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I believe the square magnitude of a 3 component vector is: (xx + yy + z*z). If you multiply a vector by its square magnitude, 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*(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, with the two qualifiers (1: the dot product is between a vector and itself, 2: the final vector's components are linearly related to the original vector's corresponding components) there may be enough to restrict to one possible solution.

anonymouscoder
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2 Answers2

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B = (|A|^2) A so |B| = |A|^3, so A = B/(|A|^2) = B/(|B|^(2/3)).

Simon Goater
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Get squared magnitude of B. Note that result SMB is equal to magnitude of A in the sixth power (or cube of SMA=squareMagnitude(A)).

B = bx +       by +       bx = 
    ax * SMA + ay * SMA + az * SMA

SMB = B.dot.B = ax^2 * SMA^2 + ay^2 * SMA^2 + az^2 * SMA^2 = 
                SMA^2*(ax^2 + ay^2 + az^2) = 
                SMA^2 * SMA = 
                SMA^3

So you can get squareMagnitude(A) as cube root of SMB and extract components of A

 SMA = SMB^(1/3) = pow(SBB, 1./3)    
 ax = bx / SMA
 ay = by / SMA
 az = bz / SMA

P.S. It is interesting - what operations do provide B value?

Concerning the second question - not very readable in comments, so I've added this:

A/(1+SMA)=B   square it
SMA/(1+2*SMA+SMA^2)=SMB
SMA = SMB+2*SMA*SMB+SMA^2*SMB
SMA^2*SMB+SMA*(2*SMB-1)+SMB = 0 
D = 4*SMB^2-4*SMB+1-4*SMB^2= 1-4*SMB  
   D should be positive, so B magnitude should be small 
SMA = (1-2*SMB + Sqrt(1-4*SMB)) /(2*SMB)
    the second solution with minus is definitely negative, so omit it
 and finally with known SMA
A = B * (1 + SMA)
MBo
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  • Thank you, this solution works perfectly. However, I am afraid my question did not present my challenge correctly. I actually need to get vector A from vector B when B is defined as: B = A / (1.0 + SMA). I am really hoping that the original vector A can be derived from B! Any thoughts? – anonymouscoder Aug 08 '23 at 01:47
  • Sadly no, in case of B = A / (1.0 + SMA) it is not possible to get A if SMA value is not known – MBo Aug 08 '23 at 02:30
  • If my math is correct, is would appear that in the last mentioned relationship, SMA/(1+SMA)^2 = SMB. Thus, you would get a quadratic relationship that make SMA = -(2SMB - 1) +- sqrt((2SMB-1)^2 - 4(SMB)^2) in the numerator and 2SMB in the denominator. With this definition of SMA, you should be able to plug it into A = B*SMA to get A in terms of B alone. You would just have 2 possible answers for each component I guess, but maybe another function or rule could rule out the undesired answer – anonymouscoder Aug 08 '23 at 03:00
  • Sorry, plug the definition of SMA into A = B(1 + SMA), not A = B*SMA – anonymouscoder Aug 08 '23 at 03:14
  • Indeed, I just tried out this approach in c# and it works – anonymouscoder Aug 08 '23 at 03:32
  • Nice! I added math in answer for the sake of completeness – MBo Aug 08 '23 at 03:48
  • Thank you! The purpose for reversing the operation is related to geometric transformations in non-euclidean space. It is for computer graphics. – anonymouscoder Aug 09 '23 at 00:18