Explain why cosine formula is always -0,5.

  • Thread starter Dafe
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  • #1
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Homework Statement


Pick any numbers that add to:
[tex] x + y + z = 0 [/tex]
Find the angle between your vector [tex]\textbf{v} = (x, y, z)[/tex]
and the vecor [tex]\textbf{w} = (z, x, y)[/tex]
Explain why [tex] \textbf{v}\bullet\textbf{w} / ||\textbf{v}||||\textbf{w}|| [/tex] is always [tex] -\frac{1}{2} [/tex]

Homework Equations



Cosine Forumla:

[tex]\frac{\textbf{v}\bullet\textbf{w}}{||\textbf{v}||||\textbf{w}||}=cos\theta[/tex]

The Attempt at a Solution



I pick:

[tex] \textbf{v} = (2, -1, -1) [/tex]
[tex] \textbf{w} = (-1, 2, -1) [/tex]

I insert the integers into the cosine forumla and get [tex] -\frac{1}{2} [/tex]

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!
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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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
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[tex]\textbf{v} \bullet \textbf{w} = xz + yz + xy [/tex]

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

[tex]||\textbf{v}|| ||\textbf{w}|| = x^2 + y^2 + z^2 [/tex]

This leads me to:

[tex]\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}[/tex]

since

[tex] x + y + z = 0 [/tex]

the answer is,

[tex] - \frac{1}{2} [/tex]

Thank you rock.freak667!
 

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