If I study a physical system, such as a car, and let it drive forward a little bit, say a distance $m$, then I can draw out the right triangle and find the car's position at $(m\cos \theta, m\sin\theta, 0)$. But why is the car's total displacement, exactly sideways after parallel parking, given by the vector field $(\sin\theta, - \cos\theta, 0)$?
I can't seem to see this. My guess is that I have some basic high school math that I just have to brush up on, but I've been trying for awhile now and thought I'd ask the question on this forum at this point.
Thanks,
Sometimes you just have to draw a picture. Based on the question, with some reinforcement in its comments, I hope this is the correct picture of what is wanted:
So the car's direction of forward travel would be in the direction of the longer vector (of length $m$) while the end result of "parallel parking" would be to move the car in the direction of the shorter vector, that is, $1$ unit distance in a direction perpendicular to the forward path and on the right side of that path.
Following the convention that angles are measured counterclockwise, and given an angle $\theta$ between the longer vector and the $x$-axis, the coordinates of the car after moving forward a distance $m$ (that is the longer vector) would be $(m \cos \theta, m \sin \theta)$.
There are a couple of ways to find the coordinates at the end of the parallel parking maneuver. By geometry, we can construct a right triangle with hypotenuse $m$ along the longer vector, one leg along the $x$-axis and angle $\theta$ between those two sides; the sides of this triangle are $m \cos \theta, m \sin \theta, m$. There is also a triangle with hypotenuse $1$ along the shorter vector and a leg along the $y$-axis, with angle $\theta$ between those two sides; this triangle is proportional to the first triangle and has sides $\cos \theta, \sin \theta, 1$. By inspection we can see that the $x$ coordinate at the end of the forward motion corresponds to the negative $y$ coordinate at the end of the parking motion, whereas the $y$ coordinate after forward motion corresponds to the $x$ coordinate after parking; that is, parking takes the car to $(\sin \theta, -\cos \theta).$
Alternatively, we can see that the second vector is in a direction at an angle $\frac\pi2$ ($90$ degrees) clockwise from the first vector, which means (by the counterclockwise convention) that it is at an angle $\theta - \frac\pi2$ relative to the $x$-axis. For the same reason the first vector takes the car to $(m \cos \theta, m \sin \theta)$, the second vector should take it to $\left(\cos\left(\theta - \frac\pi2\right), \sin\left(\theta - \frac\pi2\right)\right)$. But from trigonometry we know that
$$\begin{eqnarray} \cos\left(\theta - \frac\pi2\right) &=& \cos\theta \, \cos\frac\pi2 + \sin\theta \, \sin\frac\pi2 \\ &=& \cos\theta \cdot 0 + \sin\theta \cdot 1 \\ &=& \sin\theta \end{eqnarray}$$ and $$\begin{eqnarray} \sin\left(\theta - \frac\pi2\right) &=& \sin\theta \, \cos\frac\pi2 - \cos\theta \, \sin\frac\pi2 \\ &=& \sin\theta \cdot 0 - \cos\theta \cdot 1 \\ &=& -\cos\theta. \end{eqnarray}$$
So the coordinates actually come to $(\sin \theta, -\cos \theta).$ Since these trigonometric identities are valid at any angle, we do not have to do any extra work to confirm this formula works for forward directions into the other three quadrants, not just in the first quadrant as shown.
A third approach is to obtain the second vector by setting $m=1$ (so the first vector is unit length) and rotating the first vector through an angle $-\frac\pi2$ ($90$ degrees clockwise) using a rotation matrix, which again will come to the same result.