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April 24th, 2014, 09:28 AM   #1
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Angle Between Two Vectors Proof - RARE

Created this video to help prove why the equation below is correct:



Did not find this proof in any text book and came across it by accident when trying to solve a vector problem. Everything just fell into place. You probably won't find this proof in any a level text book in the UK.

Here is a latex version of the proof to use whilst you're watching the video:

$\displaystyle 2ab\cdot cosC={ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }\\ \\ 2ab\cdot cos\theta ={ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }\\ \\ but\quad a=\left| \underline { b } \right| ,\quad b=\left| \underline { a } \right| ,\quad c=\left| \underline { b } -\underline { a } \right| \\ \\ \therefore \quad 2\left| \underline { a } \right| \left| \underline { b } \right| \cdot cos\theta ={ \left| \underline { a } \right| }^{ 2 }+{ \left| \underline { b } \right| }^{ 2 }-{ \left| \underline { b } -\underline { a } \right| }^{ 2 }\\ \\ \therefore \quad 2\left| \underline { a } \right| \left| \underline { b } \right| \cdot cos\theta =\underline { a } \cdot \underline { a } +\underline { b } \cdot \underline { b } -\left[ \left( \underline { b } -\underline { a } \right) \left( \underline { b } -\underline { a } \right) \right] \\ \\ \therefore \quad 2\left| \underline { a } \right| \left| \underline { b } \right| \cdot cos\theta =\underline { a } \cdot \underline { a } +\underline { b } \cdot \underline { b } -\underline { a } \cdