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

[Mary O'Donovzn]

Homework Statement

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

 

[gneill]

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.

 

[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?

 

[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}}##

 

[Mary O'Donovzn]

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:
r^=r⃗ |r|=<x2−x1,y2−y1>(x2−x1)2+(y2−y1)2√OR(x2−x1)i^+(y2−y1)j^(x2−x1)2+(y2−y1)2√

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

 

[Chestermiller]

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

Chet

 

[gneill]

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

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.

 

[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.

 

[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 won't help

 

[Chestermiller]

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 won't help

Poor baby. Now, please answer my question in post #6.

Chet

 

[BiGyElLoWhAt]

Whats the actual problem? Rhat is literally just a unit vector in the direction of r. You can draw a right triangle with the hypotenuse r and pointing in the same direction, and use various methods to solve for r. Then divide that by its magnitude (which will not necessarily be one, r is different than rhat)