Question

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

• A. $(\frac{-2}{3},\frac{1}{3},\frac{-11}{3})$
• B. $(\frac{3}{7},-\frac{6}{7},\frac{19}{7})$
• C. $(\frac{2}{3},\frac{1}{3},\frac{11}{3})$
• D. $(\frac{-2}{3},\frac{-1}{3},\frac{11}{3})$

Answer 1

The correct option is A $(\frac{-2}{3},\frac{1}{3},\frac{-11}{3})$

If ($a,b,c$) is the image

Midpoint of ($a,b,c$) and ($1,2,3$) lies on the plane.

$\begin{array}{c}\frac{a+1}{2}=\frac{b+2}{2}=4\frac{c+3}{2}=0\\ i.e.a+b+4c=-15\\ thelinejoining\left(a,b,c\right)and\left(1,2,3\right)isnormaltotheplane.\\ \left(a-1\right)\hat{i}+\left(b-2\right)\hat{j}+\left(c-3\right)\hat{k}\\ so,\frac{a-1}{1}=\frac{b-2}{1}=\frac{c-3}{4}=t=\frac{a-1+b-2+4\left(c-3\right)}{1+1+16}=\frac{-30}{18}=\frac{-5}{3}\\ a=\frac{-2}{3};b=\frac{1}{3};c=\frac{-11}{3}\end{array}$