Question

The image of the point $(1,2,-1)$ with respect to the plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point $(0,7,-7),$ is

• A. $(-\frac{1}{3},-\frac{7}{3},\frac{1}{3})$
• B. $(-\frac{1}{3},\frac{2}{3},-\frac{7}{3})$
• C. $(-\frac{1}{3},0,-\frac{7}{3})$
• D. $(-\frac{1}{3},\frac{2}{3},\frac{7}{3})$

Answer 1

The correct option is B $(-\frac{1}{3},\frac{2}{3},-\frac{7}{3})$

Let the equation of plane passing through $(-1,3,-2)$ be
$a(x+1)+b(y-3)+c(z+2)=0\dots \left(1\right)$
where $(a,b,c)$ are direction ratios of normal vector to the plane.

Given that plane also passes through $(0,7,-7)$
$\Rightarrow a+4b-5c=0\dots \left(2\right)$
Direction ratio of given line is $(-3,2,1)$
$\therefore -3a+2b+c=0\dots \left(3\right)$

Solving eqns. $\left(2\right)$ and $\left(3\right),$
$a=b=c$
$\therefore$ From $\left(1\right),$ equation of plane is $x+y+z=0$

Let image point of $(1,2,-1)$ in the plane be $(p,q,r)$
$\frac{p-1}{1}=\frac{q-2}{1}=\frac{r+1}{1}=-2\left(\frac{1+2-1}{1^2+1^2+1^2}\right)\Rightarrow p=-\frac{4}{3}+1=-\frac{1}{3}\Rightarrow q=-\frac{4}{3}+2=\frac{2}{3}\Rightarrow r=-\frac{4}{3}-1=-\frac{7}{3}$
Hence $(p,q,r)=(-\frac{1}{3},\frac{2}{3},-\frac{7}{3})$