\[\mbox {proj}_{\vec {d}}\vec {v}=\begin{bmatrix}\answer {-1.1}\\\answer {3.3}\end{bmatrix}\]
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
\[\mbox {proj}_{\vec {d}}\vec {v}=\left (\frac {\vec {v}\dotp \vec {d}}{\norm {\vec {d}}^2}\right )\vec {d}=\frac {14}{5}\begin{bmatrix}2\\1\end{bmatrix}=\begin{bmatrix}28/5\\14/5\end{bmatrix}\]
Practice Problems
Find \(\mbox {proj}_{\vec {d}}\vec {v}\) if \(\vec {d}=\begin{bmatrix}-1\\3\end{bmatrix}\) and \(\vec {v}=\begin{bmatrix}1\\4\end{bmatrix}\).
Find \(\mbox {proj}_{\vec {d}}\vec {v}\) if \(\vec {d}=\begin{bmatrix}0\\2\\1\end{bmatrix}\) and \(\vec {v}=\begin{bmatrix}-1\\-4\\2\end{bmatrix}\).
\[\mbox {proj}_{\vec {d}}\vec {v}=\begin{bmatrix}\answer {0}\\\answer {-2.4}\\\answer {-1.2}\end{bmatrix}\]
![[Picture]](main.online-efd88b82181f44706e7f49236927ad7e.svg)