I have a lecture for my linear algebra and geometry module that focuses on the cross product and line and planes within Euclidean geometry. The lecture notes talk about the parametric representation of a line which is something I covered in my A-levels, however I can't make the connection in my head between what I learned at A-level and what was in the lecture leading to me not being able to understand this content. The part that I am struggling on is as follows:
The straight line through the point $P\in\Bbb R^n$ in direction of vector $\vec a$ has the parametric form: $$\vec x (t)=P+\vec a t\quad\forall t\in\Bbb R$$ Similarly, given two points $P, Q\in\Bbb R^n$, the line passing through $P$ and $Q$ has parametric form: $$\vec x (t)=P+t(Q-P)\quad\forall t\in\Bbb R$$ I was wondering if anyone could explain how this links back to earlier content that I have studied, hopefully allowing me to understand this?
$ \newcommand{\vect}[1]{\vec{\boldsymbol{\mathbf{{#1}}}}} \newcommand{\R}{\mathbb{R}} $ A good place to start would be a visual example. Suppose $\vect a = (1,2)$ and $\vect y_0= (-1,0)$. We want the line through $\vect y_0$ in the direction of $\vect a$. This is given by
$$\left\{ \vect x \in \Bbb R^2 \; \middle| \; \vect x = t \vect a + \vect y_0 \text{ for some } t \in \Bbb R \right\}$$
Here is what we get if we visualize the elements of this set for $t \in \{-3,-2,-1,0,1,2,3\}$:
What do we notice? This appears to be a line with slope $2$, and $y$-intercept $2$ as well. That means the equation of the line -- once we span over all of $t \in \R$ -- is given by
$$y = 2x+2$$
That's interesting, because $2 = a_y/a_x = 2/1$ (where $a_x$ denotes the $n$ component of $\vect a$, and similar for $y$). $y$ in particular goes through $\vect y_0$. So let's consider $\vect a$ as representing a slope through the ratio of its points, and $\vect y_0$ as a single point through which the line goes. This uniquely determines the line via its point slope form: the equation of a line with slope $m$ going through $(x',y')$ is given by
$$y - y' = m(x-x')$$
Given a vector $\vect a$ and $\vect y_0$, then
$$y - y_{0,y} = \frac{a_y}{a_x} (x - y_{0,x})$$
Plugging in $y_{0,x} = -1$, $y_{0,y} = 0$, $a_y = 2$, and $a_x = 1$, we have
$$y - 0 = \frac{2}{1} (x - (-1)) \implies y = 2x+2$$
We can work this similarly for your "line through two points" thing. Suppose we have, now, $\vect a = (1,3)$ and $\vect b = (2,1)$. The line in the green is the unique line through these two points:
Notice however, we can get a third vector from this (up to its orientation): the vector $\vect c = \vect b - \vect a$ (or the vector $\vect d = \vect a - \vect b$). Choosing $\vect c$ because why not, we then have this picture:
Notice, however, the line that we want is precisely the set of all vectors that are scalar multiples of $\vect c$ from that position. That is, we have the line is given by
$$\left\{ \vect x \in \Bbb R^2 \; \middle| \; \vect x = t \vect c + \vect a \text{ for some } t \in \Bbb R \right\}$$
Some Notes: You can use $\vect d$ in lieu of $\vect c$, and $\vect a$ in lieu of $\vect b$ in the above definition. We just want a line through $\vect a$ and $\vect b$, and $\vect c$ or $\vect d$ go in the direction necessary. Carefully examine how this is similar to our previous example! The choice of $\vect c$ versus $\vect d$, or $\vect a$ versus $\vect b$, is superfluous for this application and the end result is ultimately the same even if the definition looks different. If we were to limit $t$ to $0 \le t \le 1$, as we do when parameterizing a line segment specifically, it would be a different matter.
Anyhow, since $\vect c = \vect b - \vect a$, then we have that the set is equivalent to
$$\left\{ \vect x \in \Bbb R^2 \; \middle| \; \vect x = \vect a + t \left(\vect b - \vect a \right) \text{ for some } t \in \Bbb R \right\}$$
You can then re-analyze this through our opening example and explanation -- a line with given slope through a point, since $\vect b - \vect a$ (the thing $t$ scales) determines the slope of the line, and $\vect a$ determines the location through which the line goes.