Question

Find the orthonormal basis ${\stackrel{\mathbf{\rightharpoonup }}{\mathbf{u}}}_{\mathbf{1}}\mathbf{,}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{u}}}_{\mathbf{2}}\mathbf{,}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{u}}}_{\mathbf{3}}$ of${\mathbf{\square }}^{\mathbf{3}}$ such that

role="math" localid="1660301787314" $\mathit{s}\mathit{p}\mathit{a}\mathit{n}\mathbf{\left(}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{u}}}_{\mathbf{1}}\mathbf{\right)}\mathbf{=}\mathit{s}\mathit{p}\mathit{a}\mathit{n}\mathbf{\left(}\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\mathbf{\right)}$

and

$\mathit{s}\mathit{p}\mathit{a}\mathit{n}\mathbf{(}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{u}}}_{\mathbf{1}}\mathbf{,}{\stackrel{\mathbf{\rightharpoonup }}{\mathbf{u}}}_{\mathbf{2}}\mathbf{)}\mathbf{=}\mathit{s}\mathit{p}\mathit{a}\mathit{n}\mathbf{(}\mathbf{\left[}\begin{array}{c}\mathbf{1}\\ \mathbf{2}\\ \mathbf{3}\end{array}\mathbf{\right]}\mathbf{,}\left[\begin{array}{c}1\\ 1\\ -1\end{array}\right]\mathbf{)}$

Short answer

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

Step-by-step solution

Orthonormal basis

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

Find orthonormal basis

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

So,${\stackrel{\rightharpoonup }{u}}_1$can be obtained as:

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

First calculate localid="1660304480112" width="154">v2⇀⊥=v⇀2-(u⇀1·v⇀2)u⇀1:

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

Now, compute ${\stackrel{\rightharpoonup }{u}}_2$:

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

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

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

So, the value of third vector${\stackrel{\rightharpoonup }{u}}_3$ is:

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

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