Question

A unit vector perpendicular to P=3i-2j+2k and Q=-4i+2j-3k is

Answer 1

First find the vector perpendiculat to P and Q by finding the cross product of P and Q. So the cross product of P and Q gives

P×Q = (3i-2j+2k) × (-4i+2j-3k)
        = [(-2)×(-3) - (2×2)] i + [2×(-4) - 3×(-3)] j + [3×2 - (-2)×(-4)] k
        = [ 6 - 4] i + [-8 + 9] j + [6 - 8] k
        = 2i + j - 2k

|P×Q| = √(2² + 1² + 2²) = √9 = 3

unit vector = $\frac{P \times Q}{|P \times Q|} = \frac{2i + j - 2k}{3} = \frac{2}{3} i+ \frac{1}{3} j- \frac{2}{3} k$

Answer 2

Answer:

A unit vector perpendicular to P=3i-2j+2k and Q=-4i+2j-3k is

Explanation:

The cross product of the given vectors i-2j+3k and i+2j-3k is -4i+4j+4k. So the required vector is any vector parallel to this vector.