Homework 2 Solutions
Samir Canning, Src2165
June 4, 2015
Part A
Problem 1
(1)
a · b = (5 · 3) + (−2 · 4) = 7. Because two vectors are orthogonal if and only if
their dot product is zero, a and b are not orthogonal.
(2)
a · b = (6 · 2) + (−2 · 5) + (3 · −1) = −1. Not orthogonal.
(3)
a · b = (p · 2q) + (−p · q) + (2p · −q) = −pq. Not orthogonal.
(4)
a · b = (3 · 4) + (2 · 0) + (−1 · 5) = 7. Not orthogonal.
(5)
√
a · b = |a||b|cos(θ) = (80)(50)cos( 3π
4 ) = −2000 2. Not orthogonal.
Problem 2
(1)
To find the angle, we must find the magnitude of each vector and the dot product
a·b
√ . Therefore
. Hence cos(θ) = 1350
of the two vectors because cos(θ) = |a||b|
29
50
√
θ =arccos( 13 29 )
(2)
θ =arccos( √305√26 )
1
(3)
2
θ =arccos( 9√420 ) =arccos( 9√
)
5
Problem 3
(1)
Let α, β, and γ denote the x, y, and z direction angles respectively. Then
α =arccos( 67 ), β =arccos( 37 ) and γ =arccos( −2
7 ).
(2)
α =arccos( 13 ), β =arccos( 23 ) and γ =arccos( 23 ).
Problem 4
(1)
The general formula is proja b =
a·b
|a|2 a
Thus proja b =
14
17
< 1, 4 >
(2)
proja b =
8
81
< −1, 4, 8 >
(3)
proja b =
1
7
< 1, 2, 3 >
Problem 5
−−→
The formula for work gives us W = F · P Q = 8(6) − 6(2) + 9(12) = 144 Joules.
2