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I have an equation given in terms of three vectors $\vec{x}$, $\vec{y}$ and $\vec{z}$, all in $\mathbb{R}^n$:

$$1 - (\vec{x}\cdot\vec{y})^2 = K + 2K\vec{z}\cdot\vec{y} + (\vec{z}\cdot\vec{y})^2, K \in \mathbb{R}$$

I want to solve for both $\vec{x}$ and $\vec{z}$, given $\vec{y}$ and $K$. Now, I can make a transformation of variables $u = \vec{x}\cdot\vec{y}$, $v = \vec{z}\cdot\vec{y}$. I now get a quadratic:

$$1-u^2 = K + 2Kv + v^2$$

Now, I suspect that I may have a much better time trying to solve the quadratic parametrically than I had trying to solve the original vector equation. However, this will eventually tell me that $u$ and $v$ lie on some simpler geometric objects (probably hyperplanes, I haven't yet worked it out). I have two questions:

1) Assuming that the solutions exists in the reals for $u$ and $v$, can I just solve the change-of-variable equations (which are hyperplanes) to get $\vec{x}$ and $\vec{z}$, or do I need to intersect the solutions or something?

2) Is my change of variables valid here? I feel like I'm moving from an $n$ dimensional space into a 2 dimensional space, and something tells me I'm going to lose a lot of information in doings so. Will I be able to get a complete solution this way?

Michael Stachowsky
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The change of variables is totally fine, but you're probably not gonna get an unique solution (or a solution at all) for $u,v$. Then again if you do find a solution for $u,v$ you will not be able to solve for $\vec x, \vec y, \vec z$.

Suppose you have a solution for $v$, then you know $$v = \vec z \vec y = \sum_{k=1}^n z_k y_k$$

There is no way you could solve this equatin for one of those vectors. Even when you're just in the case $n=1$ you just have a relation, and you stil need to solve $u^2 = (\vec x \cdot \vec y)^2 = \sum_k x_k^2 y_k^2$.

flawr
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  • I'm not sure I agree with that statement. If I have a solution for $v$, then $\vec{z} = \frac{v}{||\vec{y}||^2}\vec{y} + \vec{n}$, where $\vec{n}$ is in the orthogonal complement of $\vec{y}$ is a solution. Thus it reduces to arbitrarily choosing $\vec{n}$ from said orthogonal complement, which can be done arbitrarily. Perhaps I should have mentioned that I don't want a unique solution, just an easy way to choose parameters such that the equation is satisfied. At the moment, I can't arbitrarily choose my vectors, which is hard – Michael Stachowsky Jun 16 '16 at 13:11
  • Of course, if you are not looking for an unique solution you can do that, as you have a lot of "freedom". – flawr Jun 16 '16 at 13:55