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Question

Find unit vector perpendicular to the plane of the vectors,
A=^i2^j+^k and B=2^i^k.

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Solution

The vector perpendicular to these vectors is-

n=A×B=(i2j+k)×(2ik)

n=2i+3j+4k

n=4+9+16=29

hence, unit vector is

^n=2i+3j+4k29

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