Question

The image of the point $(1,3,4)$ in the plane $2x-y+z=-3$ is:

• A. $(1,2,3)$
• B. $(3,1,-14)$
• C. $(-3,5,2)$
• D. $(1,-1,-12)$

Answer 1

The correct option is C $(-3,5,2)$

We know, for a plane with equation $ax+by+cz+d=0$ and a point $A(x_1,y_1,z_1)$ the image of $A$ in the plane is obtained as $(x^{\prime },y^{\prime },z^{\prime })$ using the formula $\frac{x^{\prime }-x_1}{a}=\frac{y^{\prime }-y_1}{b}=\frac{z^{\prime }-z_1}{c}=-2\frac{(a{x}_1+b{y}_1+c{z}_1+d)}{(a^2+b^2+c^2)}$
We have, $(1,3,4)$ and the plane $2x-y+z=-3$
Clearly, $(a,b,c)=(2,-1,1)$
Hence on solving, we have the $-2\frac{(a{x}_1+b{y}_1+c{z}_1+d)}{(a^2+b^2+c^2)}=-2\frac{2\left(1\right)-3+4+3}{(4+1+1)}$
$=-2\left(\frac{6}{6}\right)=-2$
$\therefore \frac{x^{\prime }-1}{2}=\frac{y^{\prime }-3}{-1}=\frac{z^{\prime }-4}{1}=-2$
$\Rightarrow x^{\prime }=-3,y^{\prime }=5,z^{\prime }=2$
Hence the required reflection is $(-3,5,2)$