Solutions to 9.1 Problems
(Most odd-numbered problems have been omitted here.)
12. PQ = "7 "1, 5 "1 = "8, 4 , RS = 17 "1, " 8 " 0 = 16, " 8 ,
PQ = ("8) 2 + 4 2 = 4 5 ,
RS = 16 2 + ("8) 2 = 8 5 . Because RS = "2PQ , the two vectors are parallel.
!
!
14. (1/2)v +3w = (1/2) 4,!2 + 3 1, " 3 = 5, " 8 .
20. P = (3, 2), Q = (1, 1).
!
uuur
(a) v = PQ = 1! 3, 1! 2 = !2, !1 .
! 2 + (!1)2 = 5 .
(b) v = (!2)
(d)
Diagram for
Problem 55.
!
F1
π/6
(c) –v = 2, 1 .
!2, !1
v
2
1
.
=
= !
,!
v
5
5
5
v
(e) !
=
v
π/4
2 1
.
,
5 5
F2
22. v = 3j, w = –5i, –4v = –12j, 3v – 2w = 9j – (–10i) = 10i + 9j, 4v + 7j = 12j + 7j = 19j.
27. F = i – 2j + 2i + j = 3i – j.
F =
(–4)2 +12 2 = 4 10 .
F = 9 +1 = 10 .
30. u =
2
" 1 %2 " 7 %
1 7
36. v = $ ! ' + $ !
+ = 2
' =
# 2& # 2 &
4 4
38. 4(i – 2j) + 5(i + 2j) = 9i + 2j =
28. F = F1 + F2 = (i – j) + (–5i + 13j) = –4i + 12j;
cos(3! / 4), sin(3! / 4) = "
1 1
,
.
2 2
" 1
7 %
( v = 2 $!
i!
j'
# 2 2 2 2 &
9, 2 .
48. v = 4, 7 + !1, 8 = 3, 15
" a = –3 (and b = –15).
55. See diagram above right. We have F1 = 100 cos(2! / 3), sin(2! / 3) = 100 "1 / 2, 3 / 2 =
!50, 50 3 . Let a = F2 . Then F2 = a cos(! / 4), sin(! / 4) = a / 2, a / 2
and F1 + F2 =
!50 + a / 2, 50 3 + a / 2 . Since !50 + a / 2 = 0 if the wagon remains on the y-axis, then
a = 50 2 and so F2 = 50, 50 . Therefore F1 + F2 = !50, 50 3 + 50, 50 = 0, 50(1+ 3)
!
and the magnitude of the resultant force is F1 + F2 = 50(1+ 3) .
56. F1 = 0, 100 , F2 = 120 3 / 5, 4 / 5 = 72, 96 . Let F3 = v1, v2 . For the point mass to remain at rest,
F1 + F2 + F3 = 0, i.e.
72 + v1, 196 + v2 = 0, 0
!
F3 = !72, !196 .
58. Let r be the vector representing the required heading and speed. See diagram.
w = 0, ! 3
represents the current, v = 6, 0
the actual direction and speed with
respect to land. (To cross the river in 10 minutes requires a speed of 6 mph.)
Since r + w = v, we have r = v – w = 6, 3 . Therefore the water speed of the
boat should be r = 36 + 9 = 45 ! 6.7 mph. The heading should be
tan-1(3/6) ≈ 26.6º from a line straight across the river.
r
v
w