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 $m_{1} n_{2}-m_{2} n_{1}, n_{1} l_{2}-n_{2} l_{1}, l_{1} m_{2}-l_{2} m_{1}$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) It is given that $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ are the direction cosines of two mutually perpendicular lines. Therefore,
$l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}=0$ ........$(1)$
$l_{1}^{2}+m_{1}^{2}+n_{1}^{2}=1$ ..........$(2)$
$l_{2}^{2}+m_{2}^{2}+n_{2}^{2}=1$ ...........$(3)$
Let $l, m, n$ be the direction cosines of the line which is perpendicular to the line with direction cosines $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$.
$\therefore l l_{1} + m m_{1} + n n_{1} = 0$
$l l_{2} + m m_{2} + n n_{2} = 0$
$\therefore \frac{l}{m_{1} n_{2} - m_{2} n_{1}} = \frac{m}{n_{1} l_{2} - n_{2} l_{1}} = \frac{n}{l_{1} m_{2} - l_{2} m_{1}}$
$\Rightarrow \frac{l^{2}}{(m_{1} n_{2} - m_{2} n_{1})^{2}} = \frac{m^{2}}{(n_{1} l_{2} - n_{2} l_{1})^{2}} = \frac{n^{2}}{(l_{1} m_{2} - l_{2} m_{1})^{2}}$
$= \frac{l^{2} + m^{2} + n^{2}}{(m_{1} n_{2} - m_{2} n_{1})^{2} + (n_{1} l_{2} - n_{2} l_{1})^{2} + (l_{1} m_{2} - l_{2} m_{1})^{2}}$ .........$(4)$
Since $l, m, n$ are the direction cosines of the line,$l^{2} + m^{2} + n^{2} = 1$ ........$(5)$
Using the Lagrange identity:
$(l_{1}^{2} + m_{1}^{2} + n_{1}^{2})(l_{2}^{2} + m_{2}^{2} + n_{2}^{2}) - (l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2})^{2} = (m_{1} n_{2} - m_{2} n_{1})^{2} + (n_{1} l_{2} - n_{2} l_{1})^{2} + (l_{1} m_{2} - l_{2} m_{1})^{2}$
From $(1), (2),$ and $(3),$ we obtain:
$1 \cdot 1 - 0^{2} = (m_{1} n_{2} - m_{2} n_{1})^{2} + (n_{1} l_{2} - n_{2} l_{1})^{2} + (l_{1} m_{2} - l_{2} m_{1})^{2} = 1$ .........$(6)$
Substituting $(5)$ and $(6)$ in $(4),$ we get:
$\frac{l^{2}}{(m_{1} n_{2} - m_{2} n_{1})^{2}} = \frac{m^{2}}{(n_{1} l_{2} - n_{2} l_{1})^{2}} = \frac{n^{2}}{(l_{1} m_{2} - l_{2} m_{1})^{2}} = 1$
Thus,$l = m_{1} n_{2} - m_{2} n_{1}, m = n_{1} l_{2} - n_{2} l_{1}, n = l_{1} m_{2} - l_{2} m_{1}$.

Explore More

Similar Questions

Find the position vector of a point $A$ in space such that $\overrightarrow{OA}$ is inclined at $60^{\circ}$ to $OX$ and at $45^{\circ}$ to $OY$ and $|\overrightarrow{OA}|=10$ units.

Difficult
View Solution

The direction cosines of the line passing through $P(2, 3, -1)$ and the origin $O(0, 0, 0)$ are:

If the angles made by a straight line with the coordinate axes are $\alpha, \frac{\pi}{2}-\alpha, \beta$, then $\beta$ is equal to

$\text{Assertion (A)}$: The direction ratios of line $L_1$ are $2, 5, 7$ and those of line $L_2$ are $\frac{4}{\sqrt{19}}, \frac{10}{\sqrt{19}}, \frac{14}{\sqrt{19}}$. The lines $L_1, L_2$ are parallel.
$\text{Reason (R)}$: The direction ratios of a line $L_1$ are $a_1, b_1, c_1$ and those of another line $L_2$ are $a_2, b_2, c_2$. The lines $L_1$ and $L_2$ are parallel if $a_1 a_2+b_1 b_2+c_1 c_2=0$.
The correct option among the following is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo