Question

Find the image of the point having position vector $\hat{i}+3\hat{j}+4\hat{k}$ in the plane$\overrightarrow{r}.(2\hat{i}–\hat{j}+\hat{k})+3=0$.

Answer 1

Let $(x,y,z)$ be the coordinate of the image of the point $(1,3,4)$

So, the Direction ratios of the line joining these points are $(x-1,y-3,z-4)$
$\because$ Direction ratios of the line is parallel to the normal vector of the plane
$\therefore x-1=2\times \lambda \Rightarrow x=2\lambda +1$
$y=-\lambda +3$
$z=\lambda +4$
Since, mid point of the line joining the points $(1,3,4)$ and $(x,y,z)$ lies on the plane, so coordinate of the mid point is $(\frac{x+1}{2},\frac{y+3}{2},\frac{z+4}{2})$

Putting the point into the equation of the plane
We have $(\frac{x+1}{2}\hat{i}+\frac{y+3}{2}\hat{j}+\frac{z+4}{2}\hat{k}).(2\hat{i}–\hat{j}+\hat{k})+3=0...\left(1\right)$
Putting the values of $x,y,z$ in eqn $\left(1\right)$, we get
$\left(\right(\lambda +1)\hat{i}+\frac{(-\lambda +6)}{2}\hat{j}+\frac{(\lambda +8)}{2}\hat{k}).(2\hat{i}–\hat{j}+\hat{k})+3=0$
$\Rightarrow 2(\lambda +1)+\frac{\lambda }{2}-3+\frac{\lambda }{2}+4+3=0$
$\Rightarrow \lambda =-2$
Then the coordinate of the image is $(-3,5,2)$