## Matrix Transformation ugly problem

1) Let U be a plane through the origin in R^3 with a nonzero normal vector n=[a b c]^T. Find the projection matrix of X=[x1 x2 x3]^T onto U.

I got this question from my linear algebra test today and I am dying on it.

I tried something out but ended up with a terribly ugly result in which I have no confidence of it being right.

My method:
Since (projection of X onto n) gives the perpendicular (closeest) distance from X to the plane U, I have the following inequality: (in orthongonal complement of U)
(projection of X onto n) = X - (projection of X onto U) (<-is this right?)
and then solve for (projection of X onto U) for which I can obtain the induced matrix by factoring the matrix [x1 x2 x3]^T out
and this ends up with some ugly calculations (this question only worth 5 marks, how can I take that long?)

Is there a flaw in this thinking? Is it right?

I seriously think I have missed something...Is there a very easy method to do this question? Can someone teach me? I can't sleep without it.

Thanks a lot!
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 Is there anyone good in linear algebra who can help me? thanks

Recognitions:
Gold Member
With the normal vector given as just the general <a, b, c>, I don't see any really quick way to do it. It's easy to see that a unit vector normal to the plane is <a/L, b/L, c/L> where L is the length of <a, b, c>: $\sqrt{a^2+ b^2+ c^2}$. The projection of the basis vector <1, 0, 0> onto that line is $<a^2/L^2, b/L^2, c/L^2>$ and so its projection onto the plane is [itex]<(b^2+ c^2)/L^2, -b/L^2, -c/L^2>. Similarly, the projection of <0, 1, 0> onto the plane is [itex]<-a/L^2, (a^2+ c^2)/L^2,-c/L^2) and the projection of <0, 0, 1> onto the plane is [itex]<-a/L^2, -b/L^2, (a^2+ b^2)/L^2>. The matrix representation of the projection is the matrix having those vectors as columns.