# I want to find 3 planes that each contain one and only one line from a set

The three lines intersect in the point $(1, 1, 1)$: $(1 - t, 1 + 2t, 1 + t)$, $(u, 2u - 1, 3u - 2)$, and $(v - 1, 2v - 3, 3 - v)$. How can I find three planes which also intersect in the point $(1, 1, 1)$ such that each plane contains one and only one of the three lines?

Using the equation for a plane $$a_i x + b_i y + c_i z = d_i,$$ I get $9$ equations.

Sharing equations with the lines:

$$a_1(1 - t) + b_1(1 + 2t) + c_1(1 + t) = d_1,$$ $$a_2(u) + b_2(2u - 1) + c_2(3u - 2) = d_2,$$ $$a_3(v - 1) + b_3(2v - 3) + c_3(3 - v) = d_3.$$

Intersection at $(1,1,1)$: $$a_1 + b_1 + c_1 = d_1,$$ $$a_2 + b_2 + c_2 = d_2,$$ $$a_3 + b_3 + c_3 = d_3.$$

Dot product of plane normals and line vectors is $0$ since perpendicular: $$\langle a_1, b_1, c_1 \rangle \cdot \langle -1, 2, 1 \rangle = -a_1 + 2b_1 + c_1 = 0,$$ $$\langle a_2, b_2, c_2\rangle \cdot \langle 1, 2, 3\rangle = a_2 + 2b_2 + 3c_2 = 0,$$ $$\langle a_3, b_3, c_3 \rangle \cdot \langle 1, 2, -1 \rangle = a_3 + 2b_3 - c_3 = 0.$$

I know how to find the intersection of $3$ planes using matrices/row reduction, and I know some relationships between lines and planes. However, I seem to come up with $12$ unknowns and $9$ equations for this problem. I know the vectors for the lines must be perpendicular to the normals of the planes, thus the dot product between the two should be $0$. I also know that the planes pass through the point $(1,1,1)$ and the $x,y,z$ coordinates for the parameters given in the line equations. What information am I missing? Maybe there are multiple solutions. If so, how can these planes be described with only a line and one point? Another thought was to convert the planes to parametric form, but to describe a plane with parameters normally I would have $2$ vectors and one point, but here I only have one vector and one point.

• Think about a simpler problem; the pooint $(0,0,0)$, and the $x$, $y$, and $z$ axes. Can you see that there are lots of planes that contain one of the axes and not the other two? And the intersection of a plane containing just the $x$-axis, a plane containing just the $y$, and a plane containing just the $z$, is just the origin? This suggests there are lots of solutions to your problem, and that you just try to find an example, instead of setting up equations to find all of them. – Gerry Myerson Jul 2 '12 at 5:22

Avoid writing down so many equations to be solved, but produce these planes in a forward motion: Your three lines $\ell_i$ $\ (1\leq i\leq3)$ can be given as $$\ell_i:\quad t\mapsto a+ t\ p_i \qquad(-\infty<t<\infty)$$ with $a=(1,1,1)$, $p_1=(-1,2,1)$, $p_2=(1,2,3)$, $p_3=(1,2,-1)$, where now all three lines pass the point $a$ at time $t=0$.

The plane $\pi_1$ with parametric representation $$\pi_1:\quad (u,v)\mapsto a + u p_1 +v{p_2+p_3\over 2}\qquad\bigl((u,v)\in{\mathbb R}^2\bigr)$$ contains the line $\ell_1$ and is transversal to $\ell_2$ and $\ell_3$. (In order to visualize this imagine that the $p_i$ are the three standard basis vectors.) Define $\pi_2$ and $\pi_3$ similarly.

It is easy to convert the parametric representations of the $\pi_i$ into equations, if desired.

A little background: The equation of a plane can be expressed as $\langle n, x \rangle = \pi$, where $n$ is a normal to the plane. If the plane contains the line $t \mapsto x_0 + t d$, then it is fairly straightforward to show that the plane contains the line iff $\langle n, x_0 \rangle = \pi$, and $\langle n, d \rangle = 0$.

Now let $d_1 = (-1,2-1)^T$, $d_2 = (1,2,3)^T$, $d_3 = (1,2 ,-1)^T$, and $e = (1,1,1)^T$. Then the three lines are $t \mapsto e + t d_i$, $i = 1,2,3$.

The problem is to find three normals $n_i$ such that $\langle n_i, d_j \rangle = 0$ iff $i = j$. If we can find these normals, then we can set $\pi_i = \langle n_i, e \rangle$, and the three planes will be given by $\langle n_i, x \rangle = \pi_i$.

Let $N = \begin{bmatrix}n_1 & n_2 & n_3\end{bmatrix}$, and $D = \begin{bmatrix}d_1 & d_2 & d_3\end{bmatrix}$. Then we want to find a matrix $N$ such that the following equation holds, where $*$ represents any non-zero number): $$N^T D = \Delta =\begin{bmatrix}0 & * & * \\ * & 0 & * \\ * & * & 0\end{bmatrix}.$$ Since $D$ is invertible (check!), the solution is given by $N = D^{-T} \Delta^{T}$, with the constants $\pi_i$ given by $(\pi_1, \pi_2, \pi_3)^T = N^T e$.

For example, choose $\Delta =\begin{bmatrix}0 & 4 & 4 \\ 4 & 0 & 4 \\ 4 & 4 & 0\end{bmatrix}$. Then $N = \begin{bmatrix}2 & -1 & -1 \\ 1 & 2 & 1 \\ 0 & -1 & 1\end{bmatrix}$, and $(\pi_1, \pi_2, \pi_3) = (3, 0, 1)$.