\[\mbox {proj}_{\vec {d}}\vec {v}=\begin{bmatrix}\answer {-1.1}\\\answer {3.3}\end{bmatrix}\]
Orthogonal Projections
Given a line \(l\) and a vector \(\vec {v}\) emanating from a point on \(l\), it is sometimes convenient to express \(\vec {v}\) as the sum of a vector \(\vec {v}_{\parallel }\), parallel to \(l\), and a vector \(\vec {v}_{\perp }\), perpendicular to \(l\). If you have taken a physics course, you may have seen a force vector decomposed into the sum of two components: one parallel and one perpendicular to the direction of motion.
![[Picture]](main.online-9ef076c77a152cf858fc98e5a9d9a377.svg)
Suppose \(\vec {d}\) is a direction vector for \(l\). Then \(\vec {v}_ {\parallel }=k\vec {d}\) for some scalar \(k\). Our goal is to find \(k\).
\begin{align*}\vec {v}\dotp \vec {d}&=(\vec {v}_{\parallel }+\vec {v}_{\perp })\dotp \vec {d}\\ &=(k\vec {d}+\vec {v}_{\perp })\dotp \vec {d}\\ &=k\vec {d}\dotp \vec {d}+\vec {v}_{\perp }\dotp \vec {d}\\ &=k\norm {\vec {d}}^2+0\\ &=k\norm {\vec {d}}^2 \end{align*}
We conclude that
\[k=\frac {\vec {v}\dotp \vec {d}}{\norm {\vec {d}}^2}\]
and
\[\vec {v}_{\parallel }=k\vec {d}=\left (\frac {\vec {v}\dotp \vec {d}}{\norm {\vec {d}}^2}\right )\vec {d}\]
The vector \(\vec {v}_{\parallel }=\left (\frac {\vec {v}\dotp \vec {d}}{\norm {\vec {d}}^2}\right )\vec {d}\) is called the projection of \(\vec {v}\) onto \(\vec {d}\). In our discussion, \(\vec {d}\) is a direction vector for line \(l\). So, we can also say that \(\vec {v}_{\parallel }\) is the projection of \(\vec {v}\) onto \(l\).
To find \(\vec {v}_{\perp }\), observe that \(\vec {v}_{\perp }=\vec {v}-\vec {v}_{\parallel }\).
Let \(\vec {v}\) be a vector, and let \(\vec {d}\) be a non-zero vector. The projection of \(\vec {v}\) onto \(\vec {d}\) is given
by
\[\mbox {proj}_{\vec {d}}\vec {v}=\left (\frac {\vec {v}\dotp \vec {d}}{\norm {\vec {d}}^2}\right )\vec {d}\]
Find the projection of \(\vec {v}\), shown below, onto the line given by \(y=\frac {1}{2}x-1\).
![[Picture]](main.online-207d36487d8e4704f86fd41b8a3d952b.svg)
We begin by finding vectors \(\vec {v}\) and \(\vec {d}\). The tail of \(\vec {v}\) is located at \((-2, -2)\), and the head of \(\vec {v}\) is at \((2, 4)\).
Using the “head-tail" formula we get
\[\vec {v}=\begin{bmatrix}2-(-2)\\4-(-2)\end{bmatrix}=\begin{bmatrix}4\\6\end{bmatrix}\]
The direction vector for the line \(y=\frac {1}{2}x-1\) is
\[\vec {d}=\begin{bmatrix}2\\1\end{bmatrix}\]
We find that \(\vec {v}\dotp \vec {d}=14\) and \(\norm {\vec {d}}^2=5\). Thus
\[\mbox {proj}_{\vec {d}}\vec {v}=\left (\frac {\vec {v}\dotp \vec {d}}{\norm {\vec {d}}^2}\right )\vec {d}=\frac {14}{5}\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}28/5\\14/5\end{bmatrix}\]
Distance from a Point to a Line
The shortest distance from a point to a line is the length of the perpendicular line segment dropped from the point to the line. Vector projection formula will help us find the length of such a perpendicular.
