Question

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

Answer 1

We know that,

if $\left(\alpha ,\beta ,\gamma \right)$

be the image , then mid point of $\left(\alpha ,\beta ,\gamma \right)$ and $(3,-2,1)$ must lie on $3x-y+4z=2$

$\begin{array}{c}\therefore 3\left(\frac{\alpha +3}{2}\right)-\left(\frac{\beta -2}{2}\right)+4\left(\frac{\gamma +1}{2}\right)=2\\ \Rightarrow 3\alpha +9-\beta +2+4\gamma +4=4\\ \Rightarrow 3\alpha -\beta +4\gamma =-\mathrm{11...........}\left(i\right)\end{array}$

Also, line joining $\left(\alpha ,\beta ,\gamma \right)$ and $(3,-2,1)$ should be parallel to the normal of the plane $3x-y+4z=2$

$\therefore \left(\frac{\alpha +3}{3}\right)=\left(\frac{\beta -2}{1}\right)=\left(\frac{\gamma +1}{4}\right)=\lambda$

$\begin{array}{c}\Rightarrow \alpha +3=3\lambda ;\beta -2=\lambda ;\gamma +1=4\lambda \\ \Rightarrow \alpha =3\lambda -3;\beta =\lambda +2;\gamma =4\lambda -\mathrm{1........}\left(ii\right)\end{array}$

From (i) and (ii)

$\alpha =\frac{-5}{2},\beta =\frac{13}{6},\gamma =\frac{23}{6}$

$\therefore$ The image list $\left(\frac{-5}{2},\frac{13}{6},\frac{23}{6}\right)$