Question

The equation of the plane passing through the point $(2,1,-1)$ and the line of intersection of the planes $\overrightarrow{r}.(\overrightarrow{i}+3\overrightarrow{j}-\overrightarrow{k})=0$ and $\overrightarrow{r}.(\overrightarrow{j}+2\overrightarrow{k})=0$ is

• A. $x+4y-z=0$
• B. $x+9y+11z=0$
• C. $2x+y-z+5=0$
• D. $2x-y+z=0$

Answer 1

The correct option is B $x+9y+11z=0$

The two planes are $x+3y-z=0$ and $y+2z=0$

The directional ratios of normals of given planes are $1,3,-1$ and $0,1,2$

Therefore the directional ratios of line of intersection of planes is cross product of directional ratios of those two planes

So we get $7,-2,1$ as directional ratios of line of intersection of planes

$(2\hat{i}+\hat{j}-\hat{k})\times (7\hat{i}-2\hat{j}+\hat{k})=-\hat{i}-9\hat{j}-11\hat{k}$

Therefore the directional ratios of normal of required plane is $-1,-9,-11$

therefore the equation of required plane is $x+9y+11z=0$

So the correct option is $B$