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Question

Let a and b be two vectors such that |a|=3, |b|=6 and |a+b|=7. Then the value of (3a2b).(2a+5b4a×b) is

A
306
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B
295
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C
284
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D
290
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Solution

The correct option is C 284
(3a2b)(2a+5b4a×b)
=6|a|2+15ab3a(4a×b) 4ab10|b|2+2b(4a×b)

Now 3a(4a×b)=0 and 2b(4a×b)=0 because a and b will be perpendicular to 4a×b

So the equation reduces to
(3a2b)(2a+5b4a×b)=6|a|2+15ab4ab10|b|2
=6(3)2+11ab10(6)2
=11ab306 (1)

Now |a+b|=7
|a+b|2=72
|a|2+2ab+|b|2=72
32+2ab+62=72
ab=2
Putting this value in equation (1), we get
(3a2b)(2a+5b4a×b)=11×2306=284



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