Find orthonormal basis of $\mathbb{R}^3$ with a given span of two basis vectors What is an orthonormal basis of $\mathbb{R}^3$ such that $\text{span }(\vec{u_1},\vec{u_2})=\left\{\begin{bmatrix}1\\2\\3\end{bmatrix},\begin{bmatrix}1\\1\\-1\end{bmatrix}\right\}$?
I was thinking I could use  $\text{proj }_V\vec{x}=(\vec{u_1}\cdot\vec{x})\vec{u_1}+\cdots+(\vec{u_m}\cdot\vec{x})\vec{u_m}$, or the fact that $\vec{u_1}=\frac{\vec{v_1}}{||\vec{v_1}||},\vec{u_2}=\frac{\vec{v_2}^\perp}{||\vec{v_2}^\perp||}$ to back out values for $\vec{v_1}, \vec{v_2}$ and then pick any $\vec{v_3}$, but I've only been able to solve $\vec{v_1}=\frac{1}{\sqrt{14}}\begin{bmatrix}1\\2\\3\end{bmatrix}$.
Is there a more direct way to solve for $\vec{u_3}$?

EDIT:
The text lists $\frac{1}{\sqrt{14}}\begin{bmatrix}1\\2\\3\end{bmatrix},\frac{1}{\sqrt{3}}\begin{bmatrix}1\\1\\-1\end{bmatrix},\frac{1}{\sqrt{42}}\begin{bmatrix}5\\-4\\1\end{bmatrix}$. I'm guessing they used $\text{proj }_V\vec{x}$ and took $\vec{u_2}=\frac{\vec{u_2}}{||\vec{u_2}||}$ The notation here is confusing because, as mixedmath mentioned, and as I failed to see, $\vec{u_2}$ is not a unit vector. 
 A: Here's how I might approach this problem. I write vectors horizontally and vertically here as a complete abuse of notatation, whenever I think that the meaning is clear.
I find the vector $(1, 1, -1)$ is easier to work with than the other so, why don't we choose it to start. Unfortunately, it's not a unit vector. So let $b_1 = \frac{1}{\sqrt 3} (1, 1, -1)$, which is now unit vector.
We can form $b_2$ by noting that $(1,2,3) \cdot (1,1,-1) = 0$, so they are already orthogonal. So we just need to normalize. So $b_2 = \frac{1}{\sqrt{14}} (1,2,3)$.
The fastest way to find a third vector is to find some vector not in the space spanned by $b_1, b_2$, and then to take out the components as you suggested. If you know cross-products, that's a great choice. 
But for kicks, let's see how we can not do that.
Notice that $(1,2,3) - (1,1,-1) = (0,1,4)$. I find it easier to instead work with the space spanned by $(0,1,4)$ and $(1,1,-1)$. To find a vector orthogonal to $(1,1,-1)$ with components $(x,y,z)$, we want $x + y - z = 0$, or rather $x + y = z$. To find a vector orthogonal to the first, we want $y - 4z = 0$, or rather $y = 4z$. In the latter, $x$ can be anything. Substituting into the former, we see that $x + 4z = z$, or rather $x = -3z$.
So any vector of the form $(-3, 4, 1)t$ will be orthogonal to both. Normalize, and you're done.
You might notice that this is a system of linear equations. In a linear algebra class, solving systems of linear equations is like the bread and butter of the course. So you could instead have phrased this as a question of finding the kernel of the transformation
$$ \begin{pmatrix} 1&1&-1 \\ 1&2&3 \end{pmatrix} \begin{pmatrix} x\\y\\z \end{pmatrix},$$
and then applied a classic Gaussian reduction argument. In fact, due to the way I structured my solution, you might notice that the steps I took were exactly the same as the steps you would take in a Gaussian reduction argument. $\diamondsuit$
A: Take $\vec{u}_1$ to be the first specified vector.
Take $\vec{u}_2$ to be the second specified vector, minus its orthogonal projection onto $\operatorname{Span}\left\{\vec{u}_1\right\}$.
Take $\vec{u}_3$ to be the cross-product of $\vec{u}_1$ and $\vec{u}_2$.
Rescale all three vectors so that they are length $1$.
