# Locus perpendicular to a plane in $\mathcal{R}^4$

I have solved an exercise but I'm not sure to have solved it perfectly. Could you check it? It's very important for me..

In $\mathcal{R}^4$ I have a plane $\pi$ and a point P. I have to find the locus of Q points such that line PQ is perpendicular to $\pi$.

$$\pi:\begin{cases} 3x+y-z-q+1=0\\ -x-y+z+2q=0 \end{cases}$$

$P=(0,1,1,0)$

$P$ doesn't belong to $\pi$ because it doesn't satisfy first equation.

If $PQ$ line has to be $\perp$ to $\pi$, the vector of direction of $PQ$ must be perpendicular to spanners of the plane.

Spanners of plane are: $v_1=(1, -5, 0, -2)$ and $v_2=(0, 1, 1, 0)$

Then, $PQ \cdot v_1=0$ and $PQ\cdot v_2=0$

But I can write $PQ=OQ-OP$

and so I obtain $\begin{cases} (OQ-OP)\cdot v_1=0 \\(OQ-OP)\cdot v_2=0\end{cases}$

But inner product is a bilinear form and so I can also write:

$\begin{cases} OQ\cdot v_1 - OP\cdot v_1=0 \\OQ\cdot v_2-OP\cdot v_2=0\end{cases}$

I can calculate $OP\cdot v_1$ and $OP\cdot v_2$ and, if $OQ=(x, y, z, q)$, I obtain: $\begin{cases} (x, y, z, q) \cdot (1, -5, 0, -2)=(0,1,1,0) \cdot (1,-5,0,-2) \\ (x,y,z,q) \cdot (0,1,1,0)=(0,1,1,0,) \cdot (0,1,1,0) \end{cases}$

$\begin{cases}x-5y-2q=-5 \\ y+z=2 \end{cases}$

I have solved this equations set and I have obtained that $OQ= t(1, 1/5, -1/5, 0) + s(0,-2/5, 2/5, 1) + (0, 1, 1, 0)$ (with $t, s$ belonging to $\mathbb {R}$).

And so I can say that my locus is given by $Q$ points having coordinates given by $t(1, 1/5, -1/5, 0) + s(0,-2/5, 2/5, 1) + (0, 1, 1, 0)$.

My locus is given by vectors perpendicular to $\pi$ and so I can say that my locus is the orthogonal complement of $\pi$, so it is a plane.

Is it all correct? Please, signal me all things that you think are wrong. Thank you!

EDIT:

I can develope the exercise in another way saying that if $PQ$ has to be perpendicular to $\pi$, $PQ$ has to be perpendicular to any vectors of $\pi$ and so $PQ$ belongs to orthogonal complement of $\pi$, so PQ is given by: $PQ=t(w_1)+s(w_2)$ where $w_1$ and $w_2$ are spanners of othogonal complement of $\pi$. Now I can obtain that $Q=P+t(w_1)+s(w_2)$. Is it correct? Thanks again

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@MTurgeon But I can write the plane as $\pi= P_0+V$ where V is the vector subspace solution of the homogeneous system associated to $\pi$ and $P_0$ is any point belonging to the plane.. isn't it? – sunrise Sep 26 '12 at 15:26
Yes, of course, this is why I removed my comment – M Turgeon Sep 26 '12 at 15:28
I cannot find an error. – celtschk Sep 26 '12 at 15:29
@celtschk Thanks a lot! :) I have updated the question. Could you check it out again? Many thanks! – sunrise Sep 27 '12 at 5:37
Your edit is fine, too. – celtschk Sep 30 '12 at 11:15

But, you jumped one step: how did you find your $v_1$ and $v_2$ spanner vectors of $\pi$?
I know that vectors of $\pi$ are perpendicular to (3, 1, -1, -1) and (-1,-1,1,2), so I solved the system associated to these two conditions.. :) I haven't written those steps because I'm sure that they are OK :) thank you! – sunrise Sep 26 '12 at 15:39
Still seems fine (didn't check every calculating in much detail). One more remark: when $t$ and $s$ can be arbitrary real numbers, you can rescale to avoid fractions, so instead of $(1, 1/5, -1/5, 0)$, you can use $(5,1,-1,0)$... – Berci Sep 27 '12 at 9:43
Thanks Berci, my last doubt is: when I talk about spanners of plane, I'm talking of free (geometric) vectors that span the vector space associated to this affine space 2-dimensional. But $OP$ is not free vector, because it is applied.. So, I can do inner product between $OP$ and spanners of plane, only if apply spanners in O, turning them into applied vectors.. Isn't it? – sunrise Sep 30 '12 at 15:36