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I'm looking at an exercise in which the matrix of the reflection over the plane: x-y-z=0 had to be found. The exercise is solved by using the change of basis procedure by constructing a basis B which holds the normal vector n=(1,-1-1) and two other vectors (a and b) that can be found on this plane (the author than states that these two vectors have to have the same equation as the plane, that they have to be linear independent and that dot products (n,a)=0 (n,b)=0).

The author picked these two vectors: a=(0,1,-1) b=(-1,0,-1)

Can someone explain how these two vectors were obtained because I can't figure it out.

Thank you!

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Are you asking how to see that the two vectors lie in the plane, or how the authors chose precisely those vectors among the infinity of possible vectors that line in the plane? – Henning Makholm Feb 5 '13 at 16:03
How he chose them. :) – Trom Feb 5 '13 at 16:06
up vote 2 down vote accepted

What the author meant was that you want two vectors satisfying the equation of the plane (that is, they should lie in the plane). Since the equation of the plane is $x-y-z=0$, one can choose two of the coordinates, and then solve for the third.

In your case $a$ is obtained by choosing $x=0$ and $y=1$, and plugging that into the equation of the plane in order to find out that $z=-1$. In the same way, $b$ is obtained by choosing two coordinates and solving for the third.

As a matter of fact, a better approach would be to choose orthogonal vectors in the plane, because that would facilitate an orthonormal change of basis (you can just transpose the matrix instead of inverting it the painful way).

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You know the plane passes through the origin, and you need two orthogonal vectors that span this plane. Do you see it now?

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