Question

Three planes $P_1:2x+y+z-1=0,$ $P_2:x-y+z-2=0$ and $P_3:\alpha x-y+3z-5=0$ intersect each other at point $P$ on $XOY$ plane and at point $Q$ on $YOZ$ plane, where $O$ is the origin. Then which of the following statements is (are) CORRECT?

• 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 $(\frac{1}{3},\frac{-1}{2},\frac{1}{2})$

Answer 1

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

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

$P_1:2x+y+z=1$
$P_2:x-y+z=2$
$P_3:\alpha x-y+3z=5$

If three planes intersect at more than one point, it means they intersect at a line.
Given, the three planes meet at two points. It means they have infinitely many solutions. So
$\left|\begin{array}{ccc}2& 1& 1\\ 1& -1& 1\\ \alpha & -1& 3\end{array}\right|=0$

$\Rightarrow 2(-3+1)-1(3+1)+\alpha (1+1)=0\Rightarrow \alpha =4$

Put $z=0$ in $P_1$ and $P_2,$ we get
$2x+y=1$ and $x-y=-2$
On solving, we get $x=1,y=-1$
$\therefore P$ on $XOY$ plane is $(1,–1,0)$

Put $x=0$ in $P_1$ and $P_2,$ we get
$y+z=1$ and $-y+z=2$
On solving, we get $z=\frac{3}{2}$ and $y=\frac{-1}{2}$
$\therefore Q$ on $YOZ$ plane is$(0,\frac{-1}{2},\frac{3}{2})$

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

$\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}=(0\hat{i}-\frac{1}{2}\hat{j}+\frac{3}{2}\hat{k})-(1\hat{i}-1\hat{j}+0\hat{k})=-\hat{i}+\frac{1}{2}\hat{j}+\frac{3}{2}\hat{k}$

Projection of $PQ$ on $x-$axis
$\left|\frac{\overrightarrow{PQ}\cdot \hat{i}}{|\hat{i}|}\right|=\left|\frac{(-\hat{i}+\frac{1}{2}\hat{j}+\frac{3}{2}\hat{k})\cdot \hat{i}}{|\hat{i}|}\right|=1$

Centroid of $△OPQ$ is $(\frac{0+1+0}{3},\frac{0-1-\frac{1}{2}}{3},\frac{0+0+\frac{3}{2}}{3})$
$\equiv (\frac{1}{3},\frac{-1}{2},\frac{1}{2})$