Let \(A(2, -1, 1)\) be a point in \(\RR ^3\). Suppose line \(l\) is given by parametric equations
\[x=t+3\]
\[y=-t+1\]
\[z=t-2\]
![[Picture]](main.online-3bf27056a05aaf4c5cddabb0a6ed4fbe.svg)
Find the distance from \(A\) to \(l\).
We will first construct a vector \(\vec {v}\) by picking an
arbitrary point \(B\) on \(l\) to be the tail of \(\vec {v}\) and using point \(A\) as the head of \(\vec {v}\). An
easy point to choose on line \(l\) is the point \((3, 1, -2)\) that corresponds to \(t=0\). Now we
have
\[\vec {v}=\overrightarrow {BA}=\begin{bmatrix}2-3\\-1-1\\1-(-2)\end{bmatrix}=\begin{bmatrix}-1\\-2\\3\end{bmatrix}\]
![[Picture]](main.online-4bcc32c84b11d1647a369627e7bcd9e7.svg)
The line has a direction vector
\[\vec {d}=\begin{bmatrix}1\\-1\\1\end{bmatrix}\]
We will now find the projection of \(\overrightarrow {BA}\) onto \(l\)
\[\mbox {proj}_{\vec {d}} \overrightarrow {BA}=\left (\frac {\vec {v}\dotp \vec {d}}{\norm {\vec {d}}^2}\right )\vec {d}=\frac {4}{3}\begin{bmatrix}1\\-1\\1\end{bmatrix}=\begin{bmatrix}4/3\\-4/3\\4/3\end{bmatrix}\]
![[Picture]](main.online-6c3b19661cbf8242572d6b37794c311f.svg)
Next, we find \(\vec {v}_{\perp }\).
\[\vec {v}_{\perp }=\vec {v}-\vec {v}_{\parallel }=\begin{bmatrix}-1\\-2\\3\end{bmatrix}-\begin{bmatrix}4/3\\-4/3\\4/3\end{bmatrix}=\begin{bmatrix}-7/3\\-2/3\\5/3\end{bmatrix}\]
![[Picture]](main.online-9cc9bc89e68b04b16fe2b531d37e356e.svg)
Finally, to find the distance between point \(A\) and line \(l\), we find the magnitude of \(\vec {v}_{\perp }\).
\[\norm {\vec {v}_{\perp }}=\frac {1}{3}\sqrt {49+4+25}=\frac {\sqrt {78}}{3}\]
Practice Problems
Find \(\mbox {proj}_{\vec {d}}\vec {v}\).
If \(\vec {d}=\begin{bmatrix}-1\\3\end{bmatrix}\) and \(\vec {v}=\begin{bmatrix}1\\4\end{bmatrix}\) then
If \(\vec {d}=\begin{bmatrix}0\\2\\1\end{bmatrix}\) and \(\vec {v}=\begin{bmatrix}-1\\-4\\2\end{bmatrix}\) then
\[\mbox {proj}_{\vec {d}}\vec {v}=\begin{bmatrix}\answer {0}\\\answer {-2.4}\\\answer {-1.2}\end{bmatrix}\]
Find the projection of vector \(\vec {v}\) onto line \(l\). (If entering answers in decimal form,
round to the nearest one hundredth.)
![[Picture]](main.online-4cada60ed18eed18cbc68f5ea53134ca.svg)
Answer:
\[\begin{bmatrix}\answer [tolerance=0.005]{-95/26}\\\answer [tolerance=0.005]{19/26}\end{bmatrix}\]
![[Picture]](main.online-9d8dbe59e12b4c72df0a0bc22dc706ee.svg)
Show that \(\mbox {proj}_{\vec {d}}\vec {v}\) does not depend on the length of \(\vec {d}\) by proving that \(\mbox {proj}_{\vec {d}}\vec {v}=\mbox {proj}_{k\vec {d}}\vec {v}\) for \(k\neq 0\). What does
this result mean geometrically? Illustrate your response with a diagram.
Find the radius of a circle centered at \((4, 2)\) if the line \(y=\frac {3}{2}x+3\) is tangent to the circle. Enter
your response as a fraction.
Answer:
\[r=\sqrt {\answer {196/13}}\]
The graph below shows the line \(y=\frac {3}{2}x+3\) together with a circle of radius \(1\). Change the value
of \(r\) to the radius you have found to visualize the correct answer.