$V1,V2,V3$ Orthonormal. Given $v_1,v_2,v_3$ vectors in $R^3$. Assume that ${v_1,v_2,v_1+v_3}$ is an orthonormal base.
Compute $||v_3||$.
Well, I started by saying that $(v_1+v_3) \bullet (v_1+v_3) = 1$.
And, $v_1 \bullet (v_1+v_3) = 0$
From here, I need your help. Much appreciated!
 A: You're on the right track:
$$\|v_3\|^2 = \|(v_1 + v_3) - v_1\|^2 = (v_1 + v_3, v_1 + v_3) - 2(v_1 + v_3, v_1) + (v_1, v_1) = 1 - 0 + 1 = 2,$$
since $v_1 + v_3$ and $v_3$ are orthonormal.
A: Hint: From this relation
$$
\begin{align}
0
&=v_1\cdot(v_1+v_3)\\
&=\|v_1\|^2+v_1\cdot v_3\\
&=1+v_1\cdot v_3
\end{align}
$$
we can compute $v_1\cdot v_3$. Then from
$$
\begin{align}
1
&=(v_1+v_3)\cdot(v_1+v_3)\\
&=\|v_1\|^2+2v_1\cdot v_3+\|v_3\|^2\\
&=1+2v_1\cdot v_3+\|v_3\|^2
\end{align}
$$
we can compute $\|v_3\|$.
A: Note that
$$\begin{eqnarray}v_1 \bullet v_3 &=& v_1 \bullet (v_1 + v_3 - v_1) \\
&=& v_1 \bullet (v_1+v_3) - ||v_1||^2 \\
&=& -1
\end{eqnarray}$$
and
$$\begin{eqnarray}(v_1+v_3) \bullet (v_1+v_3) &=& v_1 \bullet (v_1+v_3) + v_3 \bullet (v_1 + v_3) \\
&=& v_1 \bullet v_1 + v_1 \bullet v_3 + v_3 \bullet v_1 + v_3 \bullet v_3 \\
&=& ||v_1||^2 + 2 (v_1 \bullet v_3) + ||v_3||^2 \\
\end{eqnarray}$$
which yields that
$$\begin{eqnarray}||v_3||^2 &=& ||(v_1+v_3)||^2 - ||v_1||^2 - 2(v_1 \bullet v_3) \\
&=&  2\end{eqnarray}$$
Thus $||v_3|| = \sqrt 2$. 
