Given the vector equation of a line $L_1$, why is the direction vector written in reverse?

Question

The title of the question might make this sound a little confusing; however, I have this simple question the answer to which I can't find online.

Consider the line $L_1 = \bar a + \lambda(\bar b)$ where $\bar a$ is any point with coordinates $\left<x, y\right>$ and $\bar b$ is the direction vector. However, if I compare this to the equation of a standard line in the form of $y = mx + c$ where $c$ is my y-intercept (any point) and $m$ is the slope, I notice something a little unresting.

If I have a point $c = \left<1,2\right>$ and the slope at $m=\frac{4}{3}$, why does this equation morph to $L_1 = \left<1,2\right> + \lambda\binom{3}{4}$. The values have just switched.

Is there any specific reason to this?

Answer 1

If I have a point $c = \left<1,2\right>$ and the slope at $m=\frac{4}{3}$, why does this equation morph to $L_1 = \left<1,2\right> + \lambda\binom{3}{4}$. The values have just switched.

If by "switched" you mean that 4 was "on top" and shows up "on the bottom" (and the other way around), you need to be careful with the relation between the slope (a number) and a direction vector (a vector).

For a non-vertical line, you can derive the slope $m$ from any direction vector $(a,b)$ (where $a \ne 0$) as the ratio between the $y$- and the $x$-coordinate: $m = \tfrac{b}{a}$.

Note that although the slope is unique, the (better: a) direction vector isn't. Any non-zero multiple of a direction vector is again a direction vector, but that doesn't change the ratio of its coordinates.