Chapter 2.6 Describing Rotation in Cartesian Coordinates
In this section, we will use vectors to derive a formula that describes how to rotate points in the plane with a fixed center. We will see how this connects with motion on a circle.
Abstract Motions on a Circle
To begin, we need to understand what an arc is.
Arc
Suppose that \(p\) and \(q\) are points on a circle.
The arc from \(p\) to \(q\) is the set of all points lying counterclockwise from \(p\) and clockwise from \(q\).
The point \(p\) is the tail of the arc and the point \(q\) is the head.
Here is an example of an arc.
Example 1
The arc from \((1,0)\) to \((0,1)\) is shown below. The arc from \((0,1)\) to \((1,0)\) is shown below.
In order to develop rotation, we want to first start off by describing movement on a circle. Given two points, we want to define what it means to “add” two points. To do this, think of traveling on the circle as following an arc. Here is a picture to describe what we mean.
Given two points \(p\) and \(q\) on the unit circle, we want “addition, \(p\star q\), to describe adding \(p\) and \(q\) by starting at the point \(q\) and traveling along the same”amount” as the arc from \((1,0)\) to \(p\).
The goal now is to figure out a formula to describe addition. First, look at this next example to build up your intution.
Example 2
The point \(\left(\frac{3}{5}, \frac{4}{5}\right)\) lies on the circle. Find the coordinates of \(\left(\frac{3}{5}, \frac{4}{5}\right)\star (0,1).\)
Draw a picture like this:
The point \(\left(\frac{3}{5}, \frac{4}{5}\right)\star (0,1)\) should be the point obtained by taking \((0,1)\) and adding on the arc from \((1,0)\) to \(\left(\frac{3}{5}, \frac{4}{5}\right).\) If we draw a right triangle with verticles \((0,0)\), \(\left(\frac{3}{5}, 0\right)\) and \(\left(\frac{3}{5}, \frac{4}{5}\right)\), then when we rotate \((0,1)\) by \(\left(\frac{3}{5}, \frac{4}{5}\right)\), the resulting point will form another right triangle with the same measurements. From the picture, we can then see that the point is in quadrant II, so \(x\) will be negative.
Therefore, the point should be this:
\[\left(\frac{3}{5},\frac{4}{5}\right)\star (0,1)=\left(-\frac{4}{5},\frac{3}{5}\right).\]
In the next section, we will derive the formula for rotation.
Circle Actions and the Method of Coordinates on a Circle
Based on the geometry we observed in the previous example, this is what rotation should be defined as.
Rotation
Let \((a, b)\) and \((c, d)\) be points on the unit circle.
The sum is given by \[(a,b)\star(c,d) = (ac - bd, ad + bc).\]
Note: read \(\star\) as “star”.
We wil prove this in a bit. However, in order to aid in recalling the formula, here is a picture you can use.
The first coordinate is the difference of two products, specifically the product of the first coordinates and then we subtract by the product of the second coordinates. The second coordinate is the sum of two products, specifically one of the products is the product of the first coordinate of the first point and the second coordinate of the second point and the other product is of the second coordinate of the first point and the first coordinate of the second point.
Practice using this formula for the following example.
Example 3
Compute \(\left(\frac{2}{3}, \frac{\sqrt{5}}{3}\right)\star \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\).
\[\left(\frac{2}{3},\frac{\sqrt{5}}{3}\right)\star\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)=\left(\frac{2}{3}\cdot\frac{1}{2}-\frac{\sqrt{5}}{3}\cdot \frac{\sqrt{3}}{2},\frac{2}{3}\cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{5}}{3}\cdot \frac{1}{2}\right)\]
Now, we will go over the derivation of the formula.
To begin, let’s simplify the derivative by assuming that \(p\) is a point in the first quadrant that is not \((0,1)\).
Take \(p\) and \(q\) to have these coordinates:
\[p=(a,b)\quad\text{and}\quad q=(c,d).\]
Take \(V\) to be the vector given by \[V = \langle a, b\rangle.\] That is, this is the point that takes \((0,0)\) to \(p.\)
Because \(p\) is on the unit circle, \(V\) is a unit vector and so \[V_\perp = \langle -b, a\rangle\] is perpendicular to \(V\) and is also a unit vector.
If \(V\) moves the origin to a point \(A\), then \(V_\perp\) moves the origin to a point \(B\), where \(B\) is in the quadrant immediately counterclockwise from \(A\).
Therefore, to get \(p\star q\), translate \((0,0)\) by \[\begin{align*}aV+bV_{\perp}&=\langle ac,ad\rangle+\langle -bd,bc \rangle\\ &=\langle ac-bd,ad+bc\rangle.\end{align*}\]
So the rotated point is \[aV+bV_{\perp}+(0,0)=(ac-bd,ad+bc).\]
Now we will check that \(aV+bV_{\perp}+(0,0)\) is on the unit circle by calculating its magnitude:
\[ \begin{align*} \|\langle ac-bd,ad+bc\rangle\|&=\sqrt{(ac-bd)^2+(ad+bc)^2}\\ &=\sqrt{(ac)^2-2acbd+(bd)^2+(ad)^2+2adbc+(bc)^2}\\ &=\sqrt{a^2c^2+a^2d^2+b^2d^2+b^2c^2}\\ &=\sqrt{a^2(c^2+d^2)+b^2(c^2+d^2)}\\ &=\sqrt{a^2\cdot 1+b^2\cdot 1}&&\text{ since }q=(c,d) \text{ is on the unit circle, } c^2+d^2=1\\ &=\sqrt{a^2+b^2}\\ &=\sqrt{1}&&\text{ since }p=(a,b) \text{ is on the unit circle, } a^2+b^2=1\\ &=1. \end{align*} \]
Points on the unit circle are what we use to give rise to transformations on the plane called rotations.
