Gram-Schmidt Process yielding strange vector

Question

Let B = $\{(1,i,1),(2,-1,1+i)\}$. To find the orthonormal basis, I use the Gram-schmidt Process.

$\vec{u}_1$ = $\frac{1}{\sqrt{3}}(1,i,1)$

$\vec{u}_2$ = $\frac{1}{|(2,-1,1+i) - \frac{3 + 2i}{\sqrt{3}}(1,i,1)|}[(2,-1,1+i) - \frac{3 + 2i}{\sqrt{3}}(1,i,1)] = \frac{1}{|(2-\frac{3+2i}{\sqrt{3}}, -\frac{3+2i}{\sqrt{3}}, 1+i-\frac{3+2i}{\sqrt{3}})|}[(2-\frac{3+2i}{\sqrt{3}}, -\frac{3+2i}{\sqrt{3}}, 1+i-\frac{3+2i}{\sqrt{3}})]$

After this step, I'm fairly sure I'm doing something wrong here, I'm not sure how to evaluate $\vec{u}_2$ without it looking super ugly, or if I did the computation incorrectly. Wolfram alpha says that $\vec{u}_2 = \frac{1}{2\sqrt{6}}(3-2i,1+3i,i)$ which I have no idea how to get to. Some guidance would be helpful!

Btw, this is my first Gram-Schmidt process I've ever done, so it's very likely I did something wrong.

Answer 1

Just a minor mistake in your calculation. $u_2=b_2-\frac{<b_2,u_1>}{<u_1,u_1>}u_1$ and as $<b_2,u_1> = \frac{(3+2i)}{\sqrt{3}}$ and $u_1=\frac{1}{\sqrt{3}}(1,i,1)$, you have a $\frac{1}{\sqrt{3}}$ factor missing in your expression. Answer will be $u_2=\frac{1}{2\sqrt{6}}(3-2i,-1-3i,i)$

Answer 2

First we normalize the first vector: $$ u_1=\frac1{\sqrt3}(1,i,1) $$ Then compute the second vector: $$ \begin{align} u_2 &=\sqrt{\frac38}\left(\vphantom{\frac1{\sqrt3}}\right.(2,-1,1+i)-(2,-1,1+i)\cdot\overbrace{\frac1{\sqrt3}(1,-i,1)}^{\overline{u_1}}\overbrace{\color{#C00000}{\frac1{\sqrt3}}(1,i,1)}^{u_1}\left.\vphantom{\frac1{\sqrt3}}\right)\\ &=\frac1{2\sqrt6}(3-2i,-1-3i,i) \end{align} $$ It seems you have left out a $\color{#C00000}{\frac1{\sqrt3}}$ from part of your computation.