Let V be a non-zero inner product space in $R^3$ and let u = {$u_{1}$, $u_{2}$, $u_{3}$} be any base for V.

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

Let $u_{1} = [1, 2, 0]$ & $u_{2} = [0, 3, 1]$

Step1: $v_{1} = u_{1} = (1, 2, 0)$

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

Now $\lt u_{2}, v_{1}\gt = (0, 3, 1) * (1, 2, 0) = 6$

$||v_{1}||^2 = 1 + 4 = 5 \\ \therefore v_{2} = (0, 3, 1) - \frac{6}{5}(1, 2, 0) = \bigg(-\frac{6}{5}, \frac{3}{5}, 1 \bigg) \\ ||v_{2}||^2 = \frac{36}{25} + \frac{9}{25} + 1 = \frac{14}{5}$

Hence $v_{1}$ = (1, 2, 0) & $v_{2} = \bigg(-\frac{6}{5}, \frac{3}{5}, 1 \bigg)$ forms orthogonal basis of $R^3$

Now, the norms of these vectors are

$$||v_{1}|| = \sqrt{5} \\ ||v_{2}|| = \sqrt{\frac{14}{5}}$$

Hence orthonormal basis for $R^3$ is

$$q_{1} = \frac{v_{1}}{||v_{1}||} = \bigg(\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}, 0 \bigg) \\ q_{2} = \frac{v_{2}}{||v_{2}||} = \bigg(\frac{-6}{\sqrt{70}}, \frac{3}{\sqrt{70}}, \sqrt{\frac{5}{14}}\bigg)$$