Question

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

Answer 1

Hint: Direction cosines of a line are the cosines of the angles made by the line with positive directions of the coordinate axes. The numbers that are proportional to the direction cosines of a line are called its Direction ratios. We generally denote these direction cosines as $l$ , $m$ , $n$.

Formula used: Direction cosines of a line joining points $({x_1},{y_1},{z_1})$ and $({x_2},{y_2},{z_2})$ are given by
$\dfrac{{({x_2} - {x_1})}}{{\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} }}$, $\dfrac{{({y_2} - {y_1})}}{{\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} }}$, $\dfrac{{({z_2} - {z_1})}}{{\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2} + {{({z_2} - {z_1})}^2}} }}$.

Complete step-by-step solution:
Let ${l_1}$ , ${m_1}$, ${n_1}$ and ${l_2}$ , ${m_2}$, ${n_2}$ be the direction cosines of the two specified lines ${L_1}$ and ${L_2}$ .
Let $\widehat {{n_1}}$ and $\widehat {{n_2}}$ be the unit vectors corresponding to these lines ${L_1}$ and ${L_2}$ .
$\widehat {{n_1}}$ and $\widehat {{n_2}}$ can be expressed as:
$\overrightarrow {{n_1}} = {l_1}\widehat i + {m_1}\widehat j + {n_1}\widehat k$ and $\overrightarrow {{n_2}} = {l_2}\widehat i + {m_2}\widehat j + {n_2}\widehat k$
As the lines ${L_1}$ and ${L_2}$ are mutually perpendicular to each other so, there unit vectors will also be perpendicular to one another hence, we can write
$\widehat {{n_1}} \times \widehat {{n_2}} = \left| {\widehat {{n_1}}} \right| \cdot \left| {\widehat {{n_2}}} \right|\sin 90^\circ \cdot \widehat n$
$\widehat {{n_1}} \times \widehat {{n_2}} = \widehat n$
Now, we will find the cross product to get the direction cosines of the line perpendicular to both ${L_1}$ and ${L_2}$ i.e., $\widehat n$
$\widehat {{n_1}} \times \widehat {{n_2}} = \widehat n = \left| {\begin{array}{*{20}{c}}
  {\widehat i}&{\widehat j}&{\widehat k} \\
  {{l_1}}&{{m_1}}&{{n_1}} \\
  {{l_2}}&{{m_2}}&{{n_2}}
\end{array}} \right|$
On solving this determinant for the direction cosines of the perpendicular line $\widehat n$ , we get
$\widehat n = ({m_1}{n_2} - {m_2}{n_1})\widehat i - ({l_1}{n_2} - {l_2}{n_1})\widehat j + ({l_1}{m_2} - {l_2}{m_1})\widehat k$

Hence, the direction cosines are $({m_1}{n_2} - {m_2}{n_1})$ , $({n_1}{l_2} - {n_2}{l_1})$, $({l_1}{m_2} - {l_2}{m_1})$.

Note: If ${l_1}$ , ${m_1}$, ${n_1}$ and ${l_2}$ , ${m_2}$, ${n_2}$ are the direction cosines of two lines and $\theta $ is the angle acute angle between the two lines, then
$\cos \theta = \left| {{l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2}} \right|$
If ${a_1}$ , ${b_1}$ , ${c_1}$ and ${a_2}$ , ${b_2}$ , ${c_2}$ are the direction ratios of two lines and $\theta $ is the angle acute angle between the two lines, then
$\cos \theta = \left| {\dfrac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {a_2}^2} \cdot \sqrt {{b_1}^2 + {b_2}^2} \cdot \sqrt {{c_1}^2 + {c_2}^2} }}} \right|$ .