Question

The vector equation of the plane through the line of intersection of the planes $x+y+z=1$ and $2x+3y+4z=5$ which is perpendicular to the plane $x-y+z=0$ is :

• A. $\overrightarrow{r}\times (\hat{i}+\hat{k})+2=0$
• B. $\overrightarrow{r}\times (\hat{i}-\hat{k})+2=0$
• C. $\overrightarrow{r}\cdot (\hat{i}-\hat{k})-2=0$
• D. $\overrightarrow{r}\cdot (\hat{i}-\hat{k})+2=0$

Answer 1

The correct option is D $\overrightarrow{r}\cdot (\hat{i}-\hat{k})+2=0$

Equation of required plane is $(x+y+z-1)+\lambda (2x+3y+4z-5)=0$
$\Rightarrow (1+2\lambda )x+(1+3\lambda )y+(1+4\lambda )z+(-5\lambda -1)=0$
The above plane is perpendicular to the plane $x-y+z=0$
$\therefore 1+2\lambda -1-3\lambda +1+4\lambda =0$
$\Rightarrow \lambda =\frac{-1}{3}$

Therefore, equation of required plane is $x-z+2=0$
or, $\overrightarrow{r}\cdot (\hat{i}-\hat{k})+2=0$