Question

If three planes $P_1\equiv 2x+y+z-1=0,P_2\equiv x-y+z-2=0$ and $P_3\equiv \alpha x-y+3z-5=0$ 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, $\left|\begin{array}{ccc}2& 1& 1\\ 1& -1& 1\\ \alpha & -1& 3\end{array}\right|=0$
$\Rightarrow 2(-3+1)-1(3+1)+\alpha (1+1)=0\Rightarrow \alpha =4$
$P_1:2x+y+z=1$
$P_2:x-y+z=2$
$P_3:4x-y+3z=5$
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 -
$\frac{x-1}{4}=\frac{y+1}{-1}=\frac{z}{3}$
$\overrightarrow{PQ}=\hat{i}-\frac{1}{2}\hat{j}-\frac{3}{2}\hat{k}$
Projection of $\overrightarrow{PQ}$ on $x$-axis $\Rightarrow \left|\frac{\overrightarrow{OP}\cdot \hat{i}}{|\hat{i}|}\right|=1$
Centroid of $△OPQ$ is $\left(\frac{1}{3},\frac{-1}{2},\frac{1}{2}\right)$