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

Discussion in 'Linear and Abstract Algebra' started by anonymousinquirer, Aug 8, 2023.

1. ### anonymousinquirer

Joined:
Aug 8, 2023
Messages:
2
0
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 )
________________________________________
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?

anonymousinquirer, Aug 8, 2023
2. ### RobertSmart

Joined:
Apr 9, 2024
Messages:
25
4
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.

By the way, if you need further assistance with math assignments or want to explore more problems, you might find useful resources at website of MathsAssignmentHelp.com. You can contact them at +1 (315) 557-6473.

RobertSmart, May 13, 2024