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Consider the plane P with equation −2x−2y−z = −11 and line L with equations

x = −3−4t
y = −2−3t
z = −6−4t
Find two distinct points Q1 and Q2 on the line L at distance 3 to the plane P.

Q1 = (0, 0, 0)

Q2 = (0, 0, 0)

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1 answer

  1. Let Q be a general point on the line
    Q is (-3-4t, -2-3t, -6-4t)
    Plane : -2x -2y -z + 11 = 0

    Using the distance from a point to a plane formula
    | -2(-3-4t) - 2(-2-3t) - (-6-4t) + 11|/ √(4+4+1) = 3
    ± (6 + 8t + 4 + 6t + 6 + 4t + 11)/3 = 3
    using the positive
    18t + 27 = 9
    18t = -18
    t = -1
    then Q is (-3+4, -2+3, -6+4) ,

    Q1 is (1, 1, -2)

    using the negative result from above:
    -(6 + 8t + 4 + 6t + 6 + 4t + 11)/3 = 3
    -18t - 27 = 9
    -18t = 36
    t = -2

    Q2 is (-3+8, -2+6, -6+8)

    = Q2(5, 4, 2)

    check:
    midpoint of Q1Q2 = (3, 5/2, 0) , this point should lie on our plane
    LS = -2(3) -2(5/2) - (0)
    = -6 -5 - 0 = -11
    RS = -11

    looks good!

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