Question

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

• A. $(-1,-1,6)$
• B. $(-1,1,10)$
• C. $(2,-1,\frac{13}{2})$
• D. $(-2,-1,5)$

Answer 1

The correct option is A $(-1,-1,6)$

Let $P(\alpha ,\beta ,\gamma )$ be the image of the point $Q(1,3,4)=(x_1,y_1,z_1)$ w.r.to the plane $ax+by+cz+d=0.$
$\therefore \frac{\alpha -x_1}{a}=\frac{\beta -y_1}{b}=\frac{\gamma -z_1}{c}=-2\frac{a{x}_1+b{y}_1+c{z}_1+d}{a^2+b^2+c^2}$
$\frac{\alpha -1}{1}=\frac{\beta -3}{2}=\frac{\gamma -4}{-1}=-2\frac{1\left(1\right)+2\left(3\right)-1\left(4\right)+3}{1+4+1}=-2\times \frac{6}{6}=-2$
$\alpha =-1,\beta =-1,\gamma =6$
Image is $P(-1,-1,6)$