Question

If three planes $P_1\equiv 2x+y+z-1=0,P_2\equiv x-y+z-2=0$ and $P_3\equiv \alpha x-y+3z-5=0$ intersect 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 length of projection of $\overrightarrow{PQ}$ on $x-$axis is $1.$ unit
• 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. Centroid of the triangle $OPQ$ is $(\frac{1}{3},\frac{-1}{2},\frac{1}{2})$
• D. The value of $\alpha$ is $4.$

Answer 1

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

The value of $\alpha$ is $4.$

Three planes meet at two points it means they have infinitely many solutions.
$\therefore$$\left|\begin{array}{ccc}2& 1& 1\\ 1& -1& 1\\ \alpha & -1& 3\end{array}\right|=0$
$\Rightarrow 2(-3+1)-1(3-\alpha )+1(-1+\alpha )=0$
$\Rightarrow \alpha =4$
Thus, equation of planes are
$P_1:2x+y+z=1$
$P_2:x-y+z=2$
$P_3:4x-y+3z=5$
Since point $P$ lies on $XOY$ plane
$\therefore$ Put $z=0$ in the above planes
$\Rightarrow x=1,y=-1$
$\therefore P(1,-1,0)$
and point $Q$ lies on $YOZ$ plane
$\therefore$ Put $x=0$ in the above planes
$\Rightarrow z=\frac{3}{2},y=-\frac{1}{2}$
$\therefore Q(0,-\frac{1}{2},\frac{3}{2})$


Straight line perpendicular to plane $P_3$ and passing through $P$ is $\frac{x-1}{4}=\frac{y+1}{-1}=\frac{z}{3}.$

We have, $\overrightarrow{PQ}=\hat{i}+\frac{1}{2}\hat{j}+\frac{3}{2}\hat{k}$
Now, the length of projection of $\overrightarrow{PQ}$ on $x-$axis is $|(\hat{i}+\frac{1}{2}\hat{j}+\frac{3}{2}\hat{k})\cdot \left(\frac{\pm \hat{i}}{|\pm \hat{i}|}\right)|$
$1$ unit.

Centroid of $△OPQ$ is $(\frac{1}{3},\frac{-1}{2},\frac{1}{2}).$