91Ó°ÊÓ

Consider the vectors as follows.

$\left\{{\stackrel{→}{v}}_1,{\stackrel{→}{v}}_2,{\stackrel{→}{v}}_3\right\}$

To perform the Gram-Schmidt process to obtain orthonormal vectors.

According to the Gram-Schmidt process as follows.

localid="1660119238407" $$\begin{gathered} {\stackrel{→}{u}}_1=\frac{{\stackrel{→}{v}}_1}{\left|\left|\stackrel{→}{v}\right|\right|} \\ {\stackrel{→}{u}}_2=\frac{{\stackrel{→}{v}}_2-\left(\stackrel{→}{u}·{\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1}{\left|\left|{\stackrel{→}{v}}_2-\left(\stackrel{→}{u}·{\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1\right|\right|} \\ {\stackrel{→}{u}}_3=\frac{{\stackrel{→}{v}}_3-\left({\stackrel{→}{u}}_1·{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_1-\left({\stackrel{→}{u}}_2·{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_2}{\left|\left|{\stackrel{→}{v}}_2-\left(\stackrel{→}{u}·{\stackrel{→}{v}}_2\right){\stackrel{→}{u}}_1-\left({\stackrel{→}{u}}_2·{\stackrel{→}{v}}_3\right){\stackrel{→}{u}}_2\right|\right|} \end{gathered}$$

Since,$\left|\left|{\stackrel{→}{\mathrm{v}}}_1\right|\right|=\left|\mathrm{a}\right|=\mathrm{a}$

So, $\left|\left|{\stackrel{→}{u}}_1\right|\right|=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]={\stackrel{→}{e}}_1$.

Now, in order to computelocalid="1660119335550" ${\stackrel{→}{u}}_2$as follows.

$$\begin{gathered} {\stackrel{→}{\mathrm{u}}}_1·{\stackrel{→}{\mathrm{v}}}_2=\mathrm{b} \\ \left({\stackrel{→}{\mathrm{u}}}_1·{\stackrel{→}{\mathrm{v}}}_2\right){\stackrel{→}{\mathrm{u}}}_1=\frac{1}{2}\left[\begin{array}{c}\mathrm{b}\\ 0\\ 0\end{array}\right] \\ {\stackrel{→}{\mathrm{v}}}_2-\left({\stackrel{→}{\mathrm{u}}}_1·{\stackrel{→}{\mathrm{v}}}_2\right){\stackrel{→}{\mathrm{u}}}_1=\left[\begin{array}{c}\mathrm{b}\\ \mathrm{c}\\ 0\end{array}\right]-\left[\begin{array}{c}\mathrm{b}\\ 0\\ 0\end{array}\right] \\ {\stackrel{→}{\mathrm{v}}}_2-\left({\stackrel{→}{\mathrm{u}}}_1·{\stackrel{→}{\mathrm{v}}}_2\right){\stackrel{→}{\mathrm{u}}}_1=\left[\begin{array}{c}0\\ \mathrm{c}\\ 0\end{array}\right] \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} ||{\stackrel{→}{\mathrm{v}}}_2-\left({\stackrel{→}{\mathrm{u}}}_1·{\stackrel{→}{\mathrm{v}}}_2\right){\stackrel{→}{\mathrm{u}}}_1||=\left|\mathrm{c}\right|=\mathrm{c} \\ {\stackrel{→}{\mathrm{u}}}_2={\stackrel{→}{\mathrm{e}}}_2 \end{gathered}$$

In order to find${\stackrel{→}{\mathrm{u}}}_3$ need to compute the following.

