For given sets of basis vectors, use the Gram-Schmidt method to find an orthonormal set.

(a)$\mathit{A}\mathbf{=}\left(0, 2, 0, 0\right)\mathbf{,}\mathbf{ }\mathit{B}\mathbf{=}\left(3, -4, 0, 0\right)\mathbf{,}\mathbf{ }\mathit{C}\mathbf{=}\left(1, 2, 3, 4\right)$

(b)$\mathit{A}\mathbf{=}\left(0, 0, 0, 7\right)\mathbf{,}\mathbf{ }\mathit{B}\mathbf{=}\left(2, 0, 0, 5\right)\mathbf{,}\mathbf{ }\mathit{C}\mathbf{=}\left(3, 1, 1, 4\right)$

(c)$\mathit{A}\mathbf{=}\left(6, 0, 0, 0\right)\mathbf{,}\mathbf{ }\mathit{B}\mathbf{=}\left(1, 0, 2, 0\right)\mathbf{,}\mathbf{ }\mathit{C}\mathbf{=}\left(4, 1, 9, 2\right)$

The given set of basis vectors is

$A=\left(0, 2, 0, 0 \right), B=\left(3, -4, 0, 0 \right), C=\left(1, 2, 3, 4 \right)$

First, normalize the first vector as follows:

$$\begin{gathered} e_1=\frac{\overrightarrow{A}}{||\overrightarrow{A}||} \\ =\frac{\left(0, 2, 0, 0\right)}{\sqrt{2^2}} \\ =\frac{\left(0, 2, 0, 0\right)}{2} \\ =\left(0, 1, 0, 0\right) \end{gathered}$$

Now, projecting vector $B$on the first basis and subtracting this component of vector $B$as follows:

$$\begin{gathered} B-\left(B·e_1\right)e_1=\left(3, -4, 0, 0\right)-\left(-4\right)\left(0, 1,0, 0\right) \\ =\left(3, -4, 0, 0\right)-\left(0, -4,0, 0\right) \\ =\left(3, 0, 0, 0\right) \end{gathered}$$

Let’s normalize this vector and let it be $e_2$.

$$\begin{gathered} e_2=\frac{\left(3, 0, 0, 0\right)}{\sqrt{3^2}} \\ =\frac{\left(3, 0, 0, 0\right)}{3} \\ =\left(1, 0, 0, 0\right) \end{gathered}$$

Next, for vector $C$, project it on $e_1$and $e_2$, and subtracting those components from $C$.

$$\begin{gathered} C-\left(C·e_1\right)e_1-\left(C·e_2\right)e_2=\left(1, 2, 3, 4\right)-\left(2\right)\left(0, 1,0, 0\right)-\left(1\right)\left(1, 0,0,0\right) \\ =\left(1, 2, 3, 4\right)-\left(0, 2,0, 0\right)-\left(1, 0,0,0\right) \\ =\left(0, 0, 3, 4\right) \end{gathered}$$

Now, normalizing the above vector, then the vector $e_3$ is:

$$\begin{gathered} e_3=\frac{\left(0, 0, 3, 4\right)}{\sqrt{3^2+4^2}} \\ =\frac{\left(0, 0, 3, 4\right)}{\sqrt{9+16}} \\ =\frac{\left(0, 0, 3, 4\right)}{\sqrt{25}} \\ =\frac{\left(0, 0, 3, 4\right)}{5} \end{gathered}$$

Hence, the orthonormal set is$\left(e_1, e_2, e_3\right)=\left(\left(0, 1, 0, 0\right), \left(1, 0, 0, 0\right),\frac{\left(0, 0, 3, 4\right)}{5}\right)$

The given set of basis vectors is

$A=\left(0, 0, 0, 7 \right), B=\left(2, 0, 0, 5 \right), C=\left(3, 1, 1, 4 \right)$

First, normalize the first vector as follows:

$$\begin{gathered} e_1=\frac{\overrightarrow{A}}{||\overrightarrow{A}||} \\ =\frac{\left(0, 0, 0, 7\right)}{\sqrt{7^2}} \\ =\frac{\left(0, 0, 0, 7\right)}{7} \\ =\left(0, 0, 0, 1\right) \end{gathered}$$

Now, projecting vector $B$on the first basis and subtracting this component of vector $B$.

