# Calculating matrix for linear transformation of orthogonal projection onto plane.

The question goes like this:

"Calculate the matrix P for the linear transformation of an orthogonal projection of vectors onto the plane

$$2x_1+2x_2+x_3^{}= 0"$$

So I am thinking that projection is the way to go. What I basically will do is use the normal of the plane. Which is:

$$\left[ \begin{array}{cc|c} 2\\ 2\\ 1 \end{array} \right]$$

That would be my perpendicular part. And the vectors that I will project onto the plane will naturally be the basis vectors $$|e_1| = \left[ \begin{array}{cc|c} 1\\ 0\\ 0 \end{array} \right], |e_2| =\left[ \begin{array}{cc|c} 0\\ 1\\ 0 \end{array} \right], |e_3| =\left[ \begin{array}{cc|c} 0\\ 0\\ 1 \end{array} \right]$$

Basically, what I will do is set up an equation

$$Proj V_n + |n| = \left[ \begin{array}{cc|c} 1\\ 0\\ 0 \end{array} \right]$$

So I solve for the projection and that would be my first column of my matrix P. But I keep getting the wrong answer. Where is my thinking going wrong?

• Do you mean orthogonal projection? Nov 21, 2018 at 15:30
• Hmm, do you mean the question? The question literally is saying : " Calculate the matrix for the linear transformation of projections of vectors onto the given plane" Nov 21, 2018 at 15:32
• There are infinitely many such projections, but only one orthogonal one. So, yes, I do mean the question. Nov 21, 2018 at 15:35
• I would say that they look for the projection that is parallell to the plane. Nov 21, 2018 at 15:39
• So yes! It is correct. Orthogonal projection. I just looked it up and edited the question. Thanks for the heads up! Nov 21, 2018 at 15:59

Your notation is a bit hard to decipher, but it looks like you’re trying to decompose $$\mathbf e_1$$ into its projection onto and rejection from the plane. That’s a reasonable idea, but the equation that you’ve written down says that the projection of $$\mathbf e_1$$ is equal to $$\mathbf e_1-\mathbf n = (-1,-2,-1)^T$$. Unfortunately, this doesn’t even lie on the plane: $$2(-1)+2(-2)+1(-1)=-7$$.

The problem is that you’ve set the rejection of $$\mathbf e_1$$ from the plane to be equal to $$\mathbf n$$, when it’s actually some scalar multiple of it. I.e., the orthogonal projection $$P\mathbf e_1$$ of $$\mathbf e_1$$ onto the plane is $$\mathbf e_1-k\mathbf n$$ for some as-yet-undetermined scalar $$k$$. However, $$k\mathbf n$$ here is simply the orthogonal projection of $$\mathbf e_1$$ onto $$\mathbf n$$, which I suspect that you know how to compute.

• I overcomplicated things. My TA helped me through using the same concepts you use when computing reflection about a plane. Basically. You subtract the vector in question, in my case I started out with $$e_1$$ the projection of that vector on the normal of the plane. The result will be a vector lying on the plane. Nov 22, 2018 at 15:28
• @Synchrowave Yep. For reflections, decomposing into components perpendicular and parallel to the reflector and then reassembling is usually a much easier way to go.
– amd
Nov 22, 2018 at 19:25

Since $$(2,2,1)$$ is orthogonal to the plane, you wnat that its projection is the null vector. Now, take two linearly independent vectors from your plane. For instance, take the vectors $$(1,0,-2)$$ and $$(0,1,-2)$$. You want the each one is projected into itself.

So, take the only linear map $$P\colon\mathbb{R}^3\longrightarrow\mathbb{R}^3$$ such that

1. $$P(2,2,1)=(0,0,0)$$;
2. $$P(1,0,-2)=(1,0,-2)$$;
3. $$P(0,1,-2)=(0,1,-2)$$.

A simple computation shows that the matrix of $$P$$ with respect to the canonical basis is$$\frac19\begin{bmatrix}5 & -4 & -2 \\ -4 & 5 & -2 \\ -2 & -2 & 8\end{bmatrix}.$$