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Let L1 be the line passing through the point P1=(16, −1, −12) with direction vector →d1=[3, −1, −2]T, and let L2 be the line passing through the point P2=(4, 7, −17) with direction vector →d2=[3, 1, −2]T.

Find the shortest distance d between these two lines, and find a point Q1 on L1 and a point Q2 on L2 so that d(Q1,Q2) = d. Use the square root symbol '√' where needed to give an exact value for your answer.

I have wrecked my brain several times, but cannot solve this question whatsoever. Please help!

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  1. there are several ways to solve this type of problem.
    The easiest method in my opinion goes like this:

    A normal to both lines would be the cross product of
    <3, -1, -2> with <3, 1, -2> , the direction of the 2 given lines
    which is <2, 0, 3> , (I assume you know how to find the cross-product)

    Now consider 2 planes with that normal, the first one containing
    the point (16, -1, -12) and the second plane containing the point (4, 7, -17)

    The distance between these two parallel planes will of course be always
    the same , this is the key point using this method


    equation of first plane is 2x + 0y + 3z = c
    but (16, -1, -12) lies on it, so
    32 + 0 - 36 = c
    c = -4
    2x + 3y + 4 = 0

    All we need now is the distance between the two planes using the distance between a point and a plane
    We also have a point on the 2nd plane: namely (4, 7, -17)
    distance = | 2(4) + 0(7) + 3(-17) + 4| / √(2^2 + 0^2 + 3^2)
    = 39/√13 or 39√13/13 = 3√13

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