Let V be a non-zero inner product space in $R^2$ and let u = {$u_{1}$, $u_{2}$} be any base for V.

Following are the steps to find orthonormal basis using Gram-Schmidt process:

Let $u_{1} = [3,1]$ & $u_{2} = [2,2]$

Step1: $v_{1} = u_{1} = (3,1)$

Step 2: $v_{2} = u_{2} - proj u_{2} = u_{2} - \frac{\lt u_{2}, v_{1}\gt}{||v_{1}||^2}v_{1}$

Now $\lt u_{2}, v_{1}\gt = (2,2) * (3,1) = 6 + 2 = 8$

$||v_{1}||^2 = 9 + 1 = 10 \\ \therefore v_{2} = (2,2) - \frac{8}{10}(3,1) = \bigg(-\frac{2}{5}, \frac{1}{5}\bigg) \\ ||v_{2}||^2 = \frac{4}{25} + \frac{1}{25} + 1 = \frac{1}{5}$

Hence $v_{1}$ = (3,1) & $v_{2} = \bigg(-\frac{2}{5}, \frac{1}{5}\bigg)$ forms orthogonal basis of $R^3$

Now, the norms of these vectors are

$$||v_{1}|| = \sqrt{10} \\ ||v_{2}|| = \frac{1}{\sqrt{5}}$$

Hence orthonormal basis for $R^3$ is

$$q_{1} = \frac{v_{1}}{||v_{1}||} = \bigg(\frac{3}{\sqrt{10}}, \frac{1}{\sqrt{10}} \bigg) \\ q_{2} = \frac{v_{2}}{||v_{2}||} = \bigg(\frac{-2}{\sqrt{5}}, \frac{1}{\sqrt{5}}\bigg)$$