Orthogonal Projections

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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.

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 $y=\frac {1}{2}x-1$ .

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 $x=t+3$ $y=-t+1$ $z=t-2$

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 $\vec {v}=\overrightarrow {BA}=\begin {bmatrix}2-3\\-1-1\\1-(-2)\end {bmatrix}=\begin {bmatrix}-1\\-2\\3\end {bmatrix}$

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 $l$ $\mbox {proj}_{\vec {d}} \overrightarrow {BA}=\left (\frac {\vec {v}\dotp \vec {d}}{\norm {\vec {d}}^2}\right )\vec {d}=\frac {4}{3}\begin {bmatrix}1\\-1\\1\end {bmatrix}=\begin {bmatrix}4/3\\-4/3\\4/3\end {bmatrix}$

Next, we find $\vec {v}_{\perp }$ . $\vec {v}_{\perp }=\vec {v}-\vec {v}_{\parallel }=\begin {bmatrix}-1\\-2\\3\end {bmatrix}-\begin {bmatrix}4/3\\-4/3\\4/3\end {bmatrix}=\begin {bmatrix}-7/3\\-2/3\\5/3\end {bmatrix}$

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

Problems prob:proj1a-prob:proj1b

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.)

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 $l$ .

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.