Let a = i + j - k, b = i - k, and c = 2i - j. If d is a unit vector such that a. d = 0 and the scalar triple product [b, c, d] = 0, then d equals:

(3i - 2j + k)/sqrt(14)
(i + 2j + 3k)/sqrt(14)
(i - 2j + 3k)/sqrt(14)
(-3i + 2j - k)/sqrt(14)

Correct answer: (-3i + 2j - k)/sqrt(14)

Solution

Since [b c d] = 0, d is coplanar with b and c: d = xb + yc = x(i-k) + y(2i-j) = (x+2y)i - yj - xk. Applying a.d = 0: (1)(x+2y)+(1)(-y)+(-1)(-x) = x+2y-y+x = 2x+y = 0, so y = -2x. Substituting: d is proportional to (x-4x)i+(2x)j+(-x)k = x(-3i+2j-k). The unit vector is (-3i+2j-k)/sqrt(9+4+1) = (-3i+2j-k)/sqrt(14).

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