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Question

Let a=3^i+2^j+2^k,b=^i+2^j2^k. Then a unit vector perpendicular to both ab and a+b is :

A
13(2^i+2^j+^k)
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B
13(2^i+2^j^k)
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C
13(2^i2^j+^k)
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D
13(^i+^j+^k)
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Solution

The correct option is A 13(2^i+2^j+^k)
We have,
a=3^i+2^j+2^k,b=^i+2^j2^ka+b=4^i+4^j^a^b=2^i+4^kd=(ab)×(a×b)

Therefore,
=∣ ∣ ∣i^j^k204440∣ ∣ ∣=16^i+16^j+8^k

Unit vector perpendicular to
ab and a+b is
=(16^i+16^j+8^k)576=8(2^i+2^j+^k)24=13(2^i+2^j+^k)

Hence, this is the answer.

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