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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?

Thanks in advance.

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  • $\begingroup$ Do you mean orthogonal projection? $\endgroup$ Nov 21, 2018 at 15:30
  • $\begingroup$ 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" $\endgroup$ Nov 21, 2018 at 15:32
  • $\begingroup$ There are infinitely many such projections, but only one orthogonal one. So, yes, I do mean the question. $\endgroup$ Nov 21, 2018 at 15:35
  • $\begingroup$ I would say that they look for the projection that is parallell to the plane. $\endgroup$ Nov 21, 2018 at 15:39
  • $\begingroup$ So yes! It is correct. Orthogonal projection. I just looked it up and edited the question. Thanks for the heads up! $\endgroup$ Nov 21, 2018 at 15:59

2 Answers 2

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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.

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  • $\begingroup$ 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. $\endgroup$ Nov 22, 2018 at 15:28
  • $\begingroup$ @Synchrowave Yep. For reflections, decomposing into components perpendicular and parallel to the reflector and then reassembling is usually a much easier way to go. $\endgroup$
    – amd
    Nov 22, 2018 at 19:25
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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$$\frac19\begin{bmatrix}5 & -4 & -2 \\ -4 & 5 & -2 \\ -2 & -2 & 8\end{bmatrix}.$$

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