If ${l_1},{m_1},{n_1}$ and ${l_2},{m_2},{n_2}$ are the direction cosines of two mutually perpendicular lines, then show that the direction cosines of the line perpendicular to both of these are ${m_1}{n_2} - {m_2}{n_1},{n_1}{l_2} - {n_2}{l_1},{l_1}{m_2} - {l_2}{m_1}.$

.: Let $< l,m,n >$ be the direction cosines of the line which is perpendicular to two lines whose direction cosines are
$< {l_1},{m_1},{n_1} >$ and $< {l_2},{m_2},{m_2} >$

$\therefore$ $l{l_1} + m{m_1} + n{n_1} = 0$ ..(1)

and $l{l_2} + m{m_2} + n{n_2} = 0$ ...(2)

Solving (1) and (2) by cross-multiplication method,

we get
$\cfrac{l}{{{m_1}{n_2} - {m_2}{n_1}}} = \cfrac{m}{{{n_1}{l_2} - {n_2}{l_1}}} = \cfrac{n}{{{l_1}{m_2} - {l_2}{m_1}}}$

$= \cfrac{{\sqrt {{l^2} + {m^2} + {n^2}} }}{{\sqrt {{{({m_1}{n_2} - {m_2}{n_1})}^2} + {{({n_1}{l_2} - {n_2}{l_1})}^2} + {{({l_1}{m_2} - {l_2}{m_1})}^2}} }}$

$= \cfrac{1}{{\sin \theta }} = \cfrac{1}{{\sin 90^\circ }} = 1$

Hence, $l = {m_1}{m_2} - {m_2}{n_1},m = {n_1}{l_2} - {n_2}{l_1},n = {l_1}{m_2} - {l_2}{m_1}$