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Perpendicular planes proof

  1. Jul 15, 2007 #1
    i was wondering, how would you prove if two planes are perpendicular, analytically? don't you have to use vectors or something?
  2. jcsd
  3. Jul 15, 2007 #2


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    You sort of have to use vectors, using the cartesian equations (Ax+By+Cz+D=0), take normals of planes as (A,B,C) and multiply them together with dot product to see if the the resultant scalar is zero. For all intents and purposes, the normal of a plane is like the gradient of a line in two dimensions, so this is probably as analytical as it can get if you wish to keep it simple. Of course, you can use partial differentiation and consider the gradients of the planes, I believe, but even that borders on using vectors, because "gradient" doesn't have much meaning in three space.
  4. Jul 16, 2007 #3


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    A plane is "given" by a point in it and its normal vector.

    In particular, the plane through (x0, y0,z0) with normal vector <A, B, C> has equation A(x-x0)+ B(y-y0)+ C(z- z0)= 0 (which can also be written as Ax+ By+ Cz= Ax0+ By0+ Cz0).

    Two planes are perpendicular if and only if their normal vectors are perpendicular which means their dot product must be 0.

    The two planes Ax+ By+ Cz= P and Ux+ Vy+ Wz= Q are perpendicular if and only if AU+ BV+ CW= 0.
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