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Please, can you check the following execution is correct:

Problem text

I have a plane in affine space in $\Bbb R^4$ described by two following equations:

\begin{Bmatrix}3x+y-z-q +1=0\\ -x-y+z+2q=0\end{Bmatrix}

I have the coords of a point $P$: (0,1,1,0) Describe the locus of $Q$ points such that line $PQ$ is $\bot$ to the plane.

My solution

Now I look for the line (which $Q$ points belong to) perpendicular to the plane and passing trough point $P$. In order for a plane described by equation $ax+by+cz+dq+e=0$ and a line whose coefficients are $l,m,n,t$ to be perpendicular, this must be true:


So, the equations of line passing by generic point $P(x_0,y_0,z_0,q_0)$ is:


Then, in my specific problem I get:

$\begin{Bmatrix}x=y-1 \\ y=2-z \\ z=\frac{q}{2}+1 \end{Bmatrix}$

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up vote 2 down vote accepted

The two equations $$3x+y-z-q=0\ ,\qquad -x-y+z+2q=0$$ (I have omitted the "$+1$", as it amounts only to a translation and does not affect directions) determine two hyperplanes (to be exact: $3$-dimensional subspaces) $U_1$, $U_2$ in ${\mathbb R}^4$. Their intersection is a two-dimensional subspace $U\subset{\mathbb R}^4$. The lines $P\vee Q$ should be orthogonal to this $U$. There is a fundamental principle in linear algebra which says: "The orthogonal complement of an intersection is the span of the orthogonal complements." The orthogonal complement of $U_1$ is the $1$-dimensional subspace $\langle a\rangle\subset{\mathbb R}^4$ spanned by $a:=(3,1,-1,-1)$, and the orthogonal complement of $U_2$ is the $1$-dimensional subspace $\langle b\rangle$ spanned by $b:=(-1,-1,1,2)$. The quoted principle then implies that $U^\perp$ consists of all linear combinations $x=\lambda a+\mu b$. As $\vec {PQ}$ has to be orthogonal to $U$ it follows that $\vec {PQ}$ has to be such a linear combination. Therefore the locus of the admissible $Q$ has the following parametrization: $$Q=(0,1,1,0)+\lambda(3,1,-1,-1)+\mu(-1,-1,1,2)\qquad\bigl((\lambda,\mu)\in{\mathbb R}^2\bigr)\ .$$

(There are two errors in your argument: (a) Two linear equations in ${\mathbb R}^4$ determine a two-dimensional subspace $U$. Its orthogonal complement $U^\perp$ is not a line, but again a two-dimensional plane, because $4-2=2$. (b) A vector which is orthogonal to the intersection $U_1\cap U_2$ does not have to be orthogonal to $U_1$ and $U_2$ separately.)

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First of all, huge thanks ;) I have a doubt: if I try to solve the system of equations (adding the eq and replacing), I get the following equation $-y+ z+ (5/2)q+ 1/2= 0$, so the normal vector to the plane should be $0, -1, 1, 5/2$. Finally, the line passing by my point is described by $L=(0,1,1,0)+ t(0,-1,1,5/2)$. Is it equivalent to your execution? Thanks again – Surfer on the fall Jun 29 '12 at 20:01
@user1294101: See my edit. – Christian Blatter Jun 30 '12 at 7:45
Hi, thank you very much for your very precise answer. Excuse me if I haven't replied but I've tried to study this argument to avoid you wasting time. In my Linear Algebra books, I haven't found the fundamental principle mentioned in your answer.. :( What's its name? – Surfer on the fall Jul 11 '12 at 8:54
@user1294101: It's exercice 1 of section 69: "Natural isomorphisms" in Halmos' Finite dimensional vector spaces. – Christian Blatter Jul 13 '12 at 8:46
Perfect, thanks.. How should I prove that principle? I've tried with no results.. can you help me? thanks again – Surfer on the fall Jul 17 '12 at 12:32

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