### The Orthogonal Basis

The first orthorgonal vector to $\mathbf{V}_1$, called $\mathbf{V}_1$ is done by

$$

\mathbf{V}_2=(-V_y, V_x, 0).

$$

We can easily verify that $\mathbf{V}_1$ and $\mathbf{V}_2$ are orthogonal by taking the inner product

$$

\langle\mathbf{V}_1,\mathbf{V}_2\rangle = V_x\cdot(-V_y)+V_y\cdot V_x + V_z\cdot 0 = -V_xV_y+V_xV_y = 0.

$$

The last orthogonal vector $\mathbf{V}_3$ is created by taking the cross-product of $\mathbf{V}_1$ and $\mathbf{V}_2$ yielding

$$

\mathbf{V}_3=\mathbf{V}_1\times\mathbf{V}_2=(V_y\cdot 0 - V_z\cdot V_x,V_z\cdot (-V_y)-V_x\cdot 0,V_x\cdot V_x - V_y\cdot (-V_y)),

$$

and can be reduced to

$$

\mathbf{V}_3=(-V_x\cdot V_z, -V_y\cdot V_z, V^2_x+V^2_y).

$$

Thus $\mathbf{V}_1$, $\mathbf{V}_2$ and $\mathbf{V}_3$ now constitue an orthogonal basis. However, normalizing by brute-force is cumbersome (and error prone) so we will take a slightly more clever route.

### The Orthonormal Basis

If we instead assume that the vector $\mathbf{V}_1$ is normalized initially, we have that$$

\hat{\mathbf{V}}_1 = (V_x, V_y, V_z).

$$

I've re-used the same symbols as before to reduce an unnecessary cluttering of the notation, but do remember that elements of $\hat{\mathbf{V}}_1$ are normalized. To help us later on, we can write the length of $\hat{\mathbf{V}}_1$ as

$$

1=\sqrt{V^2_x + V^2_y + V^2_z}=> 1=V^2_x + V^2_y + V^2_z

$$

which we will use momentarily. $\hat{\mathbf{V}}_2$ is constructed as above but we have to normalize right away

$$

\hat{\mathbf{V}}_2 = (\frac{-V_y}{\sqrt{V^2_x + V^2_y}}, \frac{V_x}{\sqrt{V^2_x + V^2_y}}, 0)=(\frac{-V_y}{\sqrt{1-V^2_z}}, \frac{V_x}{\sqrt{1-V^2_z}}, 0)

$$

In the final equality sign, we used the length of $\hat{\mathbf{V}}_1$.

Now, for the cross-product. Remember that the no normalization for $\hat{\mathbf{V}}_3$ is needed because $\hat{\mathbf{V}}_1$ and $\hat{\mathbf{V}}_2$ are already normalized

$$

\hat{\mathbf{V}}_3=\hat{\mathbf{V}}_1\times\hat{\mathbf{V}}_2=(\frac{-V_z\cdot V_x}{\sqrt{1-V^2_z}},\frac{-V_z\cdot V_y}{\sqrt{1-V^2_z}},\frac{V^2_x+V^2_y}{\sqrt{1-V^2_z}}).

$$

The last element of the $\hat{\mathbf{V}}_3$ vector is something we can take care of and make it a tad prettier. By using the length of vector $\hat{\mathbf{V}}_1$ again we obtain

$$

\hat{\mathbf{V}}_3=(\frac{-V_z\cdot V_x}{\sqrt{1-V^2_z}},\frac{-V_z\cdot V_y}{\sqrt{1-V^2_z}},\frac{1-V^2_z}{\sqrt{1-V^2_z}}),

$$

and if we remember that $x/\sqrt(x)=\sqrt(x)$ we finally obtain $\hat{\mathbf{V}}_3$ as

$$

\hat{\mathbf{V}}_3=(\frac{-V_z\cdot V_x}{\sqrt{1-V^2_z}},\frac{-V_z\cdot V_y}{\sqrt{1-V^2_z}},\sqrt{1-V^2_z}).

$$

So to sum up, given a normalized vector $\hat{\mathbf{V}}_1$, we can construct an orthonormal basis using the following equations

$$

\hat{\mathbf{V}}_1 = (V_x, V_y, V_z),

$$

$$

\hat{\mathbf{V}}_2 =(\frac{-V_y}{\sqrt{1-V^2_z}}, \frac{V_x}{\sqrt{1-V^2_z}}, 0),

$$

$$

\hat{\mathbf{V}}_3=(\frac{-V_z\cdot V_x}{\sqrt{1-V^2_z}},\frac{-V_z\cdot V_y}{\sqrt{1-V^2_z}},\sqrt{1-V^2_z}).

$$

The clever thing about these last expressions is that the normalization factor $\sqrt{1-V^2_z}$ is present many times and can be precomputed for each new $\hat{\mathbf{V}}_1$.

Acknowledgements: Thanks to Lars for being awesome with the math.

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