Question

In $R^3,$ let $P_1:x-y+z=3$ and $L:\frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{1}$ be the equation of plane and straight line respectively. Let $Q$ be the image of the point $(-1,2,3)$ with respect to the plane $P_1.$ Then the equation of the plane $P_2$ passing through $Q$ and containing the line $L$ is

• A. $7x-9y-3z+8=0$
• B. $7x-9y+3z+8=0$
• C. $x-3y+z-4=0$
• D. $x-3y-z+4=0$

Answer 1

The correct option is A $7x-9y-3z+8=0$


Let the image of point $(-1,2,3)$ with respect to the plane $P_1$ be $Q(a,b,c)$.
Then using image formula, we get
$\frac{a+1}{1}=\frac{b-2}{-1}=\frac{c-3}{1}=\frac{-2(-1-2+3-3)}{3}=2$
$\Rightarrow a=1,b=0,c=5$
Therefore, the image is $Q(1,0,5)$

Since required plane contains the straight line $L:\frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{1},$
$\Rightarrow$ The plane passes through $(1,2,-1)$ and DR's $(3,2,1)$

Hence, required equation of the plane passing through the two points is given by,
$P_2:\left|\begin{array}{ccc}x-1& y-2& z+1\\ 3& 2& 1\\ 1-1& 2-0& -1-5\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}x-1& y-2& z+1\\ 3& 2& 1\\ 0& 1& -3\end{array}\right|=0$

$\Rightarrow 7x-9y-3z+8=0$