Question

A line passes through $(2,-1,3)$ and it is perpendicular to the lines $\overrightarrow{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda (2\hat{i}-2\hat{j}+\hat{k})$ and $\overrightarrow{r}=(2\hat{i}-\hat{j}-3\hat{k})+\mu (\hat{i}+2\hat{j}+2\hat{k})$. Obtain its equation in vector and Cartesian form.

Answer 1

Line L is passing through point $=(2\hat{i}-\hat{j}+3\hat{k})$

If $L_1\Rightarrow \overrightarrow{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda (2\hat{i}-2\hat{j}+\hat{k})$

$L_2\Rightarrow \overrightarrow{r}=(2\hat{i}-\hat{j}-3\hat{k})+\mu (\hat{i}+2\hat{j}+2\hat{k})$

Given that line is perpendicular to $L_1$ and $L_2$

Let the line L $=(a_1,a_2,a_3)$

The equation of L in vector form $\Rightarrow \overrightarrow{r}=(2\hat{i}-\hat{j}-3\hat{k})+p(a_1\hat{i}+a_2\hat{j}+a_3\hat{k})$

$p$ is any constant.

So by condition that $L$ is perpendicular to $L_1$

$2a_1-2a_2+a_3=0$--- (1)

$L\perp L_2$

So, $a_1+2a_2+2a_3=0$--- (2)

Solve (1) and (2)

$3a_1+3a_3=0$

$\Rightarrow a_3=-a_1$

Put it in (2)

$a_1+2a_2-2a_1=0$

$a_2=\frac{a_1}{2}$

So, $L=(a_1,\frac{a_1}{2},-a_1)$

So we can say DR of $L=(1,\frac{1}{2},-1)$

So equation of L in vector form:

$\overrightarrow{r}=(2\hat{i}-\hat{j}+3\hat{k})+k(\hat{i}-\frac{\hat{j}}{2}-\hat{k})$

Cartesian form is $\frac{x-2}{1}=\frac{y+1}{\frac{1}{2}}=\frac{z-3}{-1}.$