These transformations are referred to as rigid motions because they do not change the distance between points.
This is how we define rotation of a point \(p\), not on the unit circle, by a point \(q\) on the unit circle.
Rotation by Point on Unit Circle Around Origin
Suppose that \(p\) is a point on the unit circle and \(q\) is a point in the plane not equal to \((0,0)\).
Define \(R_p\), the rotation by \(p\) about \((0,0)\) in the following way: \[R_p(q) = p\star q\] where the operation is formally the same as before. This is to say that if \(p\) is \((a,b)\) and \(q\) is \((c,d)\), then \[R_{(a,b)}(c,d) = (ac-bd, ad+bc).\]
Note: for any \(p\) on the unit circle, \(R_p(0,0)\) is defined to be \((0,0)\)
The rotation formula is the same, but the effect is different. See this picture to understand the situation.
Although movement still takes place on a circle, it is not movement on the unit circle. Instead, it is movement on circle with radius equal to the magnitude of the the vector that moves \((0,0)\) to \(q\).
Example 4
Let \(p\) be the point \(\left(\frac{1}{3}, \frac{\sqrt{8}}{3}\right)\) on the unit circle. Let \(q\) be the point \((1,4)\). Calculate \(R_p(q)\).
Calculate to get \[R_p(1,4)=\left(\frac{1}{3},\frac{\sqrt{8}}{3}\right)\star(1,4)=\left(\frac{1}{3}\cdot 1-\frac{\sqrt{8}}{3}\cdot 4,\frac{1}{3}\cdot 4+\frac{\sqrt{8}}{3}\cdot 1\right).\]
In order to describe motion related to rotation, we need to develope further our understanding of how to add points.
First, we define the following point.
Rotation Additive Inverse
Take \(p\) to be the point \[p = (a,b).\]
Define \(p^{-1}\) (read “p inverse”) to be the point \[p^{-1} = (a, -b)\quad \text{so that} \quad p\star p^{-1} = (1,0).\]
View \(\star\) as a kind of addition operation on the unit circle, so that \(p^{-1}\) is the “additive” inverse of \(p\) and \((1,0)\) is an “additive” identity.
Here is how to visualize this concept.
Essentially, adding \(p\) and \(p^{-1}\) together produces the point \((1,0)\). This point is associated with the rotation that appears to “do nothing.”
Example 5
If \(p=\left(\frac{2}{3}, \frac{\sqrt{5}}{3}\right)\), then \(p^{-1}=\left(\frac{2}{3}, -\frac{\sqrt{5}}{3}\right)\).
The next thing we want to be able to do is know by how much we need to rotate to go from one point to the next.
That is, given points \(p\) and \(q\) on the unit circle, determine a point \(r\) on the unit circle so that \[p\star r = q.\]
Solve for \(r\) by rotating by \(p^{-1}\) to obtain the equality \[r = p^{-1}\star q.\]
Use this value for \(r\) to obtain the equalities \[p\star r = p\star (p^{-1}\star q) = (p\star p^{-1})\star q = (1,0)\star q = q.\]
Try this out with the next example.
Example 6
Find a point \(r\) so that \[\left(\tfrac{1}{2}, \tfrac{\sqrt{3}}{2}\right)\star r = \left(\tfrac{3}{5},\tfrac{4}{5}\right).\]
If \(\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\star r = \left(\frac{3}{5},\frac{4}{5} \right)\), then
\[ \begin{align*} r&=\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)\star\left(\frac{3}{5},\frac{4}{5}\right)\\ &=\left(\frac{1}{2}\cdot \frac{3}{5}+\frac{\sqrt{3}}{2}\cdot \frac{4}{5},\frac{1}{2}\cdot \frac{4}{5}-\frac{\sqrt{3}}{2}\cdot \frac{3}{5}\right) \end{align*} \]
Rotating Points about an Arbitrary Point
So far, we have focused on doing rotations around \((0,0)\).
Now, the goal is to rotate by any point with any center.
Refer to any point on the unit circle as an angle.
To rotate a point \(p\) about a point \(q\) by an angle \(a\), first translate \(q\) to the origin by the vector \(-V\), where \[V= q - (0,0).\]
Second, rotate the point \(-V + p\) around \((0,0)\) by the angle \(a\).
Finally, translate the rotated point by \(V\) to obtain the point \(V + R_a(-V + p)\).
Try this out with the next example.
Example 7
Rotate the point \((1,2)\) around the point \((3, 5)\) by the angle \(\big(\frac{3}{5}, \frac{4}{5}\big)\).
In this example, \(p=(1,2)\), \(q=(3,5)\), and \(a=\big(\frac{3}{5}, \frac{4}{5}\big)\).
Calculate to get \(V=q-O=\langle 3,5 \rangle .\)
Rotate \(-V+p\) around \(O\) by the angle \(a\):
\[R_a(-V+p)=R_a\left(\langle -3,-5 \rangle +(1,2)\right)=R_a(-2,-3)=\left(\frac{3}{5}\cdot (-2)-\frac{4}{5}\cdot (-3),\frac{3}{5}\cdot (-3)+\frac{4}{5}\cdot (-2)\right).\] Now translate by \(V\) to get final answer:
\[p=V+R_a(-2,3)=\left(3+\frac{3}{5}\cdot (-2)-\frac{4}{5}\cdot (-3),5+\frac{3}{5}\cdot (-3)+\frac{4}{5}\cdot (-2)\right).\]