View source for Dot Products and Orthogonality

We first consider orthogonal projection onto a line.
To orthogonally project a vector <math> \vec{v} </math> onto a line <math>\ell</math>,
mark the point on the line at which someone standing on that point could see <math> \vec{v} </math> by looking straight up or down (from that person's point of view).
<center>
[[Image:Linalg_projection_3.png|x200px]]
</center>
The picture shows someone who has walked out on the line
until the tip of <math>\vec{v}</math> is straight overhead.
That is, where the line is described as the span of some nonzero
vector <math>\ell=\{c\cdot\vec{s}\,\big|\, c\in\mathbb{R}\}</math>,
the person has walked out to find the coefficient <math>c_{\vec{p}}\,</math>
with the property that <math>\vec{v}-c_{\vec{p}}\cdot\vec{s}\,</math> is orthogonal
to <math>c_{\vec{p}}\cdot\vec{s}</math>.
<center>
[[Image:Linalg_projection_4.png|x200px]]
</center>
We can solve for this coefficient by noting that because
<math>\vec{v}-c_{\vec{p}}\vec{s}\,</math> is orthogonal to a scalar multiple
of <math>\vec{s}\,</math> it must be orthogonal to <math>\vec{s}\,</math> itself,
and then the consequent fact that the dot product
<math>(\vec{v}-c_{\vec{p}}\vec{s})\cdot \vec{s}\,</math> is zero gives that
<math>c_{\vec{p}}=\vec{v}\cdot \vec{s}/\vec{s}\cdot \vec{s}</math>.


;Definition 1.1
The '''orthogonal projection of <math> \vec{v} </math> onto the line
spanned by a nonzero <math> \vec{s}\, </math>''' is this vector.

:<math>
\mbox{proj}_{[\vec{s}\,]}({\vec{v}})=
\frac{ \vec{v}\cdot\vec{s} }{ \vec{s}\cdot\vec{s} }\vec{s}
</math>

;Remark 1.2:
The wording of that definition says "spanned by <math> \vec{s}\, </math>" instead the more formal "the span of the set <math> \{\vec{s}\,\} </math>". This casual first phrase is common.



;Example 1.3:
To orthogonally project
the vector <math>\binom{2}{3}</math> onto the line <math> y=2x </math>,
we first pick a direction vector for the line.
For instance,

:<math>
\vec{s}=\begin{pmatrix} 1 \\ 2 \end{pmatrix}
</math>

will do.
Then the calculation is routine.
<center>
<TABLE border=0 cellspacing=20px>
<TR>
<TD>[[Image:Linalg_projection_5.png|x200px]]
<TD><math>\frac{ \begin{pmatrix} 2 \\ 3 \end{pmatrix}\cdot\begin{pmatrix} 1 \\ 2 \end{pmatrix} }{
\begin{pmatrix} 1 \\ 2 \end{pmatrix}\cdot\begin{pmatrix} 1 \\ 2 \end{pmatrix} }
\cdot\begin{pmatrix} 1 \\ 2 \end{pmatrix}
={\displaystyle \frac{8}{5}}\cdot\begin{pmatrix} 1 \\ 2 \end{pmatrix}=\begin{pmatrix} 8/5 \\ 16/5 \end{pmatrix}</math>
</TABLE>
</center>



;Example 1.4:
In <math> \mathbb{R}^3 </math>, the orthogonal projection of a general vector

:<math>
\begin{pmatrix} x \\ y \\ z \end{pmatrix}
</math>

onto the <math> y </math>-axis is

:<math>
\frac{ \begin{pmatrix} x \\ y \\ z \end{pmatrix}\cdot\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} }{
\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\cdot\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} }
\cdot\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}=\begin{pmatrix} 0 \\ y \\ 0 \end{pmatrix}
</math>

which matches our intuitive expectation.


The picture above with the stick figure walking out on the line until
<math>\vec{v}</math>'s tip is overhead is one way to think of the orthogonal projection of
a vector onto a line.
We finish this subsection with two other ways.


;Example 1.5
A railroad car left on an east-west track without its brake is pushed by
a wind blowing toward the northeast at fifteen miles per hour;
what speed will the car reach?
<center>
[[Image:Linalg_railroad_wind.png|x170px]]
</center>
For the wind we use a vector of length <math>15</math> that points toward
the northeast.

