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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
up vote 3 down vote accepted

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

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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
@jim: Yeah! Just verified that. Thanks! – Aneesh Dogra Mar 18 '13 at 10:36
Your welcome, Aneesh. I think the parametric equations are much easier to read than the cartesian equations, and they're much better for representing the geometry here. – jim Mar 18 '13 at 10:54

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