Find the distance from the point to the given plane.
(1,−5,9), 3x+2y+6z=5

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asked by miolo
  1. vector normal to plane has direction
    3 i + 2 j + 6 k

    line normal to plane through point is
    (1, -5 , 9) + (3, 2, 6) t

    where does that hit the plane?
    3(1+3t) + 2(-5+2t) + 6(9+6t) = 5

    solve for t
    go back and use that t to get x, y, z in plane
    x = 1+3t y = -5+2t z = 9+6t
    d^2 = (X2-X1)^2 + (Y2-Y1)^2+(Z2-Z1)^2

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    posted by Damon

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