baa71c…: “Aboard the short square”

$\newenvironment {prompt}{}{} \newcommand {\DeclareMathOperator }[2]{\@OldDeclareMathOperator {##1}{##2}\immediate \write \myfile {\unexpanded {\DeclareMathOperator }{\unexpanded {##1}}{\unexpanded {##2}}}} \newcommand {\amatrix }[1]{\left [\begin {array}{@{}*{##1}{c}|c@{}} } \newcommand {\RR }[0]{\bf R} \newcommand {\adj }[0]{\mathop {\rm adj}} \newcommand {\degrees }[0]{$^{\circ }$} \newcommand {\Col }[0]{\mathop {\rm Col}} \newcommand {\diag }[0]{\mathop {\rm diag}} \newcommand {\Inn }[0]{\mathop {\rm Inn}} \newcommand {\Nul }[0]{\mathop {\rm Nul}} \newcommand {\rank }[0]{\mathop {\rm rank}} \newcommand {\rk }[0]{\mathop {\rm rk}} \newcommand {\Row }[0]{\mathop {\rm Row}} \newcommand {\proj }[0]{\mathop {\rm proj}} \newcommand {\Span }[0]{\mathop {\rm Span}} \newcommand {\spec }[0]{\mathop {\rm spec}} \newcommand {\tr }[0]{\mathop {\rm tr}} \newcommand {\vol }[0]{\mathop {\rm vol}} \newcommand {\reserved@a }[2]{} \newcommand {\reserved@a }[2]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{} \newcommand {\AppendGraphicsExtensions }[0]{\@ifundefined {Gin@extensions}{\let \Gin@extensions \@empty }{}\@ifstar {\grfext@Append \grfext@Check }{\grfext@Append \grfext@@Add }} \newcommand {\PrependGraphicsExtensions }[0]{\@ifundefined {Gin@extensions}{\let \Gin@extensions \@empty }{}\@ifstar {\grfext@Prepend \grfext@Check }{\grfext@Prepend \grfext@@Add }} \newcommand {\RemoveGraphicsExtensions }[1]{\@ifundefined {Gin@extensions}{\def \Gin@extensions {}}{\edef \grfext@tmp {\zap@space ##1 \@empty }\@for \grfext@ext :=\grfext@tmp \do {\def \grfext@next {\let \grfext@tmp \Gin@extensions \@expandtwoargs \@removeelement \grfext@ext \Gin@extensions \Gin@extensions \ifx \grfext@tmp \Gin@extensions \let \grfext@next \relax \fi \grfext@next }\grfext@next }}\grfext@Print \RemoveGraphicsExtensions } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} \newcommand {\epstopdfcall }[1]{\ifETE@InsideSetfile \expandafter \@firstoftwo \else \expandafter \@secondoftwo \fi {‘##1}{\Gin@base \Gin@ext }} \newcommand {\epstopdfDeclareGraphicsRule }[4]{\ifx \\##4\\\@PackageError {epstopdf-base}{Conversion command is missing}\@ehc \else \begingroup \@ifundefined {Gin@rule@##1}{}{\@PackageInfo {epstopdf-base}{Redefining graphics rule for ‘##1’}}\endgroup \@namedef {Gin@rule@##1}####1{{##2}{##3}{\epstopdfcall {##4}}}\fi } \newcommand {\mdf@patchamsthm }[0]{\let \mdf@deferred@thm@head \deferred@thm@head \pretocmd {\deferred@thm@head }{\@inlabelfalse }{\mdf@PackageInfo {mdframed detected package amsthm changed the theorem header of amsthm\MessageBreak }}{\mdf@PackageError {mdframed detected package amsthm changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\GetTitleString }[0]{\ifGTS@expand \expandafter \GetTitleStringExpand \else \expandafter \GetTitleStringNonExpand \fi } \newcommand {\GetTitleStringExpand }[1]{\def \GetTitleStringResult {##1}\begingroup \GTS@DisablePredefinedCmds \GTS@DisableHook \edef \x {\endgroup \noexpand \def \noexpand \GetTitleStringResult {\GetTitleStringResult }}\x } \newcommand {\GetTitleStringNonExpand }[1]{\def \GetTitleStringResult {##1}\global \let \GTS@GlobalString \GetTitleStringResult \begingroup \GTS@RemoveLeft \GTS@RemoveRight \endgroup \let \GetTitleStringResult \GTS@GlobalString } \newcommand {\GTS@DisableHook }[0]{} \newcommand {\label@hook }[0]{} \newcommand {\@currentlabelname }[0]{} \newcommand {\reserved@a }[0]{} \newcommand {\@safe@activestrue }[0]{} \newcommand {\@safe@activesfalse }[0]{} \newcommand {\nameref }[0]{\@ifstar \T@nameref \T@nameref } \newcommand {\Sectionformat }[2]{##1} \newcommand {\vnameref }[1]{\unskip ~\nameref {##1}\@vpageref [\unskip ]{##1}} \newcommand {\ref }[0]{\@ifstar \@refstar \T@ref } \newcommand {\pageref }[0]{\@ifstar \@pagerefstar \T@pageref } \newcommand {\nameref }[0]{\@ifstar \@namerefstar \T@nameref } \newcommand {\reserved@a }[2]{} \newcommand {\reserved@a }[2]{} \newcommand {\reserved@a }[0]{\AtBeginDocument } \newcommand {\reserved@a }[1]{} \newcommand {\reserved@a }[2]{}$

