# function to obtain original vector after dividing it its square magnitude?

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

1. ### anonymousinquirer

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Aug 8, 2023
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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 possible solution.

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

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Nov 6, 2021
Messages:
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Vector A= <x, y, z> has magnitude $|A|= \sqrt{x^2+ y^2+ z^2}$ so "square magnitude", as you say, $x^2+ y^2+ z^2$. Dividing a vector by its magnitude, |A|, gives a unit vector, with length 1. Dividing by $|A|^2$ gives a vector with length $\frac{1}{A}$.
It should be no surprise that multiplying by $|A|^2$ returns the original vector..

If you are given the vector B then |A|= 1/|B| so you return to A by multiplying B by $\frac{1}{|B|^2}$.

For example, If A= <2, 1, 3> then $|A|^2= 4+ 1+ 9= 14$. So your B=<2/14, 1/14, 9/14> and $|B|^2= (4+ 1+ 9)/196=\frac{14}{196}= \frac{1}{14}$ and
$\frac{1}{|B|^2}B= \14 \left<\frac{2/14, 1/14, 9/14\right>= <2, 1, 9>= A$.

HallsofIvy, Dec 31, 2023