Thomas Calculus 13th [Solutions]

886 Chapter 12 Vectors and the Geometry of Space
54. Let V1 , V2 , V3
, , V n be the vertices of a regular n-sided polygon and v i denote the vector from the center to
n
i(2 )
V i for i 1, 2, 3, , n . If S vi
and the polygon is rotated through an angle of where
n
i 1
i 1, 2, 3, , n , then S would remain the same. Since the vector S does not change with these rotations we
conclude that S 0.
55. Without loss of generality we can coordinatize the vertices of the triangle such that A(0, 0), B( b , 0) and
C( xc
, yc
) a is located at
b xc
yc
, , b is at
2 2
xc
yc
,
2 2
and c is at
xc
yc
Bb b i j , and Cc b x .
2 2
2 c i yc
j Aa Bb Cc 0
b
2 , 0 . Therefore,
xc
yc
Aa b i j,
2 2 2
56. Let u be any unit vector in the plane. If u is positioned so that its initial point is at the origin and terminal point
is at ( x, y ), then u makes an angle with i, measured in the counter-clockwise direction. Since u 1, we
have that x cos and y sin . Thus u cos i sin j . Since u was assumed to be any unit vector in the
plane, this holds for every unit vector in the plane.
12.3 THE DOT PRODUCT
NOTE:
In Exercises 1-8 below we calculate proj v u as the vector
number in column 5 divided by the number in column 2.
u cos
v
v , so the scalar multiplier of v is the
v u v u cos u cos
proj v u
1. 25 5 5 1 5 2i 4j 5k
2. 3 1 13
3. 25 15 5
4. 13 15 3
5. 2 34 3
6. 3 2 2 3
3
13 3
1
5
3
3
13
13
45
15
2
2
3 34
34
3 2
3 2
3 2
2
3 3 i 4 k
5 5
1 10 11 2
9 i j k
13
225
1
2 i 10 j 11 k
17 5j
3k
3 2
2
i
j
7. 10 17 26 21
10 17
546
10 17
26
10 17
26
5i
j
8.
1
6
30
6
30
6
1
5
1
30
1 1 , 1
5 2 3
9.
cos 1 u v 1 (2)(1) (1)(2) (0)( 1)
1 4 1 4
u v
cos cos cos 0.75
2 2 2 2 2 2
2 1 0 1 2 ( 1)
5 6 30
rad
10.
cos 1 u v 1 (2)(3) ( 2)(0) (1)(4)
1 10 1 2
u v
cos cos cos 2 2 2 2 2 2
2 ( 2) 1 3 0 4
9 25
3
0.84 rad
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2014 Pearson Education, Inc.