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Question

If a and b are vectors such that |a+b|=29 and a×(2^i+3^j+4^k)=(2^i+3^j+4^k)×b, then a possible value of (a+b).(7^i+2^j+3^k) is


A

0

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B

3

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C

4

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D

8

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Solution

The correct option is C

4


a×b=a×ca×ba×c=0a||bc) or bc=λaHere, a×(2^i+3^j+4^k)=(2^i+3^j+4^k)×b a×(2^i+3^j+4^k)(2^i+3^j+4^k)×b=0 (a+b)×(2^i+3^j+4^k)=0 a+b=λ(2^i+3^j+4^k)Since, |a+b|=29 ±λ4+9+16=29 λ=±1 a+b=±(2^i+3^j+4^k)Now, (a+b).(7^i+2^j+3^k)=±(14+6+12)=±4


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