that TPQ 1 is similar to OP1 P 2 . Then
OT
OP
1
TQ
OP
2
t OT t OP 1 so that T ( t x1 , t y1, t z 1).
Also,
s TQ s OP2 s x2, y2, z 2 . Letting Q ( x, y, z),
we have that TQ x t x1 , y t y1 , z t z1 s x2, y2,
z2
Thus x t x1 sx2, y t y1 s y2,
z t z1 sz2.
p
(Note that if Q is the midpoint, then 1 and t s 1 so that 1 1 x1 x2 y1 y2 z1 z2
x x
q
2
2 1 x
2 2 , y , z
2 2 2
so that this result agrees with the midpoint formula.)
51. (a) the midpoint of AB is M 5 , 5 , 0 and CM 5 1 i 5 1 j (0 3) k 3 i 3 j 3k
2 2
2 2 2 2
(b) the desired vector is 2 CM 2 3 i 3 j 3k i j 2k
3 3 2 2
(c) the vector whose sum is the vector from the origin to C and the result of part (b) will terminate at the
center of mass the terminal point of i j 3k i j 2k 2i 2j+ k is the point (2, 2, 1), which
is the location of the center of mass
52. The midpoint of AB is M 3 , 0, 5 and 2 CM 2 3 1 i (0 2) j 5 1 k 2 5 i 2j 7 k
2 2 3 3 2 2 3 2 2
5 i 4 j 7 k
3 3 3 . The vector from the origin to the point of intersection of the medians is
5 i 4 j 7 k OC 5 i 4 j 7 k i 2 j k 2 i 2 j 4 k.
3 3 3 3 3 3 3 3 3
53. Without loss of generality we identify the vertices of the quadrilateral such that A(0, 0, 0), B( x b , 0, 0),
C( xc
, y c , 0) and D( xd , yd , z d ) the midpoint of AB is
x b x c y
M BC , c , 0 , the midpoint of CD is
2 2
x b
PQ
M AB 2 , 0, 0 , the midpoint of BC is
x c x d y
, c y d z
M CD
,
d and the midpoint of AD is
2 2 2
x x x b c d
x d y
, d z
,
d
2 2 yc yd zd
M AD
the midpoint of M
2 2 2
ABM CD is , ,
2 4 4
which is the same as the
x x x b c d
2 2 yc yd zd
midpoint of M ADM
BC
, , .
2 4 4
Copyright
2014 Pearson Education, Inc.