Question

Find the image of the point $(1,3,4)$ with respect to the plane $2x-y+z+3=0$.

Answer 1

We're to find the image of the point $(1,3,4)$ with respect to the plane $2x-y+z+3=0$......(1).

The equation of the line passing through the point and perpendicular to the plane (1) is

$\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}$......(2).

Any point on the line is $(2r+1,-r+3,r+4)$.

If this point lies on the plane then we get,

$2(2r+1)-1(-r+3)+(r+4)+3=0$

or, $4r+r+r+2-3+4+3=0$

or, $6r=-6$

or, $r=-1$.

So the point on the plane is $(-1,4,3)$.

Let $(x,y,z)$ be the image of the point with respect to the plane $2x-y+z+3=0$.

Then $(-1,4,3)$ will be the mid-point of the line joining $(x,y,z)$ and $(1,3,4)$.

Then we get,

$\frac{x+1}{2}=-1$ and $\frac{y+3}{2}=4$ and $\frac{z+4}{2}=3$

or, $x=-3,y=5,z=2$.

So image point is $(-3,5,2)$.