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

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

  1. anonymousinquirer

    anonymousinquirer

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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
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  2. anonymousinquirer

    HallsofIvy

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

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

    For example, If A= <2, 1, 3> then [math]|A|^2= 4+ 1+ 9= 14[/math]. So your B=<2/14, 1/14, 9/14> and [math]|B|^2= (4+ 1+ 9)/196=\frac{14}{196}= \frac{1}{14}[/math] and
    [math]\frac{1}{|B|^2}B= \14 \left<\frac{2/14, 1/14, 9/14\right>= <2, 1, 9>= A[/math].
     
    HallsofIvy, Dec 31, 2023
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