How to solve dot products simultaneously?

Question

For context, I have to solve this equation for $u$:

$\mathbf{u}\cdot\mathbf{u}=1$

$\mathbf{u}\cdot\begin{pmatrix}2\\1\\-2\end{pmatrix}=0$

$\mathbf{u}\cdot\begin{pmatrix}3\\-2\\-1\end{pmatrix}=\pm 3$

They were derived from the following question (I am aware of the true method of solving this):

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However, I would like know if there is a practical method to prove this using the equations I have derived, in the more general case where I have:

$\mathbf{u}\cdot\mathbf{u}=a$

$\mathbf{u}\cdot\mathbf{v}=b$

$\mathbf{u}\cdot\mathbf{w}=c$

I would generate the matrix equation:

$\begin{pmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1&w_2&w_3\end{pmatrix}\begin{pmatrix}u_1\\u_2\\u_3\end{pmatrix}=\begin{pmatrix}a\\b\\c\end{pmatrix}$

However I don't know how I would solve this.

My question is, how would I solve a matrix equation of this form?

Answer 1

Solve the two linear equations: the result can be put in the parametric form ${\bf u} = {\bf d} + t \bf e$. Then ${\bf u} \cdot {\bf u} = a$ becomes a quadratic in $t$.

Answer 2

If $\mathbf v$ and $\mathbf w$ are not collinear, write $$\mathbf{u}=\alpha\mathbf{v}+\beta\mathbf{w}+\gamma\mathbf{v}\times\mathbf{w}$$ For a 3D vector, you are writing three components along some reference axes, not necessarily orthogonal. If you really want orthogonal axes, replace $\mathbf w$, in the second term only, with $\mathbf v\times(\mathbf v\times\mathbf w)$.