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Question

If θ is the angle between two vectors ^i2^j+3^k and 3^i2^j+^k, find sin θ.

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Solution

a=^i2^j+3^k

b=3^i2^j+^k

and θ is the angle between a and b.

|a|=12+(2)2+(3)2

=1+4+9=14

|b|=32+(2)2+12

=9+4+1=14

Now, |a|b sin θ=a×b

sin θ=a×bab=∣ ∣ ∣^i^j^k123321∣ ∣ ∣14 14

= 4^i+8^j+4^k14

= 16+64+1614=9614=4614=276

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