According to my book:
Equation of line passing through 2 points with position vectors $a$ and $b$ is
$$r = a + K(b - a)$$
My question:
If we are given 2 points how do we determine which point is to be taken as b and which as a?
For example:
Here's a question from my book:
Find the equation of line passing through A (5, 1, 6), B (3, 4, 1).
I can get 2 equations depending on the points that I select as $a$ and $b$ respectively.
If I take $a = 5i + 1j + 6k$ and $b = 3i + 4j + 1k$
I get:
$$b - a = -2i + 3j - 5k$$ $$r = 5i + 1j + 6k + K(-2i + 3j - 5k)$$ $$r = (5-2K)i + (1+3K)j + (6 - 5K)k$$
In cartesian form:
$$\frac{5 - x}{2} = \frac{y - 1}{3} = \frac{6 - z}{5}$$
If I take $a = 3i + 4j + 1k$ and $b = 5i + 1j + 6k$
I get:
$$b - a = 2i - 3j + 5k$$ $$r = 3i + 4j + 1k + K(2i - 3j + 5k)$$
$$r = (3 + 2K)i + (4 - 3K)j + (1 + 5K)k$$
In cartesian form:
$$\frac{x - 3}{2} = \frac{4 - y}{3} = \frac{z - 1}{5}$$
Do the 2 equations represent the same line?
Yes. Both these equation represent the same line. You can arrive at the other equation by some manipulation. Its just that the normal to these equation are pointed in opposite direction.
For example:
$x+y =1$ and $2x+2y = 2$ represent the same line and same normal.
$x+y =1$ and $-x-y = -1$ represent the same line but there normals are in different directions