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Question

If |a×b|2+|ab|2=144 and |a|=4, then the value of |b| is

A
1
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B
2
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C
3
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D
4
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Solution

The correct option is C 3
|a×b|2+|ab|2=144
|a|2|b|2sin2θ + |a|2|b|2cos2θ=144
|a|2|b|2=144
16|b|2=144
|b|=3

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