Find the magnitdue of a vector along a specific heading I've gotten some seemingly conflicting answers about this, so I'm hoping someone can help straighten this out for me.
Given a vector of components East/West (u) and North/South (v), how do I find the magnitude of a specific bearing.
I have an Eastward component of 3 and a Southward component of -5
Speed=sqrt(3^2+(-5)^2)
Speed=5.83

theta<-atan2(v,u)
theta<- -59.0362

In compass bearings
Direction=90-theta
Direction=149.0362°

So I have a speed of 5.83 and a direction of 149°
So given this information, now I want to know the magnitude of the component vector of 120°.
I'm not sure if I need to be looking into vector rotation, or some other trig. I've found the following different answers:
New Speed= cos(149-120)

New Speed= v*cos(180-120) + u*sin(180-120)

New Speed= u*sin(120)-v*cos(120)

Which all have different outcomes so I'm not sure which one is right. 
BTW-This is not homework, I'm trying to isolate flow speeds in a single specific direction. So even though my raw data has a range of speeds and directions, in the end I only want data on the speed of the vector at 120°
 A: You are trying to project your speed vector on a given direction.  Let $\vec s=(s_x,s_y)$ be the speed vector, and $\vec d=(\cos \theta, \sin \theta)$ be a unit vector in the direction of interest.  Then the length of the projection of $\vec s$ on $\vec d$ is $\vec s \cdot \vec d$, the dot product.
A: The difference in angle between the two vectors, call it $\theta$, is what you need to use.  Specifically, $\cos$ is used for the vector that forms a "V" shape with the original where $\theta$ is the angle between them.  Given $s$ is the old speed and $n$ is the new speed, then $n=s\cos(\theta)$, because the new speed is merely a component of the old speed.  Also, $m=s\sin(\theta)$ is the complement of the new speed (and is perpendicular to it) and the two together make up your original vector, i.e., $\sqrt{m^2+n^2}=s$.
Note that this transformation applies equally whether "newangle-oldangle" or "oldangle-newangle" is used as the value of $\theta$.  The reason this is true is twofold:


*

*$\cos \theta = \cos -\theta$, so you are guaranteed to get the same value either way

*$\sin -\theta = -\sin \theta$, so the magnitude or absolute-value obtained is the same either way.


If you needed to find a direction, then the sign of $\sin\theta$ would make a difference in which angle you were picking.  As it is, you have chosen your new angle (i.e., $120$°), so you don't need to worry about this piece.