$$\begin{gathered} {\stackrel{→}{\mathrm{u}}}_1·{\stackrel{→}{\mathrm{v}}}_3=d \\ \left({\stackrel{→}{\mathrm{u}}}_1·{\stackrel{→}{\mathrm{v}}}_3\right){\stackrel{→}{\mathrm{u}}}_1=\frac{1}{2}\left[\begin{array}{c}d\\ 0\\ 0\end{array}\right] \\ {\stackrel{→}{\mathrm{u}}}_2·{\stackrel{→}{\mathrm{v}}}_3=e \\ \left({\stackrel{→}{\mathrm{u}}}_2{\stackrel{→}{\mathrm{v}}}_3\right){\stackrel{→}{\mathrm{u}}}_2=\left[\begin{array}{c}0\\ e\\ 0\end{array}\right] \\ {\stackrel{→}{v}}_3-\left({\stackrel{→}{\mathrm{u}}}_1·{\stackrel{→}{\mathrm{v}}}_3\right){\stackrel{→}{\mathrm{u}}}_1-\left({\stackrel{→}{\mathrm{u}}}_2·{\stackrel{→}{\mathrm{v}}}_3\right){\stackrel{→}{\mathrm{u}}}_2=\left[\begin{array}{c}d\\ e\\ f\end{array}\right]-\left[\begin{array}{c}d\\ 0\\ 0\end{array}\right]-\left[\begin{array}{c}0\\ e\\ 0\end{array}\right] \\ {\stackrel{→}{v}}_3-\left({\stackrel{→}{\mathrm{u}}}_1·{\stackrel{→}{\mathrm{v}}}_3\right){\stackrel{→}{\mathrm{u}}}_1-\left({\stackrel{→}{\mathrm{u}}}_2·{\stackrel{→}{\mathrm{v}}}_3\right){\stackrel{→}{\mathrm{u}}}_2=\left[\begin{array}{c}0\\ 0\\ f\end{array}\right] \\ \left|\left|{\stackrel{→}{v}}_3-\left({\stackrel{→}{\mathrm{u}}}_1·{\stackrel{→}{\mathrm{v}}}_3\right){\stackrel{→}{\mathrm{u}}}_1-\left({\stackrel{→}{\mathrm{u}}}_2·{\stackrel{→}{\mathrm{v}}}_3\right){\stackrel{→}{\mathrm{u}}}_2\right|\right|=\left|f\right|=f \\ {\stackrel{→}{\mathrm{u}}}_3=\frac{1}{2}\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]={\stackrel{→}{e}}_3 \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} {\stackrel{→}{v}}_3-\left({\stackrel{→}{\mathrm{u}}}_1·{\stackrel{→}{\mathrm{v}}}_3\right){\stackrel{→}{\mathrm{u}}}_1-\left({\stackrel{→}{\mathrm{u}}}_2·{\stackrel{→}{\mathrm{v}}}_3\right){\stackrel{→}{\mathrm{u}}}_2=\left[\begin{array}{c}d\\ e\\ f\end{array}\right]-\left[\begin{array}{c}d\\ 0\\ 0\end{array}\right]-\left[\begin{array}{c}0\\ e\\ 0\end{array}\right] \\ {\stackrel{→}{v}}_3-\left({\stackrel{→}{\mathrm{u}}}_1·{\stackrel{→}{\mathrm{v}}}_3\right){\stackrel{→}{\mathrm{u}}}_1-\left({\stackrel{→}{\mathrm{u}}}_2·{\stackrel{→}{\mathrm{v}}}_3\right){\stackrel{→}{\mathrm{u}}}_2=\left[\begin{array}{c}0\\ 0\\ f\end{array}\right] \\ \left|\left|{\stackrel{→}{v}}_3-\left({\stackrel{→}{\mathrm{u}}}_1·{\stackrel{→}{\mathrm{v}}}_3\right){\stackrel{→}{\mathrm{u}}}_1-\left({\stackrel{→}{\mathrm{u}}}_2·{\stackrel{→}{\mathrm{v}}}_3\right){\stackrel{→}{\mathrm{u}}}_2\right|\right|=\left|f\right|=f \\ {\stackrel{→}{\mathrm{u}}}_3=\frac{1}{2}\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]={\stackrel{→}{e}}_3 \end{gathered}$$

Therefore, the orthonormal vectors are $\left\{{\stackrel{→}{e}}_1,{\stackrel{→}{e}}_2,{\stackrel{→}{e}}_3\right\}$.

The orthonormal vectors $\left\{{\stackrel{→}{e}}_1,{\stackrel{→}{e}}_2,{\stackrel{→}{e}}_3\right\}$are sketched in the graph as follows.

Thus, the orthonormal vectors $\left\{{\stackrel{→}{e}}_1,{\stackrel{→}{e}}_2,{\stackrel{→}{e}}_3\right\}$are sketched in the graph.