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A unit vector perpendicular to P=3i-2j+2k and Q=-4i+2j-3k is |
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Answer» ind the vector perpendiculat to P and Q by FINDING the cross product of P and Q. So the cross product of P and Q givesP×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 = 3unit vector = |
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