projection of a vector onto a plane in $\Bbb R^5$

Question

Hy guys, i have some question about this problem. I have three points $P(1,0,2,0,1)\; Q(0,1,0,1,0)\;R(2,0,1,0,2)$

The parametric representation of the plane $\pi$ which contains that point is

$\{P+s(Q-P)+t(R-P)\;|s,t \in \Bbb R^5 \}$

i.e

$$ \left\{ \begin{aligned} 1-s+t =x_1 \\ s=x_2 \\ 2-2s-t=x_3\\ s=x_4\\ 1-s+t=x_5 \end{aligned} \right. $$

so my cartesian representation is:

$$ \left\{ \begin{aligned} x_1-x_5=0\\ x_2-x_4=0 \end{aligned} \right. $$ that is right?? Let's continue:

the problem asks me to find an orthogonal base of the direction $\pi_0$ of the plane $\pi$.

Isn't the direction the vectors $s(Q-P)+t(R-P)$?

$ s(Q-P)+t(R-P)=\, s\left(\begin{matrix} -1\\ 1\\ -2\\ 1\\ -1\\ \end{matrix} \right) + t\left(\begin{matrix} 1\\ 0\\ -1\\ 0\\ 1\\ \end{matrix} \right) $

So, for $s=1$ and $t=1$

$\left(\begin{matrix} -1\\ 1\\ -2\\ 1\\ -1\\ \end{matrix} \right) \cdot \left(\begin{matrix} 1\\ 0\\ -1\\ 0\\ 1\\ \end{matrix} \right) =0$

These two vectors form my orthogonal basis?

The last question: find the orthogonal projection of the vector $v=\left(\begin{matrix}1\\1\\1\\1\\1\end{matrix}\right)$ onto the $\pi_0$

Thank you so much guys!!

Answer 1

Your Cartesion representation is not correct. What you got was a $3$-dimensional space, not a plane. You can add to ou pair of equations the equation $x_1+3x_2+3x_3=3$, for instance.

Yes, your two vectors (I will call them $v_1$ and $v_2$) form an orthogonal basis. The projection of $v$ onto $\pi_0$ will be equal to$$\frac{\langle v,v_1\rangle}{\|v_1\|^2}v_1+\frac{\langle v,v_2\rangle}{\|v_2\|^2}v_2.$$

Answer 2

We have $$ \begin{align} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{pmatrix} &= \begin{pmatrix} 1 \\ 0 \\ 2 \\ 0 \\ 1 \end{pmatrix} +s \left( \begin{pmatrix} 0 \\ 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} - \begin{pmatrix} 1 \\ 0 \\ 2 \\ 0 \\ 1 \end{pmatrix} \right) +t \left( \begin{pmatrix} 2 \\ 0 \\ 1 \\ 0 \\ 2 \end{pmatrix} - \begin{pmatrix} 1 \\ 0 \\ 2 \\ 0 \\ 1 \end{pmatrix} \right) \\ &= \begin{pmatrix} 1 \\ 0 \\ 2 \\ 0 \\ 1 \end{pmatrix} +s \begin{pmatrix} -1 \\ 1 \\ -2 \\ 1 \\ -1 \end{pmatrix} + t \begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \\ 1 \end{pmatrix} \\ &= \begin{pmatrix} 1 -s + t\\ s \\ 2-2s-t \\ s \\ 1-s+t \end{pmatrix} \end{align} $$ so we agree.

We need to extract three linear independent equations to describe the two-dimensional plane.

We have $s = x_2$ and $2 x_1 - x_3 = 3t$ so we get $$ x_3 = 2 - 2s - t = 2 - 2 x_2 - 2/3 x_1 + 1/3 x_3 \iff \\ 2/3 x_1 + 2 x_2 + 2/3 x_3 = 2 \iff \\ 2 x_1 + 6 x_2 + 2 x_3 = 6 $$ and thus $$ \begin{align} x_1 - x_5 &= 0 \\ x_2 - x_4 &= 0 \\ 2 x_1 + 6 x_2 + 2 x_3 &= 6 \end{align} $$