A unit vector perpendicular to both the vectors a = 2i - j - k and b = -i + 2j - k is: A. (1/sqrt(3))(i + j + k) B. (1/sqrt(3))(i - j - k) C. (1/sqrt(3))(i + j - k) D. (1/sqrt(3))(i - j + k)

GATE 1995 · Engineering Mathematics · Vector Space · medium

Answer: A. (1/sqrt(3))(i + j + k).

1. Compute a x b component by component: i-component: a2*b3 - a3*b2 = (-1)(-1) - (-1)(2) = 1 + 2 = 3. j-component: -(a1*b3 - a3*b1) = -((2)(-1) - (-1)(-1)) = -(-2 - 1) = 3. k-component: a1*b2 - a2*b1 = (2)(2) - (-1)(-1) = 4 - 1 = 3. So a x b = 3i + 3j + 3k.
2. Compute magnitude and normalise: |a x b| = sqrt(9 + 9 + 9) = sqrt(27) = 3*sqrt(3). Unit vector = (3i + 3j + 3k) / (3*sqrt(3)) = (1/sqrt(3))(i + j + k).
3. Verify perpendicularity via dot products: u = (1/sqrt(3))(1,1,1). u . a = (1/sqrt(3))(1*2 + 1*(-1) + 1*(-1)) = (1/sqrt(3))(2-1-1) = 0. u . b = (1/sqrt(3))(1*(-1) + 1*2 + 1*(-1)) = (1/sqrt(3))(-1+2-1) = 0. Both dot products are zero, confirming u is perpendicular to both a and b.