A calculus of the absurd
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Chapter 9 Coordinate geometry
"This is some kind of a plot, right?" Slothrop sucking saliva from velvet pile. "Everything is some kind of a plot, man," Bodine laughing. "And yes but, the arrows are pointing all different ways"
— Thomas Pynchon, GRAVITY ’ S RAINBOW. Usually somewhere near p. 600 (depending on edition).
I make no secret of the fact that I think geometry should be wrapped up, and placed in a compost heap. From there it might helpfully decompose and from the limited nutrients of its remains give birth to something better.
Unfortunately, today is not the day. In any 2D coordinate system, we may reference any point in that 2D space through two values. Usually, these are referred to as \(x\) and \(y\).
9.1 Distances
Consider the vector \(\vec {a}\). Let’s say that \(a=(1, 1)\). How far is \(a\) from the origin?
5959 The distance between a vector and the origin is the same thing as the magnitude of the vector.
To find the distance between the origin (the vector \((0, 0)\)) and \(\vec {a}\) we can use Pythagoras’ theorem.
\(\seteqnumber{0}{9.}{0}\)
\begin{equation}
\norm {a} = \sqrt {(a_x)^2 + (a_y)^2}
\end{equation}
How about if we want to find the distance between the vector \(\vec {a}=(1,1)\) and another vector \(\vec {b} = (2, 0)\)? What we can do is find the vector between the two points (this is written as \(\vec {ab}\)), and then use Pythagoras’ theorem in the same way we did
above?
First, we can draw a diagram:
We have no clue how to find the vector between \(\vec {a}\) and \(\vec {b}\) (\(\vec {ab}\)). To find the vector between \(\vec {a}\) and \(\vec {b}\) can draw a diagram and think about what we do know. Remember that we can read \(\vec {a}\) as "move from the origin to
\((1, 1)\)" and \(\vec {b}\) as "move from the origin to \((2, 0)\)". Then, to move from \(\vec {a}\) to \(\vec {b}\) we want to move "from \(\vec {a}\) to the origin" and "from the origin to \(\vec {b}\)". This is \(-\vec {a} + \vec {b}\). To find the distance (aka magnitude
of this vector), we just use Pythagoras’s theorem.
Therefore the distance between \(\vec {a}\) and \(\vec {b}\) is
\(\seteqnumber{0}{9.}{1}\)
\begin{align*}
\norm {-\vec {a}+\vec {b}} &= \norm { \begin{bmatrix} -1 \\ -1 \end {bmatrix} + \begin{bmatrix} 2 \\ 0 \end {bmatrix} } \\ &= \sqrt {(-1+2)^2 + (-1)^2} \\ &= \sqrt {2}
\end{align*}