The shortest distance between the lines 3(x-1) = 6(y-2) = 2(z-1) and 4(x-2) = 2(y-lambda) = (z-3) is 1/sqrt(38). Find the integral (integer) value of lambda.

3
2
5
-1

Correct answer: 3

Solution

Computing d1 x d2 = (-2, -5, 3) with |d1 x d2| = sqrt(38). The dot product of (B-A) = (1, lambda-2, 2) with (-2, -5, 3) gives 14 - 5*lambda. Setting |14 - 5*lambda| = 1 yields lambda = 13/5 (non-integer) or lambda = 3 (integer).

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