Question

The equation of line passing through $(3,-1,2)$ and perpendicular to the lines
$\overline{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda (2\hat{i}-2\hat{j}+\hat{k})$ and $\overline{r}=(2\hat{i}+\hat{j}-3\hat{k})+\mu (\hat{i}-2\hat{j}+2\hat{k})$ is

• A. $\frac{x+3}{2}=\frac{y+1}{3}=\frac{z-2}{2}$
• B. $\frac{x-3}{3}=\frac{y+1}{2}=\frac{z-2}{2}$
• C. $\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-2}{2}$
• D. $\frac{x-3}{2}=\frac{y+1}{2}=\frac{z-2}{2}$

Answer 1

The correct option is C $\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-2}{2}$

The line is passing through the given point $(3,-1,2)$ and is perpendicular to the $\overrightarrow{b_1}$ and $\overrightarrow{b_2}$.

Where $\overrightarrow{b_1}=2\hat{i}-2\hat{j}+\hat{k}$

And $\overrightarrow{b_2}=\hat{i}-2\hat{j}+2\hat{k}$

Let line perpendicular to $\overrightarrow{b_1}$ and $\overrightarrow{b_2}$ is $\overrightarrow{c}$.

$\overrightarrow{c}=\overrightarrow{b_1}\times \overrightarrow{b_2}=\left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -2& 1\\ 1& -2& 2\end{array}\right]$

$\overrightarrow{c}=\hat{i}(-4+2)-\hat{j}(4-1)+\hat{k}(-4+2)$

$\overrightarrow{c}=\hat{i}(-4+2)-\hat{j}(4-1)+\hat{k}(-4+2)$

$\overrightarrow{c}=-2\hat{i}-3\hat{j}-2\hat{k}$
Direction ratios of $\overrightarrow{c}$ are $a=-2,b=-3,c=-2$


Thus, the equation of line passing through points $(3,-2,1)$ having direction ratios $a=-2,b=-3,c=-2$ will be,

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$
$\frac{x-3}{-2}=\frac{y+1}{-3}=\frac{z-2}{-2}$

Cancel negative sign from above equation
$\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-2}{2}$