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Question : 156 of 180
Marks:
+1,
-0
Solution:
Let
a=3++2 and
b=++2 a×b=|| =(2−2)−(6−2)+(3−1)=−4+2 |a×b|=√(−4)2+(2)2=√16+4=√20 Since,
|a×b|=|a||b|sinθ If
θ is angle between
a and
b.
√20=√9+1+4√1+1+4sinθ ⇒sinθ= ⇒sinθ=√=√
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