If →A=^i+2^j−^k and →B=2^i+3^j+4^k, Then the vector which is perpendicular to both vectors →A and →B is
A
11^i+6^j−^k
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B
5^i−6^j+^k
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C
11^i−6^j−^k
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D
5^i−6^j−^k
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Solution
The correct option is C11^i−6^j−^k As we know that →A×→B will be the perpendicular to both vectors →A and →B
So, we have →A×→B=⎡⎢⎣^i^j^k12−1234⎤⎥⎦
=[4×2−(3×(−1)]^i−[1×4−(−1×2)]^j+[3×1−2×2]^k
=11^i−6^j−^k