$l_{1}, m_{1}, n_{1}, a n d l_{2}, m_{2}, n_{2}$ be the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of them are (m $_{1}$ n $_{2}$ -m $_{2}$ n $_{1}$ ),(n $_{1}$ l $_{2}$ -n $_{2}$ l $_{1}$ ),(l $_{1}$ m $_{2}$ -l $_{2}$ m $_{1}$ )

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Solution

$l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = 0$ __ (c), $l_{1}^{2} + m_{1}^{2} + n_{1}^{2} = 1$ ___ (a) $\Rightarrow l_{2}^{2} + m_{2}^{2} + n_{2}^{2} = 1$ __ (b)
then dc's of reqd line : let $(l m n) \Rightarrow l^{2} + m^{2} + n^{2} = 1$
Now APC
$l l_{1} + m m_{1} + n n_{1} = 0$
$l l_{2} + m m^{2} + n n_{2} = 0$
$\frac{l}{m_{1} n_{2} - m_{2} n_{1}} = \frac{m}{- l_{1} n_{2} + n_{1} l_{2}} = \frac{n}{l_{1} n_{2} - m_{1} l_{2}}$
If $\frac{x}{p} = \frac{y}{q} = \frac{z}{r}$ then it is also equal to
$\frac{\sqrt{x^{2} + y^{2} + z^{2}}}{\sqrt{p^{2} + q^{2} + r^{2}}}$
so
$\frac{l}{m_{1} n_{2} - m_{2} n_{1}} = \frac{m}{n_{1} m_{2} - l_{1} m_{2}} = \frac{n}{l_{1} m_{2} - l_{2} m_{1}} = \frac{\sqrt{l^{2} + m^{2} + n^{2}}}{\sqrt{(m_{1} n_{2} - m_{2} n_{1})^{2} + (n_{1} l_{2} - l_{1} n_{2})^{2} + (l_{1} m_{2} - l_{2} m_{1})^{2}}}$
$= \frac{1}{\sqrt{(l_{1}^{2} + m_{1}^{2} + n^{2})^{2} (l_{2} + m_{2} + n_{2})^{2} - (l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2})^{2}}} \Leftarrow \frac{1}{\sqrt{(m_{1} n_{2} - m_{2} n_{1})^{2} + (n_{1} l_{2} - l_{1} n_{2})^{2} + (l_{1} m_{2} - l_{2} m_{1})^{2}}}$
$\Rightarrow$ from $e q^{n}$ (a) (b) & (c)
$\Rightarrow \frac{l}{m_{1} n_{2} - m_{2} n_{1}} = \frac{m}{n_{1} l_{2} - l_{1} n_{2}} = \frac{n}{l_{1} m_{2} - m_{1} l_{2}} = 1$
$\Rightarrow l = m_{1} n_{2} - m_{2} n_{1}$ $m = n_{1} l_{2} - l_{1} n_{2}$ $n = l_{1} m_{2} - m_{1} l_{2}$

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