# Vectors in the plane.

1. Aug 24, 2009

### htk

2. Aug 24, 2009

### Dragonfall

Think about the triangle that the vector and its components form.

3. Aug 24, 2009

### HallsofIvy

Staff Emeritus
I presume you are talking about vectors in the plane since vectors in three dimensions have three "direction angles". How are you given the vector? If in x,y components, say ai+ bj, then b/a is the tangent of the angle the vector makes with the x-axis:
$$\theta= arctan(\frac{a}{b})$$.

4. Aug 24, 2009

### tiny-tim

Welcome to PF!

Hi htk! Welcome to PF!
You find the cosine of the angle …

which you do by finding the dot product.

5. Aug 25, 2009

### HallsofIvy

Staff Emeritus
Re: Welcome to PF!

His question was about a single vector. What do you want him to take the dot product with?

6. Aug 25, 2009

### tiny-tim

with whatever mysterious entity he had in mind when he specified

7. Aug 25, 2009

### Elucidus

A two-dimensional vector $\langle a,b \rangle$ will have a direction angle $\theta \text{ such that } \tan \theta = b / a$ (not (a/b)) but this does not uniquely determine $\theta$, even if it is restricted to the interval $[0, 2 \pi )$.

You also need to consider in which quadrant does the vector lie. You need to adjust the value of $\theta$ so that it falls into the correct quadrant.

For example, the vector $\langle -3, 3 \rangle$ has a direction angle so that $\tan \theta = 3 / -3 = -1 \text{ which implies } \theta = -\pi /4 + n \pi$ for an appropriate choince of integer n. Since the vector is in the second quadrant, we need to select the angle to fall there, so $\theta = 3\pi / 4$ (here n = 1).

I hope this helps.

--Elucidus

8. Aug 25, 2009

### Dragonfall

Re: Welcome to PF!

I suppose he could dot it with (1,0).

EDIT: Modulo sign.

Last edited: Aug 25, 2009