A) 16
B) 8
C) 3
D) 12
Correct Answer: C
Solution :
We know that \[{{(\mathbf{a}\times \mathbf{b})}^{2}}+{{(\mathbf{a}\,.\,\mathbf{b})}^{2}}=\,|\mathbf{a}{{|}^{2}}|\mathbf{b}{{|}^{2}}\] \[\therefore \,\,\,144=16|\mathbf{b}{{|}^{2}}\Rightarrow \,|\mathbf{b}|=3.\]You need to login to perform this action.
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