Problem source

The points $A,B$ and $C$ lie on the surface of the ground, which is an inclined plane. The point $B$ is 100m due north of $A,$ and $C$ is 60m due east of $B$. The vertical displacements from $A$ to $B,$ and from $B$ to $C$, are each 5m downwards. A plane coal seam lies below the surface and is to be located by making vertical bore-holes at $A,B$ and $C$. The bore-holes strike the coal seam at 95m, 45m and 76m below $A,B$ and $C$ respectively. Show that the coal seam is inclined at $\cos^{-1}(\frac{4}{5})$ to the horizontal. 
The coal seam comes to the surface along a line. Find the bearing of this line.

Solution source

Set up a coordinate system so that $x$ is E-W, $y$ is N-S and $z$ is the vertical direction. Also assume $B$ is the origin, then, $A = \begin{pmatrix} 0 \\ -100 \\ 5\end{pmatrix}, B = \begin{pmatrix} 0 \\ 0 \\ 0\end{pmatrix}, C= \begin{pmatrix} 60 \\ 0\\ -5\end{pmatrix},$.

The coal seam has points: $\begin{pmatrix} 0 \\ -100 \\ -90\end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ -45\end{pmatrix}, \begin{pmatrix} 60 \\ 0\\ -81\end{pmatrix},$

Therefore we can find the normal to the coal seam:

\begin{align*}
\mathbf{n} &= \left (\begin{pmatrix} 0 \\ -100 \\ -90\end{pmatrix} -   \begin{pmatrix} 0 \\ 0 \\ -45\end{pmatrix}\right ) \times  \left ( \begin{pmatrix} 60 \\ 0\\ -81\end{pmatrix} -   \begin{pmatrix} 0 \\ 0 \\ -45\end{pmatrix}\right ) \\
&= \begin{pmatrix} 0 \\ - 100 \\ -45\end{pmatrix} \times \begin{pmatrix} 60 \\ 0 \\ -36\end{pmatrix} \\
&= \begin{pmatrix} 3600 \\ -60 \cdot 45 \\ 60 \cdot 100 \end{pmatrix} \\
&= 300\begin{pmatrix} 12 \\ -9 \\ 20\end{pmatrix}
\end{align*}

To measure the incline $\theta$ to the horizontal we can take a dot with $\hat{\mathbf{k}}$, to see:

\begin{align*}
\cos \theta &= \frac{20}{\sqrt{12^2+(-9)^2+20^2} \sqrt{1^2+0^2+0^2}} \\
&= \frac{20}{25} \\
&= \frac{4}{5}
\end{align*}

Therefore the angle is $\cos^{-1} \tfrac 45$

The equation of the seam is $12x - 9y + 20z = -900$.

The equation of the surface is $5x + 3y + 60z = 0$

We can compute the direction of the overlap again with a cross product:

\begin{align*}
\mathbf{d} &= \begin{pmatrix} 12 \\ -9 \\ 20\end{pmatrix} \times \begin{pmatrix} 5 \\ 3 \\ 60\end{pmatrix} \\
&= \begin{pmatrix} -600 \\ -620 \\ 81 \end{pmatrix}
\end{align*}

To get the bearing of this vector we just need to look at the $x$ and $y$ components, so it will be $\tan^{-1} \frac{600}{620} = \tan^{-1} \frac{30}{31}$