The mirror image of the point (1,2,3) in a plane is
Question:

The mirror image of the point $(1,2,3)$ in a plane is

$\left(-\frac{7}{3},-\frac{4}{3},-\frac{1}{3}\right)$. Which of the following points lies on this plane?

1. $(1,1,1)$

2. $(1,-1,1)$

3. $(-1,-1,1)$

4. $(-1,-1,-1)$

Correct Option: , 2

Solution:

$\vec{n}=\frac{-7}{3}-1, \frac{-4}{3}-2, \frac{-1}{3}-3$

$\vec{n}=\frac{10}{3}, \frac{10}{3}, \frac{10}{3}$

$D . r$ of normal to the plane

$(1,1,1)$

Midpoint of $P$ and $Q$ is

$\left(\frac{-2}{3}, \frac{1}{3}, \frac{4}{3}\right)$

$\therefore \quad$ Equation of required

$Q\left(\frac{-7}{3}, \frac{-4}{3}, \frac{-1}{2}\right)$

plane $Q$

$\vec{r} \cdot \vec{n}=\vec{a} \cdot \vec{n}$

$\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=\frac{-2}{3}+\frac{1}{3}+\frac{4}{3}$

$\therefore$ Equation of plane is $x+y+z=1$