# Intersection of $2$ planes.

Find the intersection of a line formed by the intersection of two planes $$\vec r . \vec n_1 = p_1$$ and $$\vec r . \vec n_2=p_2$$.

I know that the line would be along $$(\vec n_1 \times \vec n_2)$$. So i need a point on the line to get the equation. I assumed a point $$C$$ such that $$\vec {OC}$$ is perpendicular to the line of intersection. I dont really know how to proceed from here. Do I have to use the equations $$\vec c . \vec n_1 = p_1$$ and $$\vec c . \vec n_2=p_2$$?

• You can use 'trial and error' (should be easy) to simply find a point that lies on both planes) if you know the equations for the two planes, ie pick values for the vector components and make sure they satisfy both equations. Nov 13, 2018 at 21:40
• @AnyAD There arent any equations given, this was the entire question. Nov 13, 2018 at 21:42
• What form of equation do you want to end up with?
– amd
Nov 13, 2018 at 21:57
• @amd Possibly a vector equation where i have used only $\vec n_1$,$\vec n_2$, their magnitudes i.e. |$\vec n_1$| and |$\vec n_1$| , $p_1$ and $p_2$. Nov 13, 2018 at 22:03
• @amd The equation would be something like $\vec c + k(\vec n_1 \times \vec n_2)$ where $k$ is an arbitrary constant. Just need a way to write $\vec c$ in the form of whatever I mentioned in my previous comment. Nov 13, 2018 at 22:06

You know that the line is of the form $$t(n_1\times n_2)+p$$ for some point $$p$$. Now to find $$p$$, we can assume it is of the form $$an_1+bn_2$$ (since the line is perpendicular to $$n_1$$ and $$n_2$$, so it must pass through the plane spanned by them somewhere). Then since $$p$$ is in both planes, we have the equations $$p\cdot n_1=a+bn_1\cdot n_2=p_1$$, and $$p\cdot n_2=an_1\cdot n_2 + b=p_2$$. This gives us the system of equations $$\newcommand\bmat{\begin{pmatrix}}\newcommand\emat{\end{pmatrix}}\bmat 1 & n_1\cdot n_2 \\ n_1\cdot n_2 & 1 \emat \bmat a \\ b\emat = \bmat p_1\\p_2\emat.$$ The determinant of the matrix is $$1-(n_1\cdot n_2)^2$$, and since $$n_1$$ and $$n_2$$ are not parallel (since the planes intersect in a line, so they are not themselves parallel), this is positive. Hence we can invert the matrix to get $$\bmat a \\ b \emat = \frac{1}{1-(n_1\cdot n_2)^2}\bmat 1 & -n_1\cdot n_2 \\ -n_1\cdot n_2 & 1\emat\bmat p_1\\p_2\emat,$$ or letting $$n_1\cdot n_2 = \alpha$$, $$a = \frac{p_1-\alpha p_2}{1-\alpha^2},$$ and $$b=\frac{p_2-\alpha p_1}{1-\alpha^2}.$$