Function to return a vector after it is divided by (1 + square magnitude)?

[anonymousinquirer]

The square magnitude (SM) of a 3 component vector is: (xx + yy + z*z). If you divide vector A by ( 1.0 + SMA), where SMA is the square magnitude of A, is there a function that you can perform to obtain the original vector?

Lets take original vector3 A = (Ax, Ay, Az). The final vector3 B = A/(1.0 + AxAx + AyAy + Az*Az).

Now, is there a function of B that will return A?

Thank you!

I originally found that the dot product of a Vector with itself equals the vectors square magnitude. However, I also read that the inverse of dot product is impossible because of many solutions. Still, I am hoping that this case contains qualifiers that may restrict the solution set to one or two possible solution.

I think that I worked out so far that SMA and SMB have the quadratic relationship that make SMA = -(2SMB - 1) +/- sqrt( (2SMB-1)^2 - 4(SMB)^2 )
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2SMB

And I believe you could punch in this definition to satisfy the relationship A = B(1 + SMA).

Would this mean you have to possible solutions for each component of A? Is my math correct?

 

[RobertSmart]

It seems you're trying to find a way to retrieve the original vector A from another vector B after dividing by a certain factor related to their square magnitudes. Your approach is quite close, but there's a simpler way to approach this problem.

Given vector A=(Ax,Ay,Az), and the final vector B= A/1+Ax^2+Ay^2+Az^2, you want to find a function to retrieve A from B.

Let's define SMA=Ax^2+Ay^2+Az^2 and SMB=Bx^2+By^2+Bz^2.

Since B=A/1+SMA, we can rearrange this to get A=B(1+SMA).

Now, substitute Bx^2+By^2+Bz^2 for SMB and Ax^2+Ay^2+Az^2 for SMA to get:

A=B(1+Ax^2+Ay^2+Az^2)

=(Bx,By,Bz)(1+Ax^2+Ay^2+Az^2)

=(Bx+Bx⋅Ax^2+By⋅Ay^2+Bz⋅Az^2, By+Bx⋅Ax^2+By⋅Ay^2+Bz⋅Az^2, Bz+Bx⋅Ax^2+By⋅Ay^2+Bz⋅Az^2)

Now, you can see that each component of A is a function of B and the original vector components Ax,Ay,Az. So, you don't have multiple solutions for each component of A, rather, each component of A is a unique function of B and the original components Ax,Ay,Az.

Your calculations seem correct, and this approach avoids the need for complex quadratic relationships.


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