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A vector a lies in the plane of b = 2i + j and c = i - j + k, such that a is equally inclined to b and d, where d = j + 2k. Find the unit vector a.

1. (i + j + k) / sqrt(3)
2. (i - j + k) / sqrt(3)
3. (2i + j) / sqrt(5)
4. None of the above

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

Solution

a lies in the plane of b = 2i+j and c = i-j+k, so a = alpha*(2i+j) + beta*(i-j+k) = (2alpha+beta)i + (alpha-beta)j + beta*k. Equal inclination to b = 2i+j (|b| = sqrt(5)) and d = j+2k (|d| = sqrt(5)) means a.b/|b| = a.d/|d|, so a.b = a.d. Computing: a.b = 2(2alpha+beta) + (alpha-beta) = 5alpha+beta. a.d = (alpha-beta) + 2*beta = alpha+beta. Setting equal: 5alpha+beta = alpha+beta → 4alpha = 0 → alpha = 0. So a = beta*(i-j+k), unit vector = (i-j+k)/sqrt(3). Wait — let me recheck: a.b = (2alpha+beta)*2 + (alpha-beta)*1 + beta*0 = 4alpha+2beta+alpha-beta = 5alpha+beta. a.d = (2alpha+beta)*0 + (alpha-beta)*1 + beta*2 = alpha-beta+2beta = alpha+beta. Setting 5alpha+beta = alpha+beta → 4alpha = 0 → alpha = 0. Then a = beta*(i-j+k). Unit vector = (i-j+k)/sqrt(3).

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