Prove that the vectors $u,v,w$ are orthonormal in $V$

Question

Let $V$ be the vector space $\Bbb R^3$ with inner product $$\langle v,w\rangle=3v_1w_1-2v_1w_2-2v_2w_1+5v_2w_2-3v_2w_3-3v_3w_2+3v_3w_3$$

where $v=v_1,v_2,v_3$ and $w=w_1,w_2,w_3$.

Prove that the vectors $u=(1,1,1)$, $v=(0,1/2,(\sqrt6-2/2\sqrt6)$, $w=(0,1/2,3+\sqrt6/6)$ are orthonormal in $V$.

Answer 1

We define: $\langle v,w\rangle=3v_1w_1-2v_1w_2-2v_2w_1+5v_2w_2-3v_2w_3 -3v_3w_2+3v_3w_3$

Where $u^\top=\begin{pmatrix}1&1&1\end{pmatrix}$, $v^\top=\begin{pmatrix}0&\frac{1}{2}&\frac{\sqrt{6}-2}{2\sqrt{6}}\end{pmatrix}$, we have $\langle u, v\rangle = 3(0)-2(\frac{1}{2})-2(0)+5(\frac{1}{2}) -3(\frac{\sqrt{6}-2}{2\sqrt{6}})-3(\frac{1}{2})+3(\frac{\sqrt{6}-2}{2\sqrt{6}})= 0$

$\langle u, u\rangle$ = 3(1) -2(1) -2(1)+5(1)-6(1)+3(1)=1, which means that $\|u\| =1$, since $0\le \|u\| = \sqrt{\langle u, u\rangle}$.

Now, do the same procedure to show $v$ and $w$ are norm 1 and calculate the inner products $\langle u,w\rangle$ and $\langle v, w \rangle$ to confirm they are equal to 0, i.e. orthogonal.