Direction Cosines of the a line perpendicular to two lines 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.
 A: $\def\ca{\cos\alpha}
\def\cb{\cos\beta}
\def\cc{\cos\gamma}
\def\l{\lambda}
\def\det{\mathrm{det}\,}
\def\VA{{\bf A}}$Let 
$$A = \left(\begin{array}{ccc}
\ca & \cb & \cc \\
\ca' & \cb' & \cc' \\
\ca'' & \cb'' & \cc''
\end{array}\right).$$
Denote the $i$th row by $\VA_i$ and let $\l = \det A$. 
Then $|\l|$ is the volume of the parallelepiped defined by $\VA_1$, $\VA_2$, and $\VA_3$.
Since $\VA_1$ is perpendicular to the other vectors and has unit magnitude, $|\l|$ is the area of the parallelogram defined by the vectors
$\VA_2$ and $\VA_3$.
Therefore, $|\l| = \sin\theta$, where $\theta$ is the angle between the two vectors.
Note in particular that $|\l|$ does not depend on $\alpha$, $\beta$, or $\gamma$.
This argument could be written to avoid the terminology of vectors, but it would be cumbersome.
By Cramer's rule
$$\begin{eqnarray*}
\ca &=& \frac{%
\left|\begin{array}{ccc}
1 & \cb & \cc \\
0 & \cb' & \cc' \\
0 & \cb'' & \cc''
\end{array}\right|}{%
\left|\begin{array}{ccc}
\ca & \cb & \cc \\
\ca' & \cb' & \cc' \\
\ca'' & \cb'' & \cc''
\end{array}\right|}
\end{eqnarray*}$$
and so 
$$\ca = \frac{1}{\l}(\cb'\cc''-\cb''\cc').$$
A: In problems like this vector cross product works nicely.
Recall that cross product of two vectors gives a vector perpendicular to both

You can take
$$a=(\cos \alpha',\cos \beta',\cos \gamma')$$
$$b=(\cos \alpha'',\cos \beta'',\cos \gamma'')$$
$$c=(\lambda\cos \alpha,\lambda\cos \beta,\lambda\cos \gamma)$$
(here $\lambda$ is just some constant dependent on $\theta$)
and then take $$c=a\times b$$
Evaluate the determinant and you get answer immediately.
Also if you have any doubt take a look at wiki page of Vector cross product.
