find the angle between the vectors i+2j+3k and 3i-j+2k explanation
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Answered by
0
Answer:
5/7
Step-by-step explanation:
Let the given vectors are
a
=
i
^
−2
j
^
+3
k
^
and
b
=3
i
^
−2
j
^
+
k
^
∣
a
∣=
1
2
+(−2)
2
+3
2
=
1+4+9
=
14
∣
∣
∣
∣
b
∣
∣
∣
∣
=
3
2
+(−2)
2
+1
2
=
9+4+1
=
14
Now,
a
⋅
b
=(
i
^
−2
j
^
+3
k
^
)(3
i
^
−2
j
^
+
k
^
)
=1.3+(−2)(−2)+3.1
=3+4+3=10
Also, we know that
a
⋅
b
=∣
a
∣∣
b
∣cosθ
∴10=
14
14
cosθ
⇒cosθ=
14
10
⇒θ=cos
−1
(
7
5
)
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