# 2 points

1. Nov 26, 2007

### tronter

1. The problem statement, all variables and given/known data

Consider two points located at $$\bold{r}_1$$ and $$\bold{r}_2$$ separated by a distance $$r = |\bold{r}_1 - \bold{r}_2|$$. Find a vector $$\bold{A}$$ from the origin to a point on the line between $$\bold{r}_1$$ and $$\bold{r}_2$$ at distance $$xr$$ from the point at $$\bold{r}_1$$, where $$x$$ is some number.

2. Relevant equations

3. The attempt at a solution

So would it be $$\frac{x}{\sqrt{2}}\bold{r}_1 + \frac{x}{\sqrt{2}}\bold{r}_2$$?

2. Nov 27, 2007

### HallsofIvy

Staff Emeritus
No, it wouldn't! For one thing, the two coefficients are the same. If x is not 1/2, then the point will not be equidistant from the two points so the equation should not be symmetric. Where did you get that $\sqrt{2}$?

3. Sep 6, 2008

### rkbgt

I have this same question on my homework problem set. I got something completely different. It takes up a the entire width of a piece of notebook paper and is two lines tall. Can anyone solve this so I can check my answer?

4. Sep 6, 2008

### HallsofIvy

Staff Emeritus
If x is just "some number", then, in general, the new point will not be between r1 and r2. If x is between 0 and 1, then it will be.

Try (1-x)r1+ xr2

Notice that if x= 0, that is just r1 itself, which "0r" from r1. If x= 1, that is just r2 which is distance "1r" from r2. Finally, if x= 1/2, that is (r1+ r2)/2, the midpoint.

5. Sep 6, 2008

### tiny-tim

Welcome to PF!

Hi rkbgt! Welcome to PF!

hmm … shouldn't take that long …

if you show us how it begins, maybe we can help …

Hint: what is the general expression for a point on the line through r1 and r2?

6. Sep 6, 2008

### rkbgt

I set up a triangle and solved for the A vector. Here's my work:

http://i147.photobucket.com/albums/r311/brkuprel/0906081435.jpg [Broken]

The labels on the triangle are vector r1, vector r2, vector A, theta, alpha, r, and xr

Last edited by a moderator: May 3, 2017
7. Sep 6, 2008

### tiny-tim

eugh! :yuck:

This is a vector problem … use vector methods … they're much quicker!

Hint: use vector addition … OA = OX1 + X1A

Last edited by a moderator: May 3, 2017
8. Sep 6, 2008

### rkbgt

Could someone just give me a 30 second explanation on how to go about these problems. The thing is, I went to a pretty crappy high school that never taught anything about vectors, and this is for my freshman honors physics class. I read the chapter, but it doesn't go over stuff this complicated (which may really not be that complicated).

What do you mean by OA = OX1 + X1A? Is O the origin? What's X1?

Thanks

9. Sep 6, 2008

### tiny-tim

X1 is the point located at $$\bold{r}_1$$.