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Let L1 be the line passing through the points Q1=(4, 2, −3) and Q2=(0, −2, 3). Find a value of k so the line L2 passing through the point P1 = P1(−11, 2, k) with direction vector →d=[3, −2, −3]T intersects with L1

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  1. Direction of L1 = <4,4,-6> or reduced to <2,2,-3>
    direction of L2 = <3,-2,-3> , that was given

    If L1 and L2 intersect they must lie on the same plane.
    The normal of that plane is the cross product of <2,2,-3> with <3,-2,-3>

    Using whatever method you learned, that normal is <12,3,10>

    so the equation of the plane is 12x + 3y + 10z = k
    but the point (4, 2, −3) lies on it, so 48 + 6 - 30 = k ----> k = 24
    ( I could have used the point (0, −2, 3) to get 0 -6 + 30 = k , k = 24)

    So the equation of the plane containing our two lines is
    12x + 3y + 10z = 24
    That also contains any point on either line, and (-11,2,k) is supposed to be on it, so
    -132 + 6 + 10k = 24
    10k = 150
    k = 15

    check my arithmetic

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    Reiny

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