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.

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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 \(\vec {v}_{\perp }=\vec {v}-\vec {v}_{\parallel }\).

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.

Practice Problems

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
\[\mbox {proj}_{\vec {d}}\vec {v}=\begin{bmatrix}\answer {-1.1}\\\answer {3.3}\end{bmatrix}\]
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.)

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Answer:

\[\begin{bmatrix}\answer [tolerance=0.005]{-95/26}\\\answer [tolerance=0.005]{19/26}\end{bmatrix}\]
Find the distance between point \(A\) and line \(l\).

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Answer: \(\sqrt {\answer {3.2}}\).

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.