Finding Mirror Images of Points on a Plane

Question

I'm having trouble with finding the coordinate: "find the coordinates of the mirror image of $A(3, 1, 2)$ when reflected in the plane $x+2y+z=1$."

I've tried multiple methods such as assuming that it is perpendicular and then using the normal to work it out, or trying to find the vector equation of the line and then working it out that way, but I can't seem to get the answer of $(1, -3, 0)$.

Any help is greatly appreciated!! Thanks!!

Answer 1

As you suggest, using the normal line will get you the solution:

(1) Construct the line normal to the plane that intersects point $A (3,1,2)$:

line $(x,y,z) = (3,1,2) + t(1,2,1)$

(Any line normal to the plane $x+2y+z=c$ will move in the direction $(1,2,1)$ )

(2) Find the point B on the normal line that intersects the plane:

$(3+t) + 2(1+2t) + (2+t) = 1$

Solving, we get $t=-1$. Hence the intersection point $B$ (on the plane) is at $(2,-1,1)$.

(3) Point $A$ is at $t=0$, and point $B$ is at $t=-1$, so the mirror image of $A$, say $A’$ will be twice the distance, at $t=-2$: $$\boxed{ A' \equiv (1,-3,0)}$$

Answer 2

We have $$ 1 = x + 2y + z = (1,2,1) \cdot (x,y,z) $$ so $(1,2,1)$ is a normal vector of the plane. We divide both sides of the equation by the norm of that vector and get $$ d = 1/\sqrt{6} = (1,2,1)/\sqrt{6} \cdot (x,y,z) = n \cdot u $$ where $n$ is a unit normal vector of the plane and $d$ is the distance of the plane to the origin.

The part of $A$ in direction of $n$ is $$ A_n = n \cdot A = (1,2,1)/\sqrt{6} \cdot (3,1,2) = 7/\sqrt{6} \\ A_\perp = A_n n = (7/6) (1,2,1) $$ We now decompose $A$ into an orthogonal and a parallel part (relative to the plane): $$ A = A_\parallel + A_\perp \Rightarrow \\ A_\parallel = A - A_\perp = (3,1,2) - (7/6) (1,2,1) = (1/6) (11, -8, 5) $$ $A_\parallel$ points from the origin to the closest point $C_0$ from $A$ to the plane $n \cdot A = 0$, which is a plane parallel to the mirror plane, but through the origin.

To make it from $C_0$ to the closest point $C$ to $A$ on the mirror plane we need to add $d n$: $$ C = C_0 + dn = A_\parallel + d n = (1/6) (11, -8, 5) + (1/\sqrt{6}) (1, 2, 1) (1/\sqrt{6}) = (1/6) (12, -6, 6) = (2, -1, 1) $$ We need the vector from $C$ to $A$: $$ v = A - C $$ and reverse it and add it to $C$ to get the mirror image $A'$: $$ A' = C - v = C - (A-C) = 2C - A = (4, -2, 2) - (3, 1, 2) = (1, -3, 0) $$ Here is a visualization:

(Large version)

Answer 3

I'll tell you an easy way out

For any point like $A(3,1,2)$ taken as $(x_1,y_1,z_1)$ And plane like $x+2y+z-1=0$ taken as $ax +by+cz+d=0$ Its mirror image will be $$(x-x1)/a=(y-y1)/b=(z-z1)/c=-2(ax1+y1+cz1+d)/a^2+b^2+c^2$$ which in this case would be $$(x-3)/1=(y-1)/2=(z-2)/1=-2(1(3)+2(1) + 1(2) -1/[(1)^2+(2)^2+(1)^2]$$ Now you just have to solve for $x,y,z$. And there you go :) (Applicable for every point)