Question

Let $L_1:\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{1}$
$L_2:\frac{x}{1}=\frac{y}{-1}=\frac{z-5}{3}$
The equation of the line perpendicular to $L_1$ and $L_2$ and passing through the point of intersection of $L_1$ and $L_2$ is

• A. $\overrightarrow{r}=\hat{i}+\hat{j}+\hat{k}+\lambda (4\hat{i}-5\hat{j}+\hat{k}),\lambda \in R$
• B. $\overrightarrow{r}=\hat{i}+\hat{j}+\lambda (4\hat{i}-5\hat{j}-3\hat{k}),\lambda \in R$
• C. $\overrightarrow{r}=-\hat{i}+\hat{j}+2\hat{k}+\lambda (4\hat{i}-5\hat{j}-3\hat{k}),\lambda \in R$
• D. $\overrightarrow{r}=-\hat{i}+\hat{j}+2\hat{k}+\lambda (\hat{i}+\hat{j}-\hat{k}),\lambda \in R$

Answer 1

The correct option is C $\overrightarrow{r}=-\hat{i}+\hat{j}+2\hat{k}+\lambda (4\hat{i}-5\hat{j}-3\hat{k}),\lambda \in R$

A point on $L_1(2\lambda +1,\lambda +2,\lambda +3)$
A point on $L_2(\mu ,-\mu ,3\mu +5)$
To find the point of intersection,
$2\lambda +1=\mu$
$\lambda +2=-\mu =-2\lambda -1$
$3\lambda =-3$
$\lambda =-1,\mu =-1$
Hence, the point of intersection is $(-1,1,2)$
The vector along the line perpendicular to $L_1$ and $L_2\left|\begin{array}{ccc}\hat{i}& \hat{j}& k\\ 2& 1& 1\\ 1& -1& 3\end{array}\right|$
$=4\hat{i}-5\hat{j}-3\hat{k}$
The equation of the line can be written in the vector form as $\overrightarrow{r}=(-\hat{i}+\hat{j}+2\hat{k})+\lambda (4\hat{i}-5\hat{j}-3\hat{k})$