Thomas Calculus 13th [Solutions]

Section 12.3 The Dot Product 889
28. No, v 1 need not equal v 2 . For example, i j i 2 j but i i j i i i j 1 0 1 and
i i 2j i i 2i j 1 2 0 1.
2
proj u u v v u u v v u v v u u v v u v v u v v u v ( u v) u v ( v v)
v v v v v v v v
29. v 2 2 2 2 2 2 2 2
2 2
( u v) ( u v)
v
2 4
v
2
v
0
30. F 2i j 3k and v = 3 i j proj F v 5 3 3 1
v F v i j i j
2 2
2 2
, is the vector parallel to v.
v 10
F proj 3 1 1 3
vF 2i j 3k i j i j 3k is the vector orthogonal to v.
2 2 2 2
31. P ( x 1, y 1) P x 1, c a
b b
x 1 and Q ( x2 , y2 ) Q x2 , c a
b b
x 2 and any two points P and Q on the line with
b 0 PQ x 2 x 1 i a a
b
x 1 x 2 j PQ v x 2 x 1 i
b
x 1 x 2 j a i b j
a x2 x a
1 b x1 x2 0 v is perpendicular to PQ for b 0. If b 0, then v ai is
b
perpendicular to the vertical line ax c . Alternatively, the slope of v is b and the slope of the line ax by c
a
is a , so the slopes are negative reciprocals the vector v and the line are perpendicular.
b
32. The slope of v is b and the slope of bx ay c is b , provided that a 0. If a 0, then v bj is parallel to
a a
the vertical line bx c . In either case, the vector v is parallel to the line bx ay c.
33. v i 2j is perpendicular to the line x 2 y c;
P (2,1) on the line 2 2 c x 2y
4
34. v 2i j is perpendicular to the line 2 x y c;
P ( 1, 2) on the line ( 2)( 1) 2 c
2x
y 0
Copyright
2014 Pearson Education, Inc.