# Simple Work On Line Question.

1. May 17, 2009

### Master J

Simple question.

So in an example question I'm lookin at finding the work between 2 points on a line.

The points are A(1, 0, 0) and B(0, 1, (pi^3)/8)

The vector of the line joining these points is then given as:

r = i + t(-i + j + [(pi^3)/8]k )

Is that correct? I can't see how that equation arises? It could just be a typo, or I'm just having a stupid day!!

Cheers!!

2. May 17, 2009

### Staff: Mentor

No, that's the parametric equation of the (straight) line connecting those points. Taking different values of t in the range [0,1] gives you different points along that line.

3. May 17, 2009

### Master J

Yes, but is it correct? Shud there be a -i in there?

Is it r = a + tb ????

4. May 18, 2009

### Staff: Mentor

It looks OK to me. Substitute t = 0 and you get point A. Substitute t = 1 and you get point B. In unit vector notation,

$$\vec A = 1 \hat i + 0 \hat j + 0 \hat k$$

$$\vec B = 0 \hat i + 1 \hat j + \frac{\pi^3}{8} \hat k$$

5. May 18, 2009

### Master J

I guess I'm confused as to that vector. How do you make that position vector from those 2 points?

6. May 18, 2009

### Nick89

The vector from B to A is simply
$$\vec{B} - \vec{A}$$