# A line ' l ' passing through origin is perpendicular to the lines

Question:

A line ' $l$ ' passing through origin is perpendicular to the lines

$l_{1}: \overrightarrow{\mathrm{r}}=(3+\mathrm{t}) \hat{\mathrm{i}}+(-1+2 \mathrm{t}) \hat{\mathrm{j}}+(4+2 \mathrm{t}) \hat{\mathrm{k}}$

$l_{2}: \overrightarrow{\mathrm{r}}=(3+2 \mathrm{~s}) \hat{\mathrm{i}}+(3+2 \mathrm{~s}) \hat{\mathrm{j}}+(2+\mathrm{s}) \hat{\mathrm{k}}$

If the co-ordinates of the point in the first octant

on ' $l_{2}$ ' at a distance of $\sqrt{17}$ from the point of intersection of ' $l$ ' and ' $l_{1}^{\prime}$ are (a, b, c), then 18(a $+b+c)$ is equal to________.

Solution:

$\ell_{1}: \overrightarrow{\mathrm{r}}=(3+\mathrm{t}) \hat{\mathrm{i}}+(-1+2 \mathrm{t}) \hat{\mathrm{j}}+(4+2 \mathrm{t}) \hat{\mathrm{k}}$

$\ell_{2}: \overrightarrow{\mathrm{r}}=(3+2 \mathrm{~s}) \hat{\mathrm{i}}+(3+2 \mathrm{~s}) \hat{\mathrm{j}}+(4+\mathrm{s}) \hat{\mathrm{k}}$

DR of $\ell_{1} \equiv(1,2,2)$

DR of $\ell_{2} \equiv(2,2,1)$

DR of $\ell$ (line $\perp$ to $\ell_{1} \& \ell_{2}$ )

$=(-2,3,-2)$

$\therefore \ell: \overrightarrow{\mathrm{r}}=-2 \mu \hat{\mathrm{i}}+3 \mu \hat{\mathrm{j}}-2 \mu \hat{\mathrm{k}}$

for intersection of $\ell \& \ell_{1}$

$3+t=-2 \mu$

$-1+2 t=3 \mu$

$4+2 t=-2 \mu$

$\Rightarrow \mathrm{t}=-1 \& \lambda=-1$

$\therefore$ Point of intersection $\mathrm{P} \equiv(2,-3,2)$

Let point on $\ell_{2}$ be $Q(3+2 s, 3+2 s, 2+s)$

Given $\mathrm{PQ}=\sqrt{17} \quad \Rightarrow(\mathrm{PQ})^{2}=17$

$\Rightarrow(2 \mathrm{~s}+1)^{2}+(6+2 \mathrm{~s})^{2}+(\mathrm{s})^{2}=17$

$\Rightarrow 9 \mathrm{~s}^{2}+28 \mathrm{~s}+20=0$

$\Rightarrow \mathrm{s}=-2,-\frac{10}{9}$

$\mathrm{s} \neq-2$ as point lies on $1^{\text {st }}$ octant.

$\therefore a=3+2\left(-\frac{10}{9}\right)=\frac{7}{9}$

$b=3+2\left(-\frac{10}{9}\right)=\frac{7}{9}$

$c=2+\left(-\frac{10}{9}\right)=\frac{8}{9}$

$\therefore 18(a+b+c)=18\left(\frac{22}{9}\right)=44$