\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus - Dot Product Application - Vector Projections

Coordinate Systems

Vectors

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Vector Functions

Partial Derivatives

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Double Integrals - 2Int

Triple Integrals - 3Int

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Vector Functions

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Double Integrals - 2Int

Triple Integrals - 3Int

Practice

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Projection
We know from basic trigonometry, the projection of vector \(\vec{A}\) onto vector \(\vec{B}\) (see figure to the right) is just \( \|\vec{A}\| \cos(\theta)\). However, what if we don't know the value of angle \(\theta\)? We can use the dot product to find the projection of one vector onto another. Let's derive the equation.

From the picture on the right, it is easy to see that the magnitude of the projection of vector \(\vec{A}\) onto vector \(\vec{B}\) is \( \|proj_{\vec{B}} \vec{A} \| = \|\vec{A}\| \cos(\theta)\). Since we don't know the angle \(\theta\), we can use the angle between vectors formula to substitute for \(\cos(\theta)\) as follows.

\(\begin{array}{rcl} \| proj_{\vec{B}} \vec{A}\| & = & \|\vec{A}\| \cos(\theta) \\ & = & \displaystyle{ \|\vec{A}\| \frac{\vec{A} \cdot \vec{B}}{\|\vec{A}\| \|\vec{B}\|} } \\ & = & \displaystyle{ \frac{\vec{A} \cdot \vec{B}}{\|\vec{B}\|} } \end{array}\)

Notice that this result is just the magnitude of the projection of vector \(\vec{A}\) onto \(\vec{B}\). If we want to find the projection vector itself, we can multiply this result by the unit vector in the direction of \(\vec{B}\) and end up with

\(\displaystyle{ proj_{\vec{B}} \vec{A} = \frac{\vec{A} \cdot \vec{B}}{\|\vec{B}\|} \frac{\vec{B}}{\|\vec{B}\|} = \frac{\vec{A} \cdot \vec{B}}{\|\vec{B}\|^2} \vec{B} }\)

This last equation is the preferred equation since calculating \(\|\vec{B}\|\) requires taking a square root, so squaring it removes that difficulty.

Here is a great video explaining vector projection in more detail.

Dr Chris Tisdell - vector projection [17mins-55secs]

video by Dr Chris Tisdell

Vector Component
Now that we have the projection vector, it is a simple matter to find the vector component of \(\vec{A}\) orthogonal to \(\vec{B}\) (the dashed line in the above picture) using basic addition of vectors. Let's call our unknown vector \(\vec{v}\). We know that \( proj_{\vec{B}} \vec{A} + \vec{v} = \vec{A}\). Solving for \(\vec{v}\), we have the vector component of \(\vec{A}\) that is orthogonal to \(\vec{B}\) which is

\( \vec{v} = \vec{A} - proj_{\vec{B}} \vec{A} \).

Study Note - - We highly recommend that you spend some going through this section carefully until you understand it well. When you get to lines and planes, you will need to understand and be able to use this concept.

Notation Note: You may find different notation for the projection vector than we used here. The notation \( proj_{\vec{B}} \vec{A} \) is fairly common but not used exclusively.

Practice

Find the projection of \(\vec{u}=\langle 25, 25\sqrt{3} \rangle\) onto \(\vec{v}=\langle 11,4 \rangle\).

Problem Statement

Find the projection of \(\vec{u}=\langle 25, 25\sqrt{3} \rangle\) onto \(\vec{v}=\langle 11,4 \rangle\).

Final Answer

\(\langle 36,13 \rangle\)

Problem Statement

Find the projection of \(\vec{u}=\langle 25, 25\sqrt{3} \rangle\) onto \(\vec{v}=\langle 11,4 \rangle\).

Solution

These videos are part of a longer video and this is just part of the example. You do not need to know the missing parts in order to extract the solution. The explanation is shown in three video clips.

1810 video

video by Larson Calculus

1810 video

video by Larson Calculus

1810 video

video by Larson Calculus

Final Answer

\(\langle 36,13 \rangle\)

close solution

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Topics You Need To Understand For This Page

Related Topics and Links

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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