Question

If three planes $P_1 \equiv 2 x+y+z-1=0, P_2 \equiv x-y+z-2=0$ and $P_3 \equiv \alpha x-y+3 z-5=0$ intersects each other at point $P$ on $XOY$ plane and at point $Q$ on $YOZ$ plane, where $O$ is the origin then identify the correct statement(s)?

• A. The value of $\alpha$ is 4 .
• B. Straight line perpendicular to plane $P _3$ and passing through $P$ is $\frac{ x -1}{4}=\frac{ y +1}{-1}=\frac{ z }{3}$.
• C. The length of projection of $\overrightarrow{ PQ }$ on $x$-axis is 1 .
• D. Centroid of the triangle $OPQ$ is $\left(\frac{1}{3}, \frac{-1}{2}, \frac{1}{2}\right)$

Answer 1

Answer: (A), (B), (C), (D)

Three planes meet at two points it means they have infinitely many solutions, so
So, $\begin{vmatrix}2 & 1 & 1 \\ 1 & -1 & 1 \\ \alpha & -1 & 3\end{vmatrix}=0$
$\Rightarrow 2(-3+1)-1(3+1)+\alpha(1+1)=0 \Rightarrow \alpha=4 $
$P_1: 2 x+y+z=1 $
$P_2: x-y+z=2$
$P_3: 4 x-y+3 z=5$
Pon XOY plane $\equiv(1,-1,0) $ (which can be obtained by putting $z =0$ in any two of the given planes.) Q on YOZ plane $\equiv\left(0, \frac{-1}{2}, \frac{3}{2}\right)$ (which can be obtained by putting $x=0$ in any two of the given planes.) $\therefore$ Straight line perpendicular to plane $P _3$ passing through $P$ is -
$\frac{x-1}{4}=\frac{y+1}{-1}=\frac{z}{3} $
$\overrightarrow{P Q}=\hat{i}-\frac{1}{2} \hat{j}-\frac{3}{2} \hat{k}$
Projection of $\overrightarrow{ PQ }$ on $x$-axis $\Rightarrow\left|\frac{\overrightarrow{ OP } \cdot \hat{ i }}{|\hat{ i }|}\right|=1$
Centroid of $\triangle O P Q$ is $\left(\frac{1}{3}, \frac{-1}{2}, \frac{1}{2}\right)$