The unit vector perpendicular to each of the vectors 2i−j+k and 3i+4j−k is
A
−3i+5j+11k
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B
(−3i+5j+11k)/155
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C
(−3i+5j+11k)/√(155)
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D
None of these
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Solution
The correct option is C(−3i+5j+11k)/√(155) We know that a×b is vector perpendicular to the plane of a and b Therefore unit vector perpendicular to the plane of a and b=a×b|a×b| Now a×b=(2i−j+k)×(3i+4j−k) =∣∣
∣∣ijk2−1134−1∣∣
∣∣=i(1−4)+j(3+2)+k(8+3)=−3i+5j+11k ⇒ unit vector parallel to a×b =a×b|a×b|=−3i+5j+11k√(32+52+112) =−3i+5j+11k√155