Problem 1
: Prove that
y
x
a
a
P
1
1
sin
cos
θ
θ
+
=
and
y
x
a
a
Q
2
2
sin
cos
θ
θ
+
=
are unit vectors
in the
xy
plane respectively making angles
θ
1
and
θ
2
with the
x
axis. By means of dot
product, obtain the formula for
(
)
1
2
cos
θ
θ
−
. By similarly formulating
P
and
Q
, obtain the
formula for
(
)
1
2
cos
θ
θ
+
.
Solution:
1
sin
cos
1
2
1
2
2
=
+
=
θ
θ
P
. Similarly,
1
2
=
Q
. So
P
and
Q
are unit vectors.
The dot product gives us
2
1
2
1
sin
sin
cos
cos
θ
θ
θ
θ
+
=
⋅
Q
P
. We also know that
(
1
2
cos
θ
θ
−
⋅
=
⋅
Q
P
Q
P
)
where
(
)
1
2
θ
θ
−
is the angle between vectors
P
and
Q
. So,
(
)
2
1
2
1
1
2
sin
sin
cos
cos
cos
θ
θ
θ
θ
θ
θ
+
=
−
.
Let
(
)
(
)
y
x
y
x
a
a
a
a
P
1
1
1
1
1
sin
cos
sin
cos
θ
θ
θ
θ
−
=
−
+
−
=
and
y
x
a
a
Q
Q
2
2
1
sin
cos
θ
θ
+
=
=
.
Then by the dot product, we have
2
1
2
1
1
1
sin
sin
cos
cos
θ
θ
θ
θ
−
=
⋅
Q
P
and
(
)
(
)
(
)
1
2
1
2
1
1
1
1
cos
cos
θ
θ
θ
θ
+
=
−
−
⋅
=
⋅
Q
P
Q
P
. So,
(
)
2
1
2
1
1
2
sin
sin
cos
cos
cos
θ
θ
θ
θ
θ
θ
−
=
+
.
Problem 2
: Decompose vector
z
y
x
a
a
a
4
5
10
+
−
into vectors parallel and perpendicular to
vector
y
x
a
a
+
3
.
Solution:
A
⊥
= A

A

A
B
A

= (
A
⋅
a
B
)
a
B
Let
z
y
x
a
a
a
A
4
5
10
+
−
=
,
y
x
a
a
B
+
=
3
and
a
B
the unit vector in the direction of
B
.
y
x
B
a
a
B
B
a
10
1
10
3
+
=
=
.
(
)
y
x
B
B
a
a
a
a
A
A
5
.
2
5
.
7

+
=
⋅
=
z
y
x
a
a
a
A
A
A
4
5
.
7
5
.
2

+
−
=
−
=
⊥
Problem 3
:
E
and
F
are vector fields given by
z
y
x
yz
x
a
a
a
E
+
+
=
2
, and
. Determine:
z
y
x
xyz
y
xy
a
a
a
F
+
−
=
2
(a) 
E
 at (1, 2, 3)
(b) the component of
E
along
F
at (1, 2, 3)
(c) a vector perpendicular to both
E
and
F
at (0, 1, 3) whose magnitude is unity.
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