Question

The equation of the plane passing through the line of intersection of the planes $\overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})=1$ and $\overrightarrow{r}\cdot (2\hat{i}+3\hat{j}-\hat{k})+4=0$ and parallel to the $x-$axis is

• A. $\overrightarrow{r}\cdot (\hat{i}+3\hat{k})+6=0$
• B. $\overrightarrow{r}\cdot (\hat{j}-3\hat{k})+6=0$
• C. $\overrightarrow{r}\cdot (\hat{j}-3\hat{k})-6=0$
• D. $\overrightarrow{r}\cdot (\hat{i}-3\hat{k})+6=0$

Answer 1

The correct option is B $\overrightarrow{r}\cdot (\hat{j}-3\hat{k})+6=0$

Equation of plane passing through line of intersection of planes $\overrightarrow{r}\cdot (\hat{i}+\hat{j}+\hat{k})=1$ and $\overrightarrow{r}\cdot (2\hat{i}+3\hat{j}-\hat{k})+4=0$ is
$\Rightarrow (x+y+z-1)+\lambda (2x+3y-z+4)=0$
$\Rightarrow (1+2\lambda )x+(1+3\lambda )y+(1-\lambda )z+(4\lambda -1)=0\cdots \left(i\right)$
$\because$ This plane is parallel to $x-$axis
$\therefore 1\cdot (1+2\lambda )+0\cdot (1+3\lambda )+0\cdot (1-\lambda )=0$
$\therefore \lambda =-\frac{1}{2}$
$\therefore$ Required equation of plane is
$-\frac{1}{2}y+\frac{3}{2}z-3=0$
$\therefore y-3z+6=0$
$\therefore \overrightarrow{r}\cdot (\hat{j}-3\hat{k})+6=0$