Find angle between the vectors , ā = 2i - 3j + k and b = 5i + 4j + 2k
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Answer:
We're asked to find the angle between two vectors, given their unit vector notations.
To do this, we can use the equation
→
A
⋅
→
B
=
A
B
cos
θ
rearranging to solve for angle,
θ
:
cos
θ
=
→
A
⋅
→
B
A
B
θ
=
arccos
⎛
⎝
→
A
⋅
→
B
A
B
⎞
⎠
where
→
A
⋅
→
B
is the dot product of the two vectors, which is
→
A
⋅
→
B
=
A
x
B
x
+
A
y
B
y
+
A
z
B
z
=
(
2
)
(
1
)
+
(
3
)
(
2
)
+
(
1
)
(
−
4
)
=
4
A
and
B
are the magnitudes of vectors
→
A
and
→
B
, which are
A
=
√
2
2
+
3
2
+
1
2
=
√
14
B
=
√
1
2
+
2
2
+
(
−
4
)
2
=
√
21
Therefore, we have
θ
=
arccos
(
4
√
14
√
21
)
=
arccos
(
4
7
√
6
)
=
76.5
o
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