Difference between revisions of "031 Review Part 2, Problem 6"
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<span class="exam">(c) Let <math style="vertical-align: -5px">L=\text{Span }\{\vec{v}\}.</math> Compute the orthogonal projection of <math style="vertical-align: -3px">\vec{y}</math> onto <math style="vertical-align: 0px">L.</math> | <span class="exam">(c) Let <math style="vertical-align: -5px">L=\text{Span }\{\vec{v}\}.</math> Compute the orthogonal projection of <math style="vertical-align: -3px">\vec{y}</math> onto <math style="vertical-align: 0px">L.</math> | ||
| − | |||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Foundations: | !Foundations: | ||
| + | |- | ||
| + | |'''1.''' The distance between the vectors <math style="vertical-align: 0px">\vec{u}</math> and <math style="vertical-align: 0px">\vec{v}</math> is | ||
| + | |- | ||
| + | | | ||
| + | ::<math>\text{dist}(\vec{u},\vec{v})=||\vec{u}-\vec{v}||.</math> | ||
| + | |- | ||
| + | |'''2.''' The orthogonal projection of <math style="vertical-align: -3px">\vec{y}</math> onto <math style="vertical-align: 0px">L</math> is | ||
|- | |- | ||
| | | | ||
| + | ::<math>\hat{y}=\text{proj}_L \vec{y}=\frac{\vec{y}\cdot \vec{v}}{\vec{v}\cdot \vec{v}}\vec{v}.</math> | ||
|} | |} | ||
| Line 29: | Line 36: | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Step 1: | !Step 1: | ||
| + | |- | ||
| + | |First, we calculate <math style="vertical-align: -4px">||\vec{v}||.</math> | ||
| + | |- | ||
| + | |We get | ||
|- | |- | ||
| | | | ||
| + | <math>\begin{array}{rcl} | ||
| + | \displaystyle{||\vec{v}||} & = & \displaystyle{\sqrt{(-1)^2+3^2+0^2}}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\sqrt{1+9}}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\sqrt{10}.} | ||
| + | \end{array}</math> | ||
|} | |} | ||
| Line 36: | Line 54: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
| − | | | + | |Now, to get a unit vector in the direction of <math style="vertical-align: -4px">\vec{v},</math> we take the vector <math style="vertical-align: 0px">\vec{v}</math> and divide by <math style="vertical-align: -4px">||\vec{v}||.</math> |
| + | |- | ||
| + | |Hence, we get the vector | ||
| + | |- | ||
| + | | <math>\begin{array}{rcl} | ||
| + | \displaystyle{\frac{1}{||\vec{v}||}\vec{v}} & = & \displaystyle{\frac{1}{\sqrt{10}}\begin{bmatrix} | ||
| + | -1 \\ | ||
| + | 3 \\ | ||
| + | 0 | ||
| + | \end{bmatrix}}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\begin{bmatrix} | ||
| + | \frac{-1}{\sqrt{10}} \\ | ||
| + | \frac{3}{\sqrt{10}} \\ | ||
| + | 0 | ||
| + | \end{bmatrix}.} | ||
| + | \end{array}</math> | ||
|} | |} | ||
| Line 43: | Line 77: | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Step 1: | !Step 1: | ||
| + | |- | ||
| + | |Using the formula in the Foundations section, we have | ||
|- | |- | ||
| | | | ||
| + | <math>\begin{array}{rcl} | ||
| + | \displaystyle{\text{dist}(\vec{v},\vec{y})} & = & \displaystyle{||\vec{v}-\vec{y}||}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\Bigg|\Bigg|\begin{bmatrix} | ||
| + | -3 \\ | ||
| + | 3 \\ | ||
| + | -5 | ||
| + | \end{bmatrix}\Bigg|\Bigg|.} | ||
| + | \end{array}</math> | ||
|} | |} | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Step 2: | !Step 2: | ||
| + | |- | ||
| + | |Continuing, we get | ||
|- | |- | ||
| | | | ||
| + | <math>\begin{array}{rcl} | ||
| + | \displaystyle{\text{dist}(\vec{v},\vec{y}) } & = & \displaystyle{\sqrt{(-3)^2+3^2+(-5)^2}}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\sqrt{9+9+25}}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\sqrt{34}.} | ||
| + | \end{array}</math> | ||
|} | |} | ||
| Line 57: | Line 111: | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Step 1: | !Step 1: | ||
| + | |- | ||
| + | |Using the formula in the Foundations section, we have | ||
|- | |- | ||
| | | | ||
| + | <math>\begin{array}{rcl} | ||
| + | \displaystyle{\hat{y}} & = & \displaystyle{\frac{\vec{y}\cdot \vec{v}}{\vec{v}\cdot \vec{v}}\vec{v}}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\frac{-1(2)+3(0)+0(5)}{(-1)(-1)+3(3)+0(0)}\begin{bmatrix} | ||
| + | -1 \\ | ||
| + | 3 \\ | ||
| + | 0 | ||
| + | \end{bmatrix}.} | ||
| + | \end{array}</math> | ||
|} | |} | ||
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Step 2: | !Step 2: | ||
| + | |- | ||
| + | |Continuing, we get | ||
|- | |- | ||
| | | | ||
| + | <math>\begin{array}{rcl} | ||
| + | \displaystyle{\hat{y}} & = & \displaystyle{\frac{-2}{20}\begin{bmatrix} | ||
