I am given the equations of 2 thin planes in a rock.
Plane 1: $\mathbf r\cdot(0,3,-1) = -1$
Plane 2: $\mathbf r\cdot(1,0,1) = 1$
The entire set (the rock) is cut by the plane: $\mathbf r\cdot(-1,3,2) = 0$
I guess that these plane equations are int he format $\mathbf r\cdot\mathbf n =c$ where $\mathbf n$ is the vector of the normal to the plane, $\mathbf r$ is the vector to some point in the plane and $c$ a constant.
I am to find the angle between these which is meant to be given by $$\cos(\text{angle}) = -\frac7{\sqrt 3\cdot\sqrt{91}}$$
I am not sure how to prove this. I would have though that you would take the dot product of the normal of planes 1 & 2 to get this equation but that doesn't work out.
If a plane $\mathcal{P}$ is defined by an equation
$$\mathbf{r}\cdot\mathbf{n}=a$$
with $\textbf{r}$ the points of the plane, then each line which is definded by two points in the plane is perpendicular to $\textbf{n}$.
$$ \textbf{r}_1\cdot\textbf{n}=a,\quad\textbf{r}_2\cdot\textbf{n}=a,\quad\rightarrow\quad(\textbf{r}_1-\textbf{r}_2)\cdot\textbf{n}=0$$
Since two planes, intersect in a line, this line must be orthogonal to both normals of both planes. The intersection line has thus the direction $\textbf{n}_1\times\textbf{n}_2$.
In the original problem you are asked to compute the angle between two intersecting lines. The directions of those two lines are:
$$\textbf{s}_1=\begin{pmatrix}0\\3\\-1\end{pmatrix}\times\begin{pmatrix}-1\\3\\2\end{pmatrix}=\begin{pmatrix}9\\1\\3\end{pmatrix}$$ $$\textbf{s}_2=\begin{pmatrix}1\\0\\1\end{pmatrix}\times\begin{pmatrix}-1\\3\\2\end{pmatrix}=\begin{pmatrix}-3\\-3\\3\end{pmatrix}$$
The vector angle can now be computed with the dotproduct:
$$\cos \theta=\frac{\textbf{s}_1\cdot\textbf{s}_2}{\left|\textbf{s}_1\right|\left|\textbf{s}_2\right|}=-\sqrt\frac{7}{39}$$