Question

Let $P_{1}:x-2y+3z=5$ and $P_{2}:2x-3y+z+4=0$ be two planes. The equation of the plane perpendicular to the line of intersection of $P_{1}=0$ and $P_{2}=0$ and passing through $\left(1,1 , 1\right)$ is

• A. $11x-5y+7z-13=0$
• B. $7x+5y+z=13$
• C. $x+2y+z-4=0$
• D. $x-2y+4z+3=0$

Answer 1

Answer: (B)

The normal vector of plane $P_{1}=0$ is $\overset{ \rightarrow }{n_{1}}=\hat{i}-2\hat{j}+3\hat{k}$
The normal vector of plane $P_{2}=0$ is $\overset{ \rightarrow }{n_{2}}=2\hat{i}-3\hat{j}+\hat{k}$
A vector parallel to the line of intersection of $P_{1}=0$ and $P_{2}=0$ is
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & -3 & 1 \end{vmatrix}$
$=7\hat{i}+5\hat{j}+\hat{k}$
It is also a normal vector for the required plane
Hence, the equation of the required plane is
$7\left(x - 1\right)+5\left(y - 1\right)+\left(z - 1\right)=0$
$7x+5y+z-13=0$