| + | -1 \\ | ||
| + | 3 \\ | ||
| + | 0 | ||
| + | \end{bmatrix}}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\begin{bmatrix} | ||
| + | \frac{1}{10} \\ | ||
| + | \frac{-3}{10} \\ | ||
| + | 0 | ||
| + | \end{bmatrix}.} | ||
| + | \end{array}</math> | ||
|} | |} | ||
| Line 71: | Line 151: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | | '''(a)''' | + | | '''(a)''' <math>\begin{bmatrix} |
| + | \frac{-1}{\sqrt{10}} \\ | ||
| + | \frac{3}{\sqrt{10}} \\ | ||
| + | 0 | ||
| + | \end{bmatrix}</math> | ||
| + | |- | ||
| + | | '''(b)''' <math>\sqrt{34}</math> | ||
|- | |- | ||
| − | | '''(b)''' | + | | '''(b)''' <math>\begin{bmatrix} |
| + | \frac{1}{10} \\ | ||
| + | \frac{-3}{10} \\ | ||
| + | 0 | ||
| + | \end{bmatrix}</math> | ||
|} | |} | ||
| − | [[031_Review_Part_2|'''<u>Return to | + | [[031_Review_Part_2|'''<u>Return to Review Problems</u>''']] |
Latest revision as of 12:34, 15 October 2017
Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{v}=\begin{bmatrix} -1 \\ 3 \\ 0 \end{bmatrix}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{y}=\begin{bmatrix} 2 \\ 0 \\ 5 \end{bmatrix}.}
(a) Find a unit vector in the direction of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{v}.}
(b) Find the distance between Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{v}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{y}.}
(c) Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L=\text{Span }\{\vec{v}\}.} Compute the orthogonal projection of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{y}} onto Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L.}
| Foundations: |
|---|
| 1. The distance between the vectors Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{u}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{v}} is |
|
| 2. The orthogonal projection of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{y}} onto Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} is |
|
Solution:
(a)
| Step 1: |
|---|
| First, we calculate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ||\vec{v}||.} |
| We get |
|
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{||\vec{v}||} & = & \displaystyle{\sqrt{(-1)^2+3^2+0^2}}\\ &&\\ & = & \displaystyle{\sqrt{1+9}}\\ &&\\ & = & \displaystyle{\sqrt{10}.} \end{array}} |
| Step 2: |
|---|
| Now, to get a unit vector in the direction of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{v},} we take the vector Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{v}} and divide by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ||\vec{v}||.} |
| Hence, we get the vector |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\frac{1}{||\vec{v}||}\vec{v}} & = & \displaystyle{\frac{1}{\sqrt{10}}\begin{bmatrix} -1 \\ 3 \\ 0 \end{bmatrix}}\\ &&\\ & = & \displaystyle{\begin{bmatrix} \frac{-1}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} \\ 0 \end{bmatrix}.} \end{array}} |
(b)
| Step 1: |
|---|
| Using the formula in the Foundations section, we have |
|
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\text{dist}(\vec{v},\vec{y})} & = & \displaystyle{||\vec{v}-\vec{y}||}\\ &&\\ & = & \displaystyle{\Bigg|\Bigg|\begin{bmatrix} -3 \\ 3 \\ -5 \end{bmatrix}\Bigg|\Bigg|.} \end{array}} |
| Step 2: |
|---|
| Continuing, we get |
|
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\text{dist}(\vec{v},\vec{y}) } & = & \displaystyle{\sqrt{(-3)^2+3^2+(-5)^2}}\\ &&\\ & = & \displaystyle{\sqrt{9+9+25}}\\ &&\\ & = & \displaystyle{\sqrt{34}.} \end{array}} |
(c)
| Step 1: |
|---|
| Using the formula in the Foundations section, we have |
|
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\hat{y}} & = & \displaystyle{\frac{\vec{y}\cdot \vec{v}}{\vec{v}\cdot \vec{v}}\vec{v}}\\ &&\\ & = & \displaystyle{\frac{-1(2)+3(0)+0(5)}{(-1)(-1)+3(3)+0(0)}\begin{bmatrix} -1 \\ 3 \\ 0 \end{bmatrix}.} \end{array}} |
| Step 2: |
|---|
| Continuing, we get |
|
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\hat{y}} & = & \displaystyle{\frac{-2}{20}\begin{bmatrix} -1 \\ 3 \\ 0 \end{bmatrix}}\\ &&\\ & = & \displaystyle{\begin{bmatrix} \frac{1}{10} \\ \frac{-3}{10} \\ 0 \end{bmatrix}.} \end{array}} |
| Final Answer: |
|---|
| (a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{bmatrix} \frac{-1}{\sqrt{10}} \\ \frac{3}{\sqrt{10}} \\ 0 \end{bmatrix}} |
| (b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{34}} |
| (b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{bmatrix} \frac{1}{10} \\ \frac{-3}{10} \\ 0 \end{bmatrix}} |