Vector equation to line equation

Question

I've not read vectors in math yet but I'm done with those in physics. I want to find out a line equation from a vector equation. Say I've 2 points in 3D space: Point A with coordinate $(a, b, c)$ and point B with coordinate $(d,e,f)$. I can write it as $a\hat{i} + b\hat{j} + c\hat{k}$.

Now for another point $(d, e, f)$, I write it: $d\hat{i} + e\hat{j} + f\hat{k}$. Now subtracting these 2 vectors would give me a resultant vector from one point to another. How to I get the line equation of that vector? i.e. in this case: $(a-d)\hat{i} + (b-e)\hat{j} + (c-f)\hat{k}$?

Answer 1

Hint: You can simply subtract the two vectors like $(a,b,c)-(d,e,f)$ to get what you have written.

Answer 2

We have two vectors $$ v_1 = (a, b, c) \\ v_2 = (d, e, f) $$ then one can describe the line through the endpoints of the $v_i$ as $$ (1-t) \, v_1 + t \, v_2 = v_1 + t (v_2 - v_1) \quad (t \in \mathbb{R}) $$

Answer 3

You don't have a 3D line definition with single equation like in the 2D case: $ax + by + c = 0$, or $y = mx + n$

In 3D, with a single equation of some linear combination of variable $x,y,z$, you can only define planes.

So in this case; ${\vec u} = (a,b,c)-(d,e,f)$ is our direction vector. And any point $(x,y,z)$ on the line can be written as

$(x,y,z) = (a,b,c) + t*{\vec u}$

$x = a + t*(a-d)$

$y = b + t*(b-e)$

$z = c + t*(c-f)$

If you want you can pull $x$ and re-write y-equation and z-equation in terms of x.