About vector form of a line passing through 2 points.

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?

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You have neglected to subtract the k component of $a$ from the k component of $b$ in both equations. –  jim Mar 18 '13 at 10:09
@jim: Oh god! arithmetic mistakes! I am so sorry! I have it fixed now! –  Aneesh Dogra Mar 18 '13 at 10:14

$x+y =1$ 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
To satisfy yourself that they are the same line, substitute some nice $z$ values into both equations, solve for $x$ and $y$, and see that you always get the same $(x,y,z)$ coordinates in both equations. –  jim Mar 18 '13 at 10:26