The question is :- if $l_1$, $m_1$, $n_1$ and $l_2$, $m_2$, $n_2$
are the direction cosines of two mutually perpendicular
lines, show that the direction cosines of the line perpendicular to both of these
are ( $m_1n2-m_2n_1),(n_1l_2-n_2l_1),(l_1m_2-l_2m_1$).
I know that for two mutually perpendicular lines
$$l_1l_2+m_1m_2+n_1n_2=0.$$
But I don't know the further what to do
Please can anyone guide me further?
Thank you.
Best Answer
Let us take $u_1=(l_1,m_1,n_1)$ as the unit vector along one line.
$u_2=(l_2,m_2,n_2)$ as the unit vector along another line.
Since $u_1\times u_2$ is a unit vector perpendicular to both $u_1$ and $u_2$, we can calculate $u_1\times u_2$ $$u_1\times u_2=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix}$$ $$=\vec{i}(m_1n_2-m_2n_1)-\vec{j}(n_2l_1-n_1l_2)+\vec{k}(l_1m_2-l_2m_1)$$ $$=\vec{i}(m_1n_2-m_2n_1)+\vec{j}(n_1l_2-n_2l_1)+\vec{k}(l_1m_2-l_2m_1)$$ Therefore, the required direction cosines are $(m_1n_2-m_2n_1),(n_1l_2-n_2l_1),(l_1m_2-l_2m_1)$