# What is r hat (^) like exactly and how do you calculate it?

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1. Oct 20, 2014

### Mary O'Donovzn

• OP warned about not using homework template
1. The problem statement, all variables and given/known data

What is r hat (^) like exactly and how do you calculate it?

I know that it is something to do with unit vectors and that it should be gotten in some form of cosi + sinj

like I had to input geometry co ordinates x1,y1 and stuff

Any help at all will be appreciated

2. Oct 20, 2014

### Staff: Mentor

A unit vector is a vector in some direction whose magnitude is unity (1). If you have some other vector in the same direction, say R, which is not of unit length, then you can create a unit vector in the same direction by calculating
$$\hat{r} = \frac{R}{|R|}$$
That is, divide the vector R by its own magnitude.

3. Oct 20, 2014

### Mary O'Donovzn

Okay so if I have a point (2,3) and a point (5,6) say ( I just made them up), do I calculate the distance between the points and divide it by its magnitude which is one?

4. Oct 20, 2014

### BiGyElLoWhAt

You don't calculate the distance between the two, you calculate a vector between the two (distance AND direction). Then you divide it by the distance. For 2 points, r hat from point 1 to point 2 is given as such:
$\hat{r} = \frac{\vec{r}}{|r|} = \frac{<x_2 - x_1 , y_2 - y_1>}{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}} OR \frac{(x_2 - x_1) \hat{i} + ( y_2 - y_1)\hat{j}}{\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}}$

5. Oct 20, 2014

### Mary O'Donovzn

okay so do I not use any cos or sin just that formula? because that would be the answer to my problem

6. Oct 20, 2014

### Staff: Mentor

Are you referring specifically to the unit vector in the radial direction of a cylindrical polar coordinate system?

Chet

Last edited: Oct 20, 2014
7. Oct 20, 2014

### Staff: Mentor

You might use the sin and cos method if you're given a vector in polar form (magnitude and angle). You would then use sin and cos to create the rectangular components of a unit vector with the same angle.

8. Oct 20, 2014

### BiGyElLoWhAt

It depends on what you have given. If you have 2 coordinates, no, you don't use cos or sin. But if you have a length and a direction, say 10m pointed at 45 degrees, then yes, you would break it into components using sins and cos's. For that you'd just use basic trig.

9. Oct 20, 2014

### Mary O'Donovzn

See I must sub in a magnitude and 2d coordinates and get out the force acting on each...Like I literally have no idea because I missed all this last year I missed whole term because I was sick and they didn't allow any books or things to like concentrate on and my teacher reckons I could have been studying (my brain would have exploded basically if I did) and she wont help

10. Oct 20, 2014