Calculating an orthonormal base given another base. 
Let $$W = \operatorname{span}(\{(1,1,1),(0,0,1)\})$$
Find an orthonormal base $B$ of $W$.

So. An orthonormal set is a base whose elements are orthogonal with each other and their length is $1$. The vector $(0,0,1)$ is already of length $1$, so we could put it into our answer.
$$B = \{(0,0,1)\}$$
Since $(0,0,1)$ isn't a scalar product of $(1,1,1)$, it seems clear that we require another vector in $B$ to properly span $W$.
I am not sure how to calculate this second vector. Let's try...
So, this second vector is $(a,b,c)$ such that:
$$(0,0,1) \cdot (a,b,c) = 0\\
(a,b,c)\cdot(a,b,c) = 1$$
For the first condition, $c$ must be $0$. So...
$$(0,0,1) \cdot (a,b,0) = 0\\
(a,b,0)\cdot(a,b,0) = 1$$
Now the first condition is fulfilled and we can ignore it. Good. Now we need to worry only about $a$ and $b$ in the second condition:
$$(a,b,0)\cdot(a,b,0) = 1$$
Which is to say, the equation:
$$a\cdot a + b \cdot b = 1$$
$$a^2 = 1 - b^2$$
$$a = \sqrt{1 - b^2}$$
$$a = 1 - b$$
$$a + b = 1$$
Well. $$a = b = \frac{1}{2}$$
Therefore
$$B = \left\{(0,0,1) , \left( \frac{1}{2} , \frac{1}{2} , 0 \right) \right\}$$
But this is wrong because $\left( \frac{1}{2} , \frac{1}{2} , 0 \right)$ is not of length $1$...... By looking at it, it is easy to say that the answer should be $\left( \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} , 0 \right)$, but I am not sure where is the problem in my procedure.
Is there a more standard way to solve this sort of problem? I feel like it would get very difficult with larger bases.

Hint:  
Take $\langle 0, 0, 1 \rangle$ to be one of the vectors since it is already of length $1$.  Now simply take the second vector and subtract its projection onto $\langle 0, 0, 1 \rangle$ from it.  Finally, if the resulting vector isn't of length $1$, then simply normalize it.
Draw yourself a picture to see what's going on.  Can you see why you end up with two orthogonal vectors that still span the original space?  You can also show orthogonality algebraically, but its less instructive than drawing a picture.
