I have two 3 dimensional points. $A [x_1, y_1, z_1]$ and $B [x_2, y_2, z_2]$.
I need to find a transformation matrix which when multiplied to $A$ will give me $B$ and when multiplied by $B$ give me $A$.
The transformation needs to be a reflection against the plane that's perpendicular to the middle of the $AB$ segment and passing through the midpoint of the $AB$.
Sorry if I am not using the right terminology, I studied math 20 years ago so it's a little rusty.
Practical application of the question is to find a color matrix to swap two colors with arbitrary Red, Green and Blue components.
Thank you.
Best Answer
If you can find the reflection through the plane perpendicular to the middle of the $AB$ segment, then it automatically sends $A$ to $B$ and $B$ to $A$.
How do we do this reflection? First, suppose that the midpoint of $A$ and $B$ is the origin: $A=-B$. Now, note the mathematical concept of a projection:
Here, $u,v,w,x$ are being multiplied by a projection matrix $P$: one that sends them to their closest point on the line $m$. It turns out that the projection of a vector Now suppose that $m$ is the line from $A$ to the origin.
If we subtract the projection of the line $x$ on to $OA$ from $x$ twice, then we get the reflection of $x$ in the plane through $O$ perpendicular to $OA$. Now, if $a$ is the vector of the point $a$, then the projection of the vector $x$ on to the line $a$ is given by $(x.a)x$ (do you know how to work out the dot product?) If you know properties of the dot product, you can work this out for yourself. So the transformation that reflects $x$ in this plane is given by:
$$ x\mapsto x-2(x.a)x $$
This is a linear transformation, so it has a matrix, if you want to work it out: to work out the matrix, substitute in $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ for $x$, and the vectors you get out will be the columns of the matrix.
What if $A\neq-B$? Then, as I said in my comment above, there is no matrix that will encode the reflection, as any matrix must send the origin to itself. So we have to use a trick called conjugation. First, find the midpoint $c=\frac12(a+b)$, and then let $a'=a-c$ and $b'=b-c$, and for any point $x$, let $x'=x-c$. So we are 'changing our world view' in some sense: we are taking the point $c$ and mapping it to the origin. Then $a'+b'=0$, so we can encode the reflection in the plane between them as a matrix $P'$.
Then, if $x$ is an arbitrary vector, the reflection of $x'$ in the plane between $a'$ and $c'$ is given by $P'x'=P'(x-c)$. If we now add $c$ to everything, we see that the reflection of $x$ in the plane between $a$ and $b$ is given by $P'(x-c)+c$, which we can rearrange to $P'x+d$, where $d=c-P'c$. So the transformation isn't given by a matrix, but it is given by a matrix multiplication, followed by adding a constant vector.
The process of carrying out an operation (subtracting $c$), then carrying out another operation (the reflection) and then doing the inverse of the first operation (adding $c$ back again) is known as conjugation, and is very important in mathematics. Think of it as a temporary change of 'world view': we pretend that $c$ is the origin so we can do our reflection, and then we map everything back again at the end.