Mathematics - Vector Question with Solution | TestHub

$l_1:\overrightarrow{r}=\left(3+t\right)\hat{\mathrm{i}}+\left(-1+2t\right)\hat{\mathrm{j}}+\left(4+2t\right)\hat{\mathrm{k}}$

$l_2:\overrightarrow{r}=\left(3+2s\right)\hat{\mathrm{i}}+\left(3+2s\right)\hat{\mathrm{j}}+\left(2+s\right)\hat{\mathrm{k}}$

$\mathrm{DR}$ of $l_1\equiv \left(1,2,2\right)$

$\mathrm{DR}$ of $l_2\equiv \left(2,2,1\right)$

$l_1\times l_2=\left|\begin{array}{ccc}i& j& k\\ 1& 2& 2\\ 2& 2& 1\end{array}\right|$

$=i\left(2-4\right)-j\left(1-4\right)+k\left(2-4\right)$

$=-2i+3j-2k$

Hence, $\mathrm{DR}$ of $l$ ( line $\perp$ to $l_1 \& l_2$)

$=\left(-2,3,-2\right)$

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

for intersection of $l \& l_1$

$3+t=-2\mu$

$-1+2t=3\mu$

$4+2\mathrm{t}=-2\mu$

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

$\therefore$ Point of intersection $P\equiv \left(2, -3, 2\right)$

Let point on $l_2$ be $Q\left(3+2s, 3+2s, 2+s\right)$

Given $PQ=\sqrt{17} \Rightarrow {\left(PQ\right)}^2=17$

$\Rightarrow {\left(2s+1\right)}^2+{\left(6+2s\right)}^2+{\left(s\right)}^2=17$

$\Rightarrow 9s^2+28s+20=0$

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

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

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

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

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

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