26. Find the sine of the angle between the vectors
(i)3 i + j + 2 k and 2 i - 2 j + 4 k
(ii) 2 i - j + k and 3 i + 4 j - k
Answers
Explanation:
a
=
3
i
−
2
j
+
k
8888
b
=
−
2
i
+
2
j
+
4
k
The magnitude of a vector
a
with components
x
1
,
x
2
,
x
3
is defined as:
|
|
a
|
|
=
√
(
x
1
)
2
+
(
x
2
)
2
+
(
x
3
)
2
This is just the distance formula found in coordinate geometry.
We can find the angle between two vectors by using the dot product
For vectors
a
and
b
a
⋅
b
=
|
|
a
|
|
⋅
|
|
b
|
|
⋅
cos
(
θ
)
The multiplication is carried out in a different way than multiplying brackets in algebra. We usually multiply brackets in algebra:
(
a
+
b
+
c
)
(
d
+
e
+
f
)
=
a
d
+
a
e
+
a
f
+
b
d
+
b
e
+
b
f
+
c
d
+
c
e
+
c
f
In the dot product we multiply corresponding components and sum the results:
(
a
1
+
b
1
+
c
1
)
(
a
2
+
b
2
+
c
2
)
=
a
1
a
2
+
b
1
b
2
+
c
1
c
2
Question 1:
|
|
a
|
|
=
√
(
3
)
2
+
(
−
2
)
2
+
(
1
)
2
=
√
14
|
|
b
|
|
=
√
(
−
2
)
2
+
(
2
)
2
+
(
4
)
2
=
√
24
=
2
√
6
Angle between
a
and
b
:
a
⋅
b
=
(
3
−
2
+
1
)
(
−
2
+
2
+
4
)
=
−
6
−
6
=
√
14
⋅
2
√
6
⋅
cos
(
θ
)
cos
(
θ
)
=
−
6
√
14
⋅
2
√
6
=
−
3
√
84
=
−
3
√
84
84
θ
=
arccos
(
−
3
√
84
84
)
=
109.11
∘
( 2 d.p.)
Question 2:
a
+
1
2
b
=
3
i
−
2
j
+
k
+
1
2
(
−
2
i
+
2
j
+
4
k
)
8888888
=
3
i
−
2
j
+
k
−
i
+
j
+
2
k
8888888
=
2
i
−
j
+
3
k
For clarity let
v
=
a
+
1
2
b
=
2
i
−
j
+
3
k
The projection of
v
onto
a
is given by:
v
⋅
a
|
|
a
|
|
=
(
2
−
1
+
3
)
(
3
−
2
+
1
)
=
11
√
14
=
11
√
14
14
This is the resolved part of vector
v
in the direction of vector
a
Question 3
If vectors are prpendicular, then the angle between them is
90
∘
cos
(
90
∘
)
=
0
So for vectors
a
and
b
:
a
⋅
b
=
0
i.e.
a
⋅
b
=
|
|
a
|
|
⋅
|
|
b
|
|
⋅
0
=
0
( cos(90)=0)
We can use this to test which vectors are perpendicular to
a
.
c
=
−
i
−
4
j
+
2
k
a
⋅
c
=
(
3
−
2
+
1
)
(
−
1
−
4
+
2
)
=
7
not perpendicular
d
=
−
3
i
+
k
a
⋅
d
=
(
3
−
2
+
1
)
(
−
3
+
0
+
1
)
=
−
8
not perpendicular
e
=
2
i
+
2
j
−
2
k
a
⋅
e
=
(
3
−
2
+
1
)
(
2
+
2
−
2
)
=
0
perpendicular
So
e
is perpendicular to
a