# Explain why cosine formula is always -0,5.

1. Aug 10, 2009

### Dafe

1. The problem statement, all variables and given/known data
Pick any numbers that add to:
$$x + y + z = 0$$
Find the angle between your vector $$\textbf{v} = (x, y, z)$$
and the vecor $$\textbf{w} = (z, x, y)$$
Explain why $$\textbf{v}\bullet\textbf{w} / ||\textbf{v}||||\textbf{w}||$$ is always $$-\frac{1}{2}$$

2. Relevant equations

Cosine Forumla:

$$\frac{\textbf{v}\bullet\textbf{w}}{||\textbf{v}||||\textbf{w}||}=cos\theta$$

3. The attempt at a solution

I pick:

$$\textbf{v} = (2, -1, -1)$$
$$\textbf{w} = (-1, 2, -1)$$

I insert the integers into the cosine forumla and get $$-\frac{1}{2}$$

As for the question why it is always -(1/2), I am not sure where to start.

If you guys could push me in the right direction I would really appreciate it.

Thanks!

2. Aug 10, 2009

### rock.freak667

Well I'd write out what v.w works out as and what |v||w| works out as.

EDIT: Hint: remember what (x+y+z)2 equals

3. Aug 10, 2009

### Dafe

$$\textbf{v} \bullet \textbf{w} = xz + yz + xy$$

I can write this as:
$$\textbf{v} \bullet \textbf{w} = \frac{1}{2} (x + y + z)^2 - \frac{1}{2} (x^2 + y^2 + z^2)$$

$$||\textbf{v}|| ||\textbf{w}|| = x^2 + y^2 + z^2$$

$$\frac{\textbf{v} \bullet \textbf{w}}{||\textbf{v}|| ||\textbf{w}||} = \frac{\frac{1}{2} (x + y + z)^2}{(x^2 + y^2 + z^2)} - \frac{1}{2}$$

since

$$x + y + z = 0$$

$$- \frac{1}{2}$$