How do you verify that $\{\vec {u}_1,\vec {u}_2,\vec {u}_3\}$ is an orthogonal set?

Show that $\vec {u}_1\cdot \vec {u}_2\cdot \vec {u}_3 = \vec {0}$ Show that $\vec {u}_1\cdot \vec {u}_2\cdot \vec {u}_3 = 0$ Show that $\vec {u}_1\cdot \vec {u}_2=0, \vec {u}_1\cdot \vec {u}_3 = 0$ and $\vec {u}_2\cdot \vec {u}_3 =0$ Show that $\vec {u}_1\cdot \vec {u}_2=\vec {0}, \vec {u}_1\cdot \vec {u}_3 = \vec {0}$ and $\vec {u}_2\cdot \vec {u}_3 =\vec {0}$

Use the vectors below to compute the following. $\vec {y} = \begin {bmatrix} 1\\-1\\3\end {bmatrix}, \hspace {20pt} \vec {u}_1 = \begin {bmatrix} 0\\2\\-1\end {bmatrix}, \hspace {10pt} \vec {u}_2 = \begin {bmatrix} 1\\1\\2\end {bmatrix}$ $\frac {\vec {y}\cdot \vec {u}_1}{\vec {u}_1\cdot \vec {u}_1} =\answer {-1}$ Simplify.
$\frac {\vec {y}\cdot \vec {u}_2}{\vec {u}_2\cdot \vec {u}_2} =\answer {1}$ Simplify.
Compute the othogonal projection of $\vec {y}$ onto $\Span \{\vec {u}_1,\vec {u}_2\}$ .
$\hat {\bf y} = \begin {bmatrix} \answer {1}\\\answer {-1}\\\answer {3}\end {bmatrix}$
Compute $\vec {z}$ such that $\vec {y} = \hat {\bf y} +\vec {z}$ and $\vec {z}\in (\Span \{\vec {u}_1,\vec {u}_2\} )^\perp$ .
$\vec {z} = \begin {bmatrix} \answer {0}\\\answer {0}\\\answer {0}\end {bmatrix}$

In the previous question, we find that $\vec {y} = \hat {\bf y}$ . That is that the vector is its own orthogonal projection onto $\Span \{\vec {u}_1,\vec {u}_2\}$ . Which of the following statements must be true?

$\vec {y}$ is in $\Span \{\vec {u}_1,\vec {u}_2\}$ . $\vec {y}$ is not in $\Span \{\vec {u}_1,\vec {u}_2\}$ .

Use the vectors below to compute the following. $\vec {y} = \begin {bmatrix} 3\\2\\5\end {bmatrix}, \hspace {20pt} \vec {u}_1 = \begin {bmatrix} -3\\4\\1\end {bmatrix}, \hspace {10pt} \vec {u}_2 = \begin {bmatrix} 1\\2\\-5\end {bmatrix}$ $\frac {\vec {y}\cdot \vec {u}_1}{\vec {u}_1\cdot \vec {u}_1} =\frac {\answer {2}}{\answer {13}}$ Simplify.
$\frac {\vec {y}\cdot \vec {u}_2}{\vec {u}_2\cdot \vec {u}_2} =\frac {\answer {-3}}{\answer {5}}$ Simplify.
Compute the othogonal projection of $\vec {y}$ onto $\Span \{\vec {u}_1,\vec {u}_2\}$ .
$\hat {\bf y} = \frac {1}{65}\begin {bmatrix} \answer {-69}\\\answer {-38}\\\answer {205}\end {bmatrix}$
Compute $\vec {z}$ such that $\vec {y} = \hat {\bf y} +\vec {z}$ and $\vec {z}\in (\Span \{\vec {u}_1,\vec {u}_2\} )^\perp$ .
$\vec {z} = \frac {1}{65}\begin {bmatrix} \answer {264}\\\answer {168}\\\answer {120}\end {bmatrix}$

What vector represents the closest point to $\vec {y}$ in $\Span \{\vec {u}_1,\vec {u}_2\}$ ?

$\hat {\bf y}$ $\vec {z}$ $\vec {y}$