Question 566766
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I don't have your textbook, so I don't have your drawing.  Use your head for something besides a hat rack next time you post a question about a drawing that we can't possibly see.  Remember, this is Algebra.com -- it is not the Psychic Hot Line.


Be that as it may, the only way this will work is if point P is one endpoint of a long diagonal of the prism and point Q is the other end point and the prism is arranged so that the three edges of the prism that are orthogonal at point Q are each coincident with one of the three coordinate axes.


The next thing to realize is that a point with non-zero coordinates specified by an ordered triple is NOT in the coordinate PLANE.  Planes are 2 dimensional and this is three-space.  You are not in Kansas anymore, Toto.


Presuming that the *[tex \Large x] (horizontal) and *[tex \Large z] (vertical) axes are in the plane of your paper and the *[tex \Large y] axis extends out from or into the paper, then the described prism must have one surface in the *[tex \Large xz] plane, one in the *[tex \Large xy] plane, and the third in the *[tex \Large yz] plane.


Any point that is in the *[tex \Large xy] plane will have a *[tex \Large z] coordinate of zero.  Any point in the *[tex \Large xz] plane will have a *[tex \Large y] coordinate of zero.  Any point in the *[tex \Large yz] plane will have an *[tex \Large x] coordinate of zero.  If a point is actually on the *[tex \Large x] axis, the *[tex \Large y] and *[tex \Large z] coordinates will both be zero...and the pattern continues.  All coplanar points in a plane parallel to the *[tex \Large xy] axis will have equal *[tex \Large z] coordinates...and so on.


That is sufficient information to derive all eight ordered triples representing the vertices of the described rectangular prism.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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