MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.02
Fall 2007
Problem Set 1 Solutions
Problem 1: Two Vectors
Given two vectors,
ˆ
ˆ
ˆ
(4
3
5 )
=
−
+
A
i
j
k
G
and
ˆ
ˆ
ˆ
(7
4
4 )
=
+
+
B
i
j
k
G
, evaluate the following:
(a)
;
2
+
A
B
G
G
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
2
2(4i
3j
5k)
(7
4
4 )
15
2
1
+
=
−
+
+
+
+
=
−
+
A
B
i
G
G
4
j
k
i
j
k
(b)
;
3
−
A
B
G
G
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
3
(4i
3j
5k)
3(7
4
4 )
17
15
7
−
=
−
+
−
+
+
= −
−
−
A
B
i
j
k
i
j
k
G
G
(c)
;
⋅
A B
G
G
Since
and
ˆ ˆ
ˆ ˆ
ˆ
ˆ
1
⋅
=
⋅
=
⋅
=
i i
j j
k k
ˆ ˆ
ˆ
ˆ
ˆ
ˆ
0
⋅
=
⋅
=
⋅
=
i j
j k
k i
, the dot product is
( )( )
(
)( )
( )( )
4
7
3
4
5
4
36
⋅
=
+ −
+
=
A B
G
G
(d)
;
×
A
B
G
G
With
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
,
and
,
×
=
×
=
×
=
i
j
k
j
k
i
k
i
j
the cross product
×
A
B
G
G
is given by
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
4
3
5
32
19
37
7
4
4
×
=
−
= −
+
+
i
j
k
A
B
i
j
k
G
G
(e)
What is the angle between
and
A
G
B
G
?
The dot product of
A
and
is
G
B
G
cos
θ
⋅
=
A B
A B
G
G
G
G
where
θ
is the angle between the two
vectors. With:
(
)
2
2
2
(4)
3
(5)
50
5
2
A
=
=
+ −
+
=
=
A
G
2
2
2
(7)
(4)
(4)
81
9
B
=
=
+
+
=
=
B
G
,
and using the result from part (c), we obtain
30
cos
0.6
52.5 .
5
2 9
θ
θ
⋅
=
=
=
⇒
=
⋅
A B
A B
G
°
G
G
G
PS01 Solution - 1

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