How to calculate the intersection of two planes?

Solution 1:

You need to solve the two equations $$ x + 2y + z - 1 = 0 \\ 2x + 3y - 2z + 2 = 0. $$

Notice that, these are two equations in three variables, so you have a free variable say $z=t$, then we have

$$ x + 2y = 1-t \\ 2x + 3y = 2t-2. $$

Solving the last system gives

$$ \left\{ x=-7+7\,t,y=4-4\,t \right\} .$$

Then the parametrized equation of the line is given by

$$ (x,y,z)= (-7+7t, 4-4t,t)=(-7,4,0)+(7,-4,1)t . $$

Solution 2:

The first plane has normal vector $\begin{pmatrix}1\\2\\1\end{pmatrix}$ and the second has normal vector $\begin{pmatrix}2\\3\\-2\end{pmatrix}$, so the line of intersection must be orthogonal to both of these. We know that the unique vector orthogonal to two linearly independent vectors $v_1,v_2$ is $v_1\times v_2$, so the direction vector of the line of intersection is $$\begin{pmatrix}1\\2\\1\end{pmatrix}\times \begin{pmatrix}2\\3\\-2\end{pmatrix}=\begin{pmatrix}-7\\4\\-1\end{pmatrix}$$Next, we need to find a particular point on the line. Here, since a line that isn't parallel to any coordinate plane passes through all three, you can check if it is parallel to one by using the directional vector. Since here, the line passes through all three planes, we can try $y=0$(since it passes through the $x-z$ plane) and solve the resulting system of linear equations:$$\begin{align}x+z-1&=&0\\2x-2z+2&=&0\end{align}$$ giving $x=0, z=1$, thus the line of intersection is $\lbrace{\begin{pmatrix}-7t\\4t\\1-t\end{pmatrix}:t\in \Bbb R\rbrace}$

Solution 3:

While this problem has a great textbook answer, as @walcher explained, I don't think it's very elegant. This is because, the solution depends on picking an arbitrary point, which lacks geometric intuition. Ideally, we'd like this point to have some meaning, such as being close to the planes, or the line or etc.

For that, I'd like to remind you of a solution by John Krumm, which remains unnoticed by many. Let $\mathbf{p}=\{p_x,p_y,p_z\}$ and $\mathbf{n}=\{n_x,n_y,n_z\}$ compose the plane $\mathbf{P}=\{\mathbf{n},\mathbf{p}\}$. Let there be two planes $P_1$ and $P_2$, for which we'd like to compute the intersecting line $\mathbf{l}$. It's trivial to compute the direction as the cross-product: $$\mathbf{l}_d=\mathbf{n}_1 \times \mathbf{n}_2$$

If we additionally desire that the resulting point $\mathbf{p}$ is as close to the chosen point $\mathbf{p}_0$ as possible, we could write a distance : $$\lVert \mathbf{p}-\mathbf{p}_0 \rVert = (p_x-p_{0x})^2 + (p_y-p_{0y})^2 + (p_z-p_{0z})^2$$ Incorporating the other points in the similar fashion, and writing this constraint using Lagrange multipliers into an objective function results in: $$J=\lVert \mathbf{p}-\mathbf{p}_0 \rVert+\lambda(\mathbf{p}-\mathbf{p}_1)^2 + \mu(\mathbf{p}-\mathbf{p}_2)^2$$

Using the standard Lagrange framework (omitting the details), one establishes a nice matrix, in the form: $$ \mathbf{M}= \left[ {\begin{array}{ccc} 2 & 0 & 0 & n_{1x} & n_{2x}\\ 0 & 2 & 0 & n_{1y} & n_{2y}\\ 0 & 0 & 2 & n_{1z} & n_{2z} \\ n_{1x} & n_{1y} & n_{1z} & 0 & 0\\n_{2x} & n_{2y} & n_{2z} & 0 & 0 \end{array} } \right] $$ This matrix can now be used in a system of linear equations to solve for the unknown point, $\mathbf{p}$, as well as the Lagrange multipliers, $\{\lambda, \mu\}$.: $$ \mathbf{M}\left[ {\begin{array}{c} p_x \\ p_y \\ p_z \\ \lambda \\ \mu \end{array} } \right] = \left[ {\begin{array}{c} 2p_{0x} \\ 2p_{0y} \\ 2p_{0z} \\ \mathbf{p}_1 \cdot \mathbf{n}_1 \\ \mathbf{p}_2 \cdot \mathbf{n}_2 \end{array} } \right] $$ While the multipliers are not of particular interest they would be interesting for understanding configuration of points, or for different parameterizations.

I think this is a pretty neat approach giving a nice and simple method, with a geometrically interpretable results. I post the MATLAB code at my blog.