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.

  • $\begingroup$ 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


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.

  • $\begingroup$ 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

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$$\frac19\begin{bmatrix}5 & -4 & -2 \\ -4 & 5 & -2 \\ -2 & -2 & 8\end{bmatrix}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.