Answer
$B'=\{(\frac{4}{5},-\frac{3}{5},0),(\frac{3}{5},\frac{4}{5},0),(0,0,1)\}$ is orthonormal basis for $R^3$.
Work Step by Step
Let $B=\{ v_1=(4,-3,0),v_2=(1,2,0), v_3=(0,0,4) \}$.
Applying the Gram-Schmidt orthonormalization process produces
$$w_1=v_1=(4,-3,0)$$
$$w_2=v_2- \frac{v_2\cdot w_1}{w_1\cdot w_1}w_1=(1,2,0)+ \frac{2}{25}(4,-3,0)=(\frac{33}{25},\frac{44}{25},0)$$
$$w_3=v_3- \frac{v_3\cdot w_1}{w_1\cdot w_1}w_1- \frac{v_3\cdot w_2}{w_2\cdot w_2}w_2\\
=(0,0,4)-0\frac{•}{•}(4,-3,0)-0(\frac{33}{25},\frac{44}{25},0) =(0,0,4).$$
Normalizing $w_1$, $w_2$ and $w_3$ produces the orthonormal set
$$u_1=\frac{w_1}{\|w_1\|}=\frac{1}{5}(4,-3,0)=(\frac{4}{5},-\frac{3}{5},0) $$
$$u_2=\frac{w_2}{\|w_2\|}=\frac{5}{11}(\frac{33}{25},\frac{44}{25},0)=(\frac{3}{5},\frac{4}{5},0)$$
$$u_3=\frac{w_3}{\|w_3\|}=\frac{1}{4}(0,0,4)=(0,0,1)$$
Hence, $B'=\{(\frac{4}{5},-\frac{3}{5},0),(\frac{3}{5},\frac{4}{5},0),(0,0,1)\}$ is orthonormal basis for $R^3$.
