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Question

Let 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+b).(7^i+2^j+3^k)=


A

0

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B

±4

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C

±29

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D

±129

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Solution

The correct option is B

±4


a×(2^i+3^j+4^k)=(2^i+3^j+4^k)×b

(a+b) is parallel to (2^i+3^j+4^k)

a+b=λ(2^i+3^j+4^k)

|a+b|=λ|2^i+3^j+4^k|

29=λ(±29)λ=±1

a+b=±(2^i+3^j+4^k)

±(2^i+3^j+4^k).(7^i+2^j+3^k)=±4


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