Question

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

Answer 1

Let the given point be $A(3,5,7)$ and image on the plane of $A$ is $P$

Since, $AP$ is perpendicular to the plane $2x+y+z=0$

So, direction ratios of the line $AP$ are $2,1,1$

therefore,

equation of the line $AP$ is

$\frac{x-3}{2}=\frac{y-5}{1}=\frac{z-7}{1}$

Let

$\frac{x-3}{2}=\frac{y-5}{1}=\frac{z-7}{1}=r$

$x=2r+3,y=r+5,z=r+7$

General point on the line $AP$ is $(2r+3,r+5,r+7)$

Let the coordinates of point $P$ be$(2r_1+3,r_1+5,r_1+7)$

Now, mid-point of $AP=(\frac{2r_1+3+3}{2},\frac{r_1+5+5}{2},\frac{r_1+7+7}{2})$

$(r_1+3,\frac{r_1+10}{2},\frac{r_1+14}{2})$

Since mid-point of $AP$ lies on the plane

Therefore,

$2(r_1+3)+\frac{r_1+10}{2}+\frac{r_1+14}{2}=0$

$3r_1+18=0$

$r_1=-6$

Hence,

image of the point $(3,5,7)$ on the plane $2x+y+z=0$ is $(-9,-1,1)$.