Question

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

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

Answer 1

The correct option is A $(-3,4,1)$

$\begin{array}{c}P:\left(1,2,3\right)letimageQ\left(x,y,z\right)\\ Equationoflineperpendiculartoplaneand\\ pas\sin gthroughP\left(1,2,3\right)is\\ \frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{1}=\lambda \\ \left(2\lambda +1,-\lambda +2,\lambda +3\right)\\ \therefore 2\left(2\lambda +1\right)-\left(-\lambda +2\right)+\left(\lambda +3\right)+3=0\\ 4\lambda +2+\lambda -2+\lambda +3+3=0\\ 6\lambda =-6\\ \lambda =-1\\ \therefore PointN:\left(-1,3,2\right)\\ Nismid-pointofPQ.\\ \frac{1+x_1}{2}=-1,\frac{2+y_1}{2}=3,\frac{3+z_1}{2}=2\\ x_1=-3,y_1=4,z_1=1\\ \left(-3,4,1\right)\end{array}$