# Using Law of Cosines To Prove Vector Algebra

• Aug 19th 2009, 02:02 PM
RedKMan
Using Law of Cosines To Prove Vector Algebra
Use the law of cosines $c^2 = a^2 + b^2 = 2ab \cos \theta$, where a, b and c are the lengths of the sides of a triangle and theta is the angle between sides a and b. To show:-

$U_x V_x + U_y V_y + U_z V_z = ||U|| ||V|| \cos \theta$

Hint: Set $c^2 = ||U - V||, a^2 = ||U||^2$ and $b^2 = ||V||^2$. Use the dot product properties.

$||U-V|| = ||U||^2 + ||V||^2 = 2||U|| ||V|| \cos \theta = U \cdot V = ||U|| ||V|| \cos \theta$
• Aug 19th 2009, 02:14 PM
Plato
Quote:

Originally Posted by RedKMan
Use the law of cosines $c^2 = a^2 + b^2 = 2ab \cos \theta$, where a, b and c are the lengths of the sides of a triangle and theta is the angle between sides a and b. To show:-

$U_x V_x + U_y V_y + U_z V_z = ||U|| ||V|| \cos \theta$

Hint: Set $c^2 = ||U - V||, a^2 = ||U||^2$ and $b^2 = ||V||^2$. Use the dot product properties.

$||U-V|| = ||U||^2 + ||V||^2 = 2||U|| ||V|| \cos \theta = U \cdot V = ||U|| ||V|| \cos \theta$

Using the hint:
$c^2 = \left( {U - V} \right) \cdot \left( {U - V} \right) = U \cdot U - 2U \cdot V + V \cdot V = \left\| U \right\|^2 - 2U \cdot V + \left\| V \right\|^2$

Brief latex tutorial: $$||U|| ||V|| \cos(\theta)$$ gives $||U|| ||V|| \cos(\theta)$
• Aug 22nd 2009, 03:55 AM
RedKMan
I've been trying to work out how you have arrived at this answer for a few days now and I'll be honest I don't get it or understand it at all.
• Aug 22nd 2009, 09:40 AM
Plato
Quote:

Originally Posted by RedKMan
I've been trying to work out how you have arrived at this answer for a few days now and I'll be honest I don't get it or understand it at all.

Are you saying that you do not understand $c^2 = \left( {U - V} \right) \cdot \left( {U - V} \right) = U \cdot U - 2U \cdot V + V \cdot V = \left\| U \right\|^2 - 2U \cdot V + \left\| V \right\|^2$?
If that is the case, you need to have a good one on one with your instructor.

One the other hand, if you are saying that you do not understand how to finish, ok.
Note that $U \cdot V = \left\| U \right\|\left\| V \right\|\cos (\theta )$ where $\theta$ is the angle between $U~\&~V$.
Then from $c^2 = \left\| U \right\|^2 - 2U \cdot V + \left\| V \right\|^2$ we get $c^2 = a^2+b^2 - 2ab\cos(\theta).$
• Aug 22nd 2009, 11:52 AM
RedKMan
OK, now I get it. Thanks. I don't have an instructor.