Given a basis $e_1, e_2, e_3$ for $\mathbb{R}^3$ such that $||e_1||=1$, $||e_1||=\sqrt{2}$, $||e_1||=\sqrt{3}$ and
$e_1\cdot e_2 = 1$, $e_1\cdot e_3 = 1$, $e_2\cdot e_3 = 1$
I'm asked to find an orthonormal basis $B$ of $\mathbb{R}^3$ in terms of $e_1, e_2, e_3$. I try to use Gram-Schimdt but get stuck.
My attempt:
Let $v_2=e_2-\text{proj}_{e_1}e_2=e_2-(e_1\cdot e_2)e_1=e_2-e_1$. Since the length of the projection is $1$, and the length of $e_2$ is $\sqrt{2}$, we have that $||e_2-e_1||=1$. Now let
$$v_3=e_3-\text{proj}_{e_1,e_2}e_3=e_3-(e_3e_1)e_1-(e_3e_2)e_2=e_3-e_1-2e_2$$
the length of $\text{proj}_{e_1,e_2}e_3$ is $$\sqrt{||\text{proj}_{e_1}e_3||^2+||\text{proj}_{e_2}e_3||^2}=\sqrt{(e_3e_1)^2e_1+(e_3e_2)^2e_2}=\sqrt{1+4}=\sqrt{5}$$
yet this certainly can't be since $||e_3||=\sqrt{3}$.
Take $f_1=e_1$. Now, let\begin{align}a_2&=e_2-(e_2.f_1)f_1\\&=e_2-e_1\end{align}and let\begin{align}f_2&=\frac{a_2}{\|a_2\|}\\&=\frac{e_2-e_1}{\|e_2-e_1\|}\\&=e_2-e_1.\end{align}Finally, let\begin{align}a_3&=e_3-(e_3.f_1)f_1-(e_3.f_2)f_2\\&=e_3-(e_3.e_1)e_1-\bigl(e_3.(e_2-e_1)\bigr)(e_2-e_1)\\&=e_3-e_1\end{align}and let\begin{align}f_3&=\frac{a_3}{\|a_3\|}\\&=\frac{e_3-e_1}{\|e_3-e_1\|}\\&=\frac{e_3-e_1}{\sqrt2}.\end{align}Then $B=\{f_1,f_2,f_3\}$.