Answer
$B'=\{(\frac{3}{5},\frac{4}{5}),(\frac{4}{5},-\frac{3}{5})\}$ is orthonormal basis for $R^2$.
Work Step by Step
Let $B=\{ v_1=(3,4),v_2=(1,0) \}$.
Applying the Gram-Schmidt orthonormalization process produces
$$w_1=v_1=(3,4)$$
$$w_2=v_2- \frac{v_2\cdot w_1}{w_1\cdot w_1}w_1=(1,0)-\frac{3}{25}(3,4)=(\frac{16}{25},-\frac{12}{25}).$$
Normalizing $w_1$and $w_2$ produces the orthonormal set
$$u_1=\frac{w_1}{\|w_1\|}=(\frac{3}{5},\frac{4}{5})$$
$$u_2=\frac{w_2}{\|w_2\|}=\frac{5}{4}(\frac{16}{25},-\frac{12}{25})=
(\frac{4}{5},-\frac{3}{5}).$$
Hence, $B'=\{(\frac{3}{5},\frac{4}{5}),(\frac{4}{5},-\frac{3}{5})\}$ is orthonormal basis for $R^2$.