$$\begin{gathered} B-\left(B·e_1\right)e_1=\left(2, 0, 0, 5\right)-\left(5\right)\left(0, 0,0, 1\right) \\ =\left(2, 0, 0, 5\right)-\left(0, 0,0, 5\right) \\ =\left(2, 0, 0, 0\right) \end{gathered}$$

Le’ normalize this vector and let it be $e_2$.

$$\begin{gathered} e_2=\frac{\left(2, 0, 0, 0\right)}{\sqrt{2^2}} \\ =\frac{\left(2, 0, 0, 0\right)}{2} \\ =\left(1, 0, 0, 0\right) \end{gathered}$$

Next, for vector $C$, project it on $e_1$and $e_2$, and subtracting those components from $C$as follows:

$$\begin{gathered} C-\left(C·e_1\right)e_1-\left(C·e_2\right)e_2=\left(3, 1, 1, 4\right)-\left(4\right)\left(0, 0,0, 1\right)-\left(3\right)\left(1, 0,0,0\right) \\ =\left(3, 1, 1, 4\right)-\left(0, 0,0, 4\right)-\left(3, 0,0,0\right) \\ =\left(0, 1, 1, 0\right) \end{gathered}$$

Now, normalizing the above vector, then the vector $e_3$ is:

$$\begin{gathered} e_3=\frac{\left(0, 1, 1, 0\right)}{\sqrt{1^2+1^2}} \\ =\frac{\left(0, 1, 1, 0\right)}{\sqrt{1+1}} \\ =\frac{\left(0, 1, 1, 0\right)}{\sqrt{2}} \end{gathered}$$

Hence, the orthonormal set is$\left(e_1, e_2, e_3\right)=\left(\left(0, 0, 0, 1\right), \left(1, 0, 0, 0\right),\frac{\left(0, 1, 1, 0\right)}{\sqrt{2}}\right)$

The given set of basis vectors is

$A=\left(6, 0, 0, 0 \right), B=\left(1, 0, 2, 0 \right), C=\left(4, 1, 9, 2 \right)$

First, normalize the first vector as follows:

$$\begin{gathered} e_1=\frac{\overrightarrow{A}}{||\overrightarrow{A}||} \\ =\frac{\left(6, 0, 0, 0\right)}{\sqrt{6^2}} \\ =\frac{\left(6, 0, 0, 0\right)}{6} \\ =\left(1, 0, 0, 0\right) \end{gathered}$$

Now, projecting vector $B$on the first basis and subtracting this component of vector $B$.

$$\begin{gathered} B-\left(B·e_1\right)e_1=\left(1, 0, 2, 0\right)-\left(1\right)\left(1, 0,0, 0\right) \\ =\left(1, 0, 2, 0\right)-\left(1, 0,0, 0\right) \\ =\left(0, 0, 2, 0\right) \end{gathered}$$

Let’s normalize this vector and let it be $e_2$.

$$\begin{gathered} e_2=\frac{\left(0, 0, 2, 0\right)}{\sqrt{2^2}} \\ =\frac{\left(0, 0, 2, 0\right)}{2} \\ =\left(0, 0, 1, 0\right) \end{gathered}$$

Next, for vector $C$, project it on $e_1$and $e_2$, and subtracting those components from $C$gives:

$$\begin{gathered} C-\left(C·e_1\right)e_1-\left(C·e_2\right)e_2=\left(4, 1, 9, 2\right)-\left(4\right)\left(1, 0,0, 0\right)-\left(9\right)\left(0, 0,1,0\right) \\ =\left(4, 1, 9, 2\right)-\left(4, 0,0, 0\right)-\left(0, 0,9,0\right) \\ =\left(0, 1, 0, 2\right) \end{gathered}$$

Now, normalizing the above vector, the vector $e_3$is:

$$\begin{gathered} e_3=\frac{\left(0, 1, 0, 2\right)}{\sqrt{1^2+2^2}} \\ =\frac{\left(0, 1, 0, 2\right)}{\sqrt{1+4}} \\ =\frac{\left(0, 1, 0, 2\right)}{\sqrt{5}} \end{gathered}$$

Hence, the orthonormal set is $\left(e_1, e_2, e_3\right)=\left(\left(1, 0, 0, 0\right), \left(0, 0, 1, 0\right),\frac{\left(0, 1, 0, 2\right)}{\sqrt{5}}\right)$