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

• A. $(1,-1,1)$
• B. $(-1,-1,1)$
• C. $(1,1,1)$
• D. $(-1,-1,-1)$

Answer 1

The correct option is A

$(1,-1,1)$

Step 1: Draw a diagram

The coordinates of a point $(1,2,3)$

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

P’ is the image of P w.r.t. plane the Q will be the midpoint of pp’ which will lie on the plane.

$\begin{array}{c}\therefore Q\equiv \left(\frac{1-\frac{7}{3}}{2},\frac{2-\frac{4}{3}}{2},\frac{3-\frac{1}{3}}{2}\right)\\ \equiv \left(\frac{-\frac{4}{3}}{2},\frac{\frac{2}{3}}{2},\frac{\frac{8}{3}}{2}\right)\\ \equiv \left(-\frac{4}{6},\frac{2}{6},\frac{8}{3}\right)\\ \equiv \left(\frac{-2}{3},\frac{1}{3},\frac{4}{3}\right)\end{array}$

Step 2: Finding which points lying on the plane:

Direction ratios of normal to the plane.

$\begin{array}{c}P-P\text{'}\equiv 1+\frac{7}{3},2+\frac{4}{3},3+\frac{1}{3}\\ \equiv \frac{3+7}{3},\frac{6+4}{3},\frac{9+1}{3}\\ \equiv \frac{10}{3},\frac{10}{3},\frac{10}{3}\\ \equiv 1,1,1\end{array}$

Direction ratios of normal to the plane is $1,1,1$

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

Thus, it clearly $(1,-1,1)$ lies on the plane $x+y+z=1$.

Hence, option A is the correct answer.