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Question

Find the angle between the vectors ˆi2ˆj+3ˆk and 3ˆi2ˆj+ˆk.

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Solution

Here, let a=ˆi2ˆj+3ˆk and b=3ˆi2ˆj+ˆk.
Then,
a.b=(ˆi2ˆj+3ˆk).(3ˆi2ˆj+ˆk)
=(1.3)+(2)(2)+(3.1)
=3+4+3=10
|a|=1+4+9=14
and, |b|=9+4+1=14
Therefore,
cosθ=101414=1014=57

Thus, θ=cos1(57)

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