Question

Find the image of the point $P(3,5,7)$ in the plane $2x+y+z=0$.

• A. $(-9,-1,1)$
• B. $(9,1,1)$
• C. $(1,9,1)$
• D. $(1,-9,1)$

Answer 1

The correct option is A $(-9,-1,1)$

Let $P(\alpha ,\beta ,\gamma )$ be the image of the point $Q(3,5,7)$, so midpoint of $PQ$ lie on the given plane.

$\Rightarrow 2\left(\frac{\alpha +3}{2}\right)+\left(\frac{\beta +5}{2}\right)+\left(\frac{\gamma +7}{2}\right)=0$

$\therefore 2\alpha +\beta +\gamma +18=0:...\left(i\right)$

Also $PQ$ is perpendicular to the normal to the plane,

$\Rightarrow \frac{\alpha -3}{2}=\frac{\beta -5}{1}=\frac{\gamma -7}{1}...\left(ii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$, we get

$\alpha =-9,\beta =-1,\gamma =1.$

Therefore, image is $(-9,-1,1)$.