If $\alpha' ,\beta' ,\gamma'$ and $\alpha'' ,\beta'' ,\gamma'' $ are the direction angles of two lines, we have to find $\alpha ,\beta ,\gamma $ such that they are the direction angles of a third line perpendicular to both.
MY SOLUTION
I understand there are three equations like this:
- $\sum \cos \alpha\cdot \cos \alpha ' = 0 $
- $\sum \cos \alpha\cdot \cos \alpha '' = 0 $
- $\sum \cos \alpha\cdot \cos \alpha = 1 $
$\begin{bmatrix} \cos \alpha & \cos \beta & \cos \gamma\\ \cos \alpha' & \cos \beta' & \cos \gamma' \\ \cos \alpha'' & \cos \beta'' & \cos \gamma'' \end{bmatrix}\begin{bmatrix} \cos \alpha\\ \cos \beta\\ \cos\gamma \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$
By Cramer's rule, I arrive at: $\cos \alpha = \frac{\cos \beta'}{\cos \beta' \cos \gamma'' - \cos \beta'' \cos \gamma'}.$
But the answer given is somewhat different, given as $\lambda \cdot \cos \alpha = \cos \beta' \cos \gamma'' - \cos \beta'' \cos \gamma'.$
Where am I going wrong?
EDIT
The RHS should be $[1, 0, 0]^T$ rather than $[0, 0, 1]^T$ as pointed by Inquest. But my answer eludes me even more.