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 $\overrightarrow{n_1}=\hat{i}-2\hat{j}+3\hat{k}$
The normal vector of plane $P_2=0$ is $\overrightarrow{n_2}=2\hat{i}-3\hat{j}+\hat{k}$
A vector parallel to the line of intersection of $P_1=0$ and $P_2=0$ is
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 3\\ 2& -3& 1\end{array}\right|$
$=7\hat{i}+5\hat{j}+\hat{k}$
It is also a normal vector for the required plane
Hence, the equation of the required plane is
$7\left(x-1\right)+5\left(y-1\right)+\left(z-1\right)=0$
$7x+5y+z-13=0$