91Ó°ÊÓ

The orthonormal basis is given by ${\stackrel{\mathbf{â‡\P Ä}}{\mathbf{u}}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\left|\left|{\stackrel{â‡\P Ä}{v}}_1\right|\right|}{\stackrel{\mathbf{â‡\P Ä}}{\mathbf{v}}}_{\mathbf{1}}$and ${\stackrel{\mathbf{â‡\P Ä}}{\mathbf{u}}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\left|\left|{{\stackrel{â‡\P Ä}{v}}_1}^{âŠ\yen }\right|\right|}{{\stackrel{\mathbf{â‡\P Ä}}{\mathbf{v}}}_2}^{\mathbf{âŠ\yen }}$where ${\stackrel{\mathbf{â‡\P Ä}}{{\mathbf{v}}_{\mathbf{2}}}}^{\mathbf{âŠ\yen }}\mathbf{=}{\stackrel{\mathbf{â‡\P Ä}}{\mathbf{v}}}_{\mathbf{2}}\mathbf{-}\mathbf{(}{\stackrel{\mathbf{â‡\P Ä}}{\mathbf{u}}}_{\mathbf{1}}\mathbf{·}{\stackrel{\mathbf{â‡\P Ä}}{\mathbf{v}}}_{\mathbf{2}}\mathbf{)}{\stackrel{\mathbf{â‡\P Ä}}{\mathbf{u}}}_{\mathbf{1}}$.

The Gram-Schmidt process has to be used to find the orthonormal basis for the given vectors. The given vectors are ${\stackrel{â‡\P Ä}{v}}_1=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$and ${\stackrel{â‡\P Ä}{v}}_2=\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]$.

So,${\stackrel{â‡\P Ä}{u}}_1$can be obtained as:

$\begin{array}{rcl}{\stackrel{â‡\P Ä}{u}}_1& =& \frac{1}{\left|\left|{\stackrel{â‡\P Ä}{v}}_1\right|\right|}{\stackrel{â‡\P Ä}{v}}_1\\ & =& \frac{1}{\sqrt{1^2+2^2+3^2}}\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\\ & =& \frac{1}{\sqrt{14}}\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\end{array}$

First calculate localid="1660304480112" width="154">v2⇶Ä⊥=v⇶Ä2-(u⇶Ä1·v⇶Ä2)u⇶Ä1:

$\begin{array}{rcl}{\stackrel{â‡\P Ä}{{\mathrm{v}}_2}}^{âŠ\yen }& =& {\stackrel{â‡\P Ä}{\mathrm{v}}}_2-({\stackrel{â‡\P Ä}{\mathrm{u}}}_1·{\stackrel{â‡\P Ä}{\mathrm{v}}}_2){\stackrel{â‡\P Ä}{\mathrm{u}}}_1\\ & =& \left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]-\left(\frac{1}{\sqrt{14}}\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]·\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]\right)\frac{1}{\sqrt{14}}\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\\ & =& \left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]-0\left(\frac{1}{\sqrt{14}}\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\right)\\ & =& \left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]\end{array}$

Now, compute ${\stackrel{â‡\P Ä}{u}}_2$:

role="math" localid="1660305643084" $\begin{array}{rcl}{\stackrel{â‡\P Ä}{\mathrm{u}}}_2& =& \frac{1}{\left|\left|{{\stackrel{â‡\P Ä}{\mathrm{v}}}_2}^{âŠ\yen }\right|\right|}{{\stackrel{â‡\P Ä}{\mathrm{v}}}_2}^{âŠ\yen }\\ & =& \frac{1}{\sqrt{1^2+1^2+{\left(-1\right)}^2}}\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]\\ & =& \frac{1}{\sqrt{3}}\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]\end{array}$

Now, we have to compute a third vector${\stackrel{â‡\P Ä}{v}}_3$ which is orthogonal to${\stackrel{â‡\P Ä}{u}}_1$ and ${\stackrel{â‡\P Ä}{u}}_2$. So, we have to find${\stackrel{â‡\P Ä}{v}}_3=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]$ such that${\stackrel{â‡\P Ä}{u}}_1·{\stackrel{â‡\P Ä}{v}}_3=0$ and ${\stackrel{â‡\P Ä}{u}}_2·{\stackrel{â‡\P Ä}{v}}_3=0$. So, we can write:

role="math" localid="1660305258291" $\begin{array}{rcl}\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]·\left[\begin{array}{c}a\\ b\\ c\end{array}\right]& =& 0\\ a+2b+3c& =& 0\\ a& =& -2b-3c\end{array}$

And

$\begin{array}{rcl}\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]·\left[\begin{array}{c}a\\ b\\ c\end{array}\right]& =& 0\\ a+b+c& =& 0\\ -2b+3c+b-c& =& 0\\ -b-4c& =& 0\\ b& =& -4c\end{array}$

So, we have:

$$\begin{gathered} a=-2b-3c \\ a=-2\left(-4c\right)-3c \\ a=5c \end{gathered}$$

So, we can say that the value ofrole="math" localid="1660305467143" ${\stackrel{â‡\P Ä}{v}}_3$ is:

$\begin{array}{rcl}{\stackrel{â‡\P Ä}{v}}_3& =& \left[\begin{array}{c}a\\ b\\ c\end{array}\right]\\ & =& \left[\begin{array}{c}5c\\ -4c\\ c\end{array}\right]\\ & =& c\left[\begin{array}{c}5\\ -4\\ 1\end{array}\right]\end{array}$

So, the value of third vector${\stackrel{â‡\P Ä}{u}}_3$ is:

$\begin{array}{rcl}{\stackrel{â‡\P Ä}{u}}_3& =& \frac{1}{\left|\left|{\stackrel{â‡\P Ä}{v}}_3\right|\right|}{\stackrel{â‡\P Ä}{v}}_3\\ & =& \frac{1}{\sqrt{5^2+{(-4)}^2+1^2}}\left[\begin{array}{c}5\\ -4\\ 1\end{array}\right]\\ & =& \frac{1}{\sqrt{42}}\left[\begin{array}{c}5\\ -4\\ 1\end{array}\right]\end{array}$

The orthonormal basis is $\left\{{\stackrel{â‡\P Ä}{\mathrm{u}}}_1,{\stackrel{â‡\P Ä}{\mathrm{u}}}_2,{\stackrel{â‡\P Ä}{\mathrm{u}}}_3\right\}=\left\{\frac{1}{\sqrt{14}}\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right],\frac{1}{\sqrt{3}}\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right],\frac{1}{\sqrt{42}}\left[\begin{array}{c}-5\\ 4\\ -1\end{array}\right],\right\}$.