# Prove the following

Question:

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

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

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

If the co-ordinates of the point in the first octant on ${ }^{\prime} \ell_{2}{ }^{\prime}$ at the distance of $\sqrt{17}$ from the point of

intersection of $^{\prime} \ell^{\prime}$ and $^{\prime} \ell_{1}{ }^{\prime}$ are $(\mathrm{a}, \mathrm{b}, \mathrm{c})$, then $18(\mathrm{a}+\mathrm{b}+\mathrm{c})$ is equal to

Solution:

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

$\ell_{1}: \frac{x-3}{1}=\frac{y+1}{2}=\frac{z-4}{2} \Rightarrow$ D. $R$. of $\ell_{1}=1,2,2$

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

$\ell_{2}: \frac{x-3}{2}=\frac{y-3}{2}=\frac{z-2}{1} \Rightarrow$ D. $R$. of $\ell_{2}=2,2,1$

$\mathrm{D} \cdot \mathrm{R}$, of $\ell$ is $\perp$ to $\ell_{1} \& \ell_{2}$

$\therefore \quad$ D.R. of $\ell \|\left(\ell_{1} \times \ell_{2}\right) \quad \Rightarrow \quad\langle-2,3,-2\rangle$

$\therefore$ Equation of $\ell: \frac{x}{2}=\frac{y}{-3}=\frac{z}{2}$

Solving $\ell \& \ell_{1}$

$(2 \lambda,-3 \lambda, 2 \lambda)=(\mu+3,2 \mu-1,2 \mu+\mu)$

$\Rightarrow \quad 2 \lambda=\mu+3$

$-3 \lambda=2 \mu-1$

$2 \lambda=2 \mu+4$

$\mu=-1$

$\mu=-1$

$\mathrm{P}(2,-3,2)\{$ intersection point $\}$

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

Now, $\mathrm{PQ}=\sqrt{(2 v+3-2)^{2}+(2 v+3+3)^{2}+(v+2-2)^{2}}=\sqrt{17}$

$\Rightarrow(2 v+1)^{2}+(2 v+6)^{2}+(v)^{2}=17$

$\Rightarrow 9 v^{2}+28 v+36+1-17=0$

$\Rightarrow 9 v^{2}+28 v+20=0$

$\Rightarrow 9 v^{2}+18 v+10 v+20=0$

$\Rightarrow(9 v+10)(v+2)=0$

$\Rightarrow v=-2($ rejected $),-\frac{10}{9}($ accepted $)$

$\mathrm{Q}\left(3-\frac{20}{9}, 3-\frac{20}{9}, 2-\frac{10}{9}\right)$

$\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)$

$\therefore \quad 18(a+b+c)$

$=18\left(\frac{7}{9}+\frac{7}{9}+\frac{8}{9}\right)$

$=44$