Explain why cosine formula is always -0,5.

[Dafe]

Homework Statement

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}

Homework Equations

Cosine Forumla:

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

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!

 

[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

 

[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

This leads me to:

\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

the answer is,

- \frac{1}{2}

Thank you rock.freak667!