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Find the distance of the point with position vector -i + 2j + 6k from the straight line passing through the point (2, 3, -4) and parallel to the vector 6i + 3j - 4k.

1. 7
2. 4*sqrt(3)
3. 2*sqrt(13)
4. 6

Correct answer: 7

Solution

Let A = (2,3,-4), P = (-1,2,6), d = (6,3,-4). AP = P-A = (-3,-1,10). AP x d = |i j k; -3 -1 10; 6 3 -4| = i((-1)(-4)-(10)(3)) - j((-3)(-4)-(10)(6)) + k((-3)(3)-(-1)(6)) = i(4-30) - j(12-60) + k(-9+6) = (-26)i + 48j + (-3)k. |AP x d| = sqrt(676 + 2304 + 9) = sqrt(2989). |d| = sqrt(36+9+16) = sqrt(61). Distance = sqrt(2989/61) = sqrt(49) = 7.

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