On the topic of the vector equation of a line

Question

I have no problem with the premise of these notes here. However, I had a conceptual question. $P$, like $B$ or $A$ with lines $r$, $b$, and $a$ respectively, are all arbitrary points on the line. Why, then, could I not model the equation like this?:

$$b = a + t(r - a)$$

If this is not possible, I think I'm missing a key insight here, so I'd appreciate any confirmation on the sorts.

Answer 1

This is equally correct. However we use $r$ as a convention .

Basic to get right is $\vec{r}=\vec{a}+k(\vec{b})$

Here, $a$ is usually, any point's position vector lying on the line. while $b$ is the direction ratio of the line (denoted as $b-a$ in your case)

Answer 2

You can. This becomes especially obvious if you rewrite the first equation as $$ \mathbf{r} - \mathbf{a} = t(\mathbf{b} - \mathbf{a}) $$ from which it clearly follows that $$ \mathbf{b} - \mathbf{a} = \frac{1}{t}(\mathbf{r} - \mathbf{a}) $$ which naturally leads to your equation.

Answer 3

Yes you can do that, though it's not as nice to work with. The reason is that we take $A$ and $B$ to be fixed points on the line, and let $P$ be arbitrary. That way by choosing specific values of $t$, we can write any point on the line in terms of information we already know, which is more convenient that having the unknown wrapped up in the RHS.

Answer 4

I think the key thing to note here is that the points $A$ and $B$ are constants. They were "given" to you at the beginning of the problem. The point $P$ is called "any" point on the line that $A$ and $B$ are contained in. This makes $P$ a "variable", and hence we write $$ \mathbf{r} = \mathbf{a} + t(\mathbf{b}-\mathbf{a}) $$ which is supposed to be true for all possible choices of $t$. You could, in principle, decide to reverse the roles of $\mathbf b$ and $\mathbf r$ and so write $$ \mathbf{b} = \mathbf{a} + t(\mathbf{r} - \mathbf{a}) $$ but then you have to remember that $B$ is now representing the "variable" point that can slide along the line, and $A$ and $P$ are now the "fixed" points that are used as a reference for building the equation.

Answer 5

Note that both $r=a+t(b-a)$ and $b = a + t(r - a)$ can be written in the same symmetrical form that describes collinearity: $\lambda \cdot r+ \mu \cdot a + \nu \cdot b = 0\,$ with $\,\lambda+\mu+\nu=0\,$.