Question

The image of the point $(1,3,4)$ with respect to the plane $2\mathrm{x}-\mathrm{y}+\mathrm{z}+3=0$ is

• A. $\left(-1,4,3\right)$
• B. $\left(-3,5,2\right)$
• C. $\left(1,3,4\right)$
• D. $\left(-1,-3,-4\right)$

Answer 1

The correct option is B

$\left(-3,5,2\right)$

The explanation for the correct answer.

Step 1: Find the equation of a line through a point $\mathrm{P}(1,3,4)$ and perpendicular to the plane $2\mathrm{x}-\mathrm{y}+\mathrm{z}+3=0$.

The direction ratios of the normal of the plane are $2,-1,1$.

Equation of line is given by $\frac{\mathrm{x}-{\mathrm{x}}_1}{\mathrm{a}}=\frac{\mathrm{y}-{\mathrm{y}}_1}{\mathrm{b}}=\frac{\mathrm{z}-{\mathrm{z}}_1}{\mathrm{c}}$.

Therefore the equation of the line is $\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-3}{-1}=\frac{\mathrm{z}-4}{1}=\mathrm{k}\left(\mathrm{say}\right)$.

Step 2: Find the point of line on the plane.

The general point on this line is $(2\mathrm{k}+1,-\mathrm{k}+3,\mathrm{k}+4)$.

For some value $\mathrm{k}$, let $\mathrm{N}(2\mathrm{k}+1,-\mathrm{k}+3,\mathrm{k}+4)$ be a point of the line lying in the plane. Then,

$$\begin{gathered} \Rightarrow 2\left(2\mathrm{k}+1\right)-(-\mathrm{k}+3)+(\mathrm{k}+4)+3=0 \\ \Rightarrow 6\mathrm{k}+6=0 \\ \Rightarrow 6\mathrm{k}=-6 \\ \Rightarrow \mathrm{k}=-1 \end{gathered}$$

Therefore coordinates $\mathrm{N}$ are $(-2+1,1+3,-1+4)=(-1,4,3)$.

Step 3: Find the image of the point $\mathrm{P}(1,3,4)$

Let $\mathrm{Q}\left(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }\right)$ be the coordinates of the image of $\mathrm{P}(1,3,4)$.

And $\mathrm{N}\left(-1,4,3\right)$ is the midpoint of $\mathrm{PQ}$.

$$\begin{gathered} \frac{1+\mathrm{\alpha }}{2}=-1,\frac{3+\mathrm{\beta }}{2}=4,\frac{4+\mathrm{\gamma }}{2}=3 \\ \mathrm{\alpha }=-2-1,\mathrm{\beta }=8-3,\mathrm{\gamma }=6-4 \\ \mathrm{\alpha }=-3,\mathrm{\beta }=5,\mathrm{\gamma }=2 \end{gathered}$$

Hence the image of the $\mathrm{P}(1,3,4)$ is $\mathrm{Q}\left(-3,5,2\right)$.

Therefore option $\left(B\right)$ is the correct answer.