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The closest point in the plane

  1. Nov 4, 2004 #1
    Let x + y + 3 z = 7 represent a plane. (it does)

    We find the closest point to origin in this plane by [d/[n]^2] * n. In this case n = (1,1,3); d = 7; [n]^2 = 1^2 + 1^2 + 3^2 = 11; then the vector that gives us the closest point is: (7/11, 7/11, 21/11)

    I dont understand WHY this operation gives us the closest point and Strang's book doesn't really explain. I'd appreciate if you help.
  2. jcsd
  3. Nov 4, 2004 #2


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    The line from the origin to the "closest point" in the plane is perpendicular to the plane (to see that, remember that the hypotenuse of a right triangle is always longer than the two legs).

    A vector perpendicular to the given plane is <1, 1, 3>. Any line perpendicular to the plane must be of the form x= t+ a, y= t+ b, z= 3t+ c. Take t= 0 to be the value when the line passes through the origin: x= 0+ a= 0 , y= 0+ b= 0, z= 0+ c= 0 so the line is x= t, y= t, z= 3t.

    Where does that line intersect the given plane? Replacing x, y, z, by t, t, 3t, respectively in the equation of the plane, t+ t+ 3(3t)= 11t= 7 so t= 7/11. That gives
    x= 7/5, y= 7/5, z= 21/5 or (7/5, 7/5, 21/5) as the nearest point just as the formula does.

    More generally, suppose the equation of the plane is Ax+ By+ Cz= d. The line perpendicular to that plane AND passing through the origin is x= At, y= Bt, z= Ct. Putting those into the equation of the plane, we had A2t+ B2t+ C2t= d so t= d/(A2+ B2+ C2) or, using your notation,t= d/[n]2 for the value of t where that normal line intersects the plane. Putting that into the equations of the line gives x= [d/[n]2]A, y= [d/[n]2]B, z= [d/[n]2]C, once again, exactly the formula's [d/[n]2]*n.
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