Question

The image point of $(1,3,4)$ in the plane $2x-y+z+3=0$ is

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

Answer 1

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

Let $P(1,3,4)$ be the point.
Let $N$ be the point on the plane
Equation of the plane is $2x-y+z+3=0$
Thus, equation of line $PN$ is
$\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=\lambda$ (let)
$\Rightarrow x=2\lambda +1,y=-\lambda +3,z=\lambda +4$
Substituting the above value in the equation of the plane, we have
$2(2\lambda +1)-(3-\lambda )+(\lambda +4)+3=0$
$\Rightarrow 6\lambda +6=0$
$\Rightarrow \lambda =-1$
Thus, coordinates of $N$ are
$x=2(-1)+1=-1$
$y=3+1=4$
$z=-1+4=3$

Let $P^{\prime }(x_1,y_1,z_1)$ be the image of $P.$
Then, $N$ is the mid point of $P{P}^{\prime }$
$\therefore -1=\frac{1+x_1}{2};4=\frac{3+y_1}{2};3=\frac{4+z_1}{2}$
$\Rightarrow x_1=-3,y_1=5,z_1=2$
$P^{\prime }\equiv (-3,5,2)$

Alternate :
If $(x_2,y_2,z_2)$ is the image of $(x_1,y_1,z_1)$ in the plane $ax+by+cz+d=0,$ then
$\frac{x_2-x_1}{a}=\frac{y_2-y_1}{b}=\frac{z_2-z_1}{c}=\frac{-2(a{x}_1+b{y}_1+c{z}_1+d)}{a^2+b^2+c^2}$
Hence, for the given problem
$\frac{x_2-1}{2}=\frac{y_2-3}{-1}=\frac{z_2-4}{1}=\frac{-2(2\times 1-1\times 3+1\times 4+3)}{2^2+(-1{)}^2+(1{)}^2}$
$\Rightarrow x_2=-3,y_2=5,z_2=2$