:<math>
\vec{v}=\begin{pmatrix} 15\sqrt{1/2} \\ 15\sqrt{1/2} \end{pmatrix}
</math>

The car can only be affected by the part of the wind blowing in the
east-west direction&mdash; the part of <math>\vec{v}</math> in the direction
of the <math>x</math>-axis is this (the picture has the same perspective as the
railroad car picture above).
<center>
<TABLE border=0px cellspacing=20px>
<TR>
<TD>[[Image:Linalg_railroad_wind_direction.png|x150px]]
<TD><math>\displaystyle \vec{p}=\begin{pmatrix} 15\sqrt{1/2} \\ 0 \end{pmatrix}</math>
</TABLE>
</center>
So the car will reach a velocity of <math> 15\sqrt{1/2} </math> miles per hour toward the east.


Thus, another way to think
of the picture that precedes the definition is that it shows
<math>\vec{v}</math> as decomposed into two parts, the part with the line (here, the part
with the tracks, <math>\vec{p}</math>),
and the part that is orthogonal to the line
(shown here lying on the north-south axis).
These two are "not interacting" or "independent",
in the sense that the east-west car is not at all affected by the
north-south part of the wind (see [[#exer:PerpAreInd|Problem 5]]<!--\ref{exer:PerpAreInd}-->).
So the orthogonal projection of <math> \vec{v} </math>
onto the line spanned by <math> \vec{s}\, </math> can be thought of as
the part of <math> \vec{v}\, </math> that lies in the direction of <math> \vec{s} </math>.

Finally, another useful way to think of the orthogonal projection
is to have the person stand not on the line, but on the vector that is to be
projected to the line.
This person has a rope over the line and pulls it tight,
naturally making the rope orthogonal to the line.
<center>
[[Image:Linalg_projection_6.png|x200px]]
</center>
That is, we can think of the projection <math> \vec{p}\, </math> as being the vector
in the line that is closest to <math> \vec{v} </math>
(see [[#exer:ProjMinimizesDist|Problem 11]]<!--\ref{exer:ProjMinimizesDist}-->).


;Example 1.6:
A submarine is tracking a ship moving along the line <math> y=3x+2 </math>.
Torpedo range is one-half mile.
Can the sub stay where it is, at the origin on the chart below,
or must it move to reach a place where the ship will pass within range?
<center>
[[Image:Linalg_submarine_direction.png|x200px]]
</center>
The formula for projection
onto a line does not immediately apply because the line doesn't pass through
the origin, and so isn't the span of any <math>\vec{s}</math>.
To adjust for this, we start by shifting the entire map down two units.
Now the line is <math>y=3x</math>, which is a subspace, and we can project to get
the point <math>\vec{p}</math> of closest approach, the point on the
line through the origin closest to

:<math>
\vec{v}=\begin{pmatrix} 0 \\ -2 \end{pmatrix}
</math>

the sub's shifted position.

:<math>
\vec{p}=\frac{\begin{pmatrix} 0 \\ -2 \end{pmatrix}\cdot\begin{pmatrix} 1 \\ 3 \end{pmatrix}}{
\begin{pmatrix} 1 \\ 3 \end{pmatrix}\cdot\begin{pmatrix} 1 \\ 3 \end{pmatrix}}
\cdot \begin{pmatrix} 1 \\ 3 \end{pmatrix}=\begin{pmatrix} -3/5 \\ -9/5 \end{pmatrix}
</math>

The distance between  <math>\vec{v}</math> and <math>\vec{p}</math> is approximately <math> 0.63 </math> miles and so the sub must move to get in range.


This subsection has developed a natural projection map: orthogonal projection
onto a line.
As suggested by the examples, it is often called for in applications.
The next subsection shows how the definition of orthogonal
projection onto a line gives us a way to calculate especially convienent bases
for vector spaces, again something that is common in applications.
The final subsection completely generalizes projection, orthogonal or not,
onto any subspace at all.

== Licensing == 
Content obtained and/or adapted from:
* [https://en.wikibooks.org/wiki/Linear_Algebra/Orthogonal_Projection_Onto_a_Line Orthogonal Projection Onto a Line, Wikibooks: Linear Algebra] under a CC BY-SA license