Answer
$B'=\{(\frac{51}{25},\frac{68}{25}),(\frac{3}{5},\frac{4}{5}))\}$ is orthonormal basis for $R^2$.
Work Step by Step
Let $B=\{ v_1=(4,-3),v_2=(3,2) \}$.
Applying the Gram-Schmidt orthonormalization process produces
$$w_1=v_1=(4,-3)$$
$$w_2=v_2- \frac{v_2\cdot w_1}{w_1\cdot w_1}w_1=(3,2)-\frac{6}{25}(4,-3)=(\frac{51}{25},\frac{68}{25}).$$
Normalizing $w_1$and $w_2$ produces the orthonormal set
$$u_1=\frac{w_1}{\|w_1\|}=(\frac{4}{5},-\frac{3}{5})$$
$$u_2=\frac{w_2}{\|w_2\|}=\frac{5}{17}(\frac{51}{25},\frac{68}{25}) =(\frac{3}{5},\frac{4}{5}).$$
Hence, $B'=\{(\frac{51}{25},\frac{68}{25}),(\frac{3}{5},\frac{4}{5}))\}$ is orthonormal basis for $R^2$.
