Mathematics - Vector Question with Solution | TestHub

$L_1:\overrightarrow{r}=\hat{i}+\lambda \left(-\hat{i}+2\hat{j}+2\hat{k}\right), \lambda \in R$

Let a general point on line $L_1$ be $A\left(1-\lambda , 2\lambda , 2\lambda \right)$

Also $L_2:\overrightarrow{\mathrm{r}}=\mathrm{\mu }\left(2\hat{\mathrm{i}}-\hat{\mathrm{j}}+2\hat{\mathrm{k}}\right)$ 
let a general point on line $L_2$ be $B\left(2\mu , -\mu , 2\mu \right)$
Then  $\overrightarrow{AB}=\left(2\mu +\lambda -1\right)\hat{i}+\left(-\mu -2\lambda \right) \hat{j}+\left(2\mu -2\lambda \right)\hat{k}$
Now, $\overrightarrow{AB}$ is perpendicular to both $L_1$ and $L_2$
Hence, $\left(a\right)$ $\overrightarrow{AB}\cdot \left(-\hat{i}+2\hat{j}+2\hat{k}\right)=0$ ( If $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular then $\overrightarrow{a}\cdot \overrightarrow{b}=0$ )
& $\left(b\right)$ $\overrightarrow{AB}\cdot \left(2\hat{i}-\hat{j}+2\hat{k}\right)=0$

Using 
$\left(a\right)$ $\overrightarrow{AB}\cdot \left(-i+2j+2k\right)=0$
$\left(-1\right)\left(2\mu +\lambda -1\right)+\left(-\mu -2\lambda \right)\left(2\right)+\left(2\mu -2\lambda \right)2=0$
$-2\mu -\lambda +1-2\mu -4\mu +4\mu -4\lambda =0$
$9\lambda =1$
$\lambda =\frac{1}{9} \dots \left(\mathrm{i}\right)$
From $\left(b\right)$ we have      $\overrightarrow{AB}\cdot \left(2\hat{i}-\hat{j}+2\hat{k}\right)=0$
$\Rightarrow \left(2\mu +\lambda -1\right)\left(2\right)+\left(-\mu -2\lambda \right)\left(-1\right)+\left(2\mu -2\lambda \right)\left(2\right)=0$
$\Rightarrow 4\mu +2\lambda -2+\mu +2\lambda +4\mu -4\lambda =0$
$\Rightarrow 9\mu =2$
$\Rightarrow \mu =\frac{2}{9} \dots \left(\mathrm{ii}\right)$
Using value of $\lambda$ and $\mu \Rightarrow A\left(\frac{8}{9},\frac{2}{9},\frac{2}{9}\right)$ and $B\left(\frac{4}{9},-\frac{2}{9},\frac{4}{9}\right)$
$\overrightarrow{AB}=-\frac{2}{9}\left(2\hat{i}+2\hat{j}-\hat{k}\right)$
Now, using point $A$ and $\overrightarrow{AB}$
Equation of line $=\left(\frac{8}{9}\hat{i}+\frac{2}{9}\hat{j}+\frac{2}{9}\hat{k}\right)+\gamma \left(-\frac{2}{9}\left(2\hat{i}+2\hat{j}-\hat{k}\right)\right)$
Option $\left(B\right)$ :
$\overrightarrow{r}=\frac{2}{9}\left(4\hat{i}+\hat{j}+\hat{k}\right)+t\left(2\hat{i}+2\hat{j}-\hat{k}\right),t\in R$
Using point $B$ and $\overrightarrow{AB}$
Option $\left(C\right)$ :
Equation of line $\overrightarrow{r}=\frac{2}{9}\left(2\hat{i}-\hat{j}+2\hat{k}\right)+t\left(2\hat{i}+2\hat{j}-\hat{k}\right), t\in R$
Using midpoint of $A$ and $B$ , $\left(\frac{2}{3},0,\frac{1}{3}\right)$
Option $\left(A\right)$ :
$\overrightarrow{r}=\frac{1}{3}\left(2\hat{i}+\hat{k}\right)+t\left(2\hat{i}+2\hat{j}-\hat{k}\right),t\in R$