Question

The image of the point $(2,-1,1)$ by the plane $3x+4y-5z=0$ is

• A. $(-2,1,-1)$
• B. $(\frac{2}{3},\frac{-1}{4},\frac{-1}{5})$
• C. $(\frac{59}{25},\frac{-13}{25},\frac{2}{5})$
• D. none of these

Answer 1

Answer: (C)

$(\frac{59}{25},\frac{-13}{25},\frac{2}{5})$

Let $P(\alpha ,\beta ,\gamma )$ be the image of the point $Q(2,-1,1)$, so midpoint of $PQ$ lie on the given plane
$\Rightarrow 3\left(\frac{\alpha +2}{2}\right)+4\left(\frac{\beta -1}{2}\right)-5\left(\frac{\gamma +1}{2}\right)=0$
$\therefore 2\alpha -\beta +\gamma +9=0:...\left(i\right)$
Also $PQ$ is perpendicular to the normal to the plane,
$\Rightarrow \frac{\alpha -2}{3}=\frac{\beta +1}{4}=\frac{\gamma -1}{-5}...\left(ii\right)$
Solving (i) and (ii), we get
$\alpha =\frac{59}{25},\beta =-\frac{-13}{25},\gamma =\frac{2}{5}.$
Therefore, image is $(\frac{59}{25},-\frac{13}{25},\frac{2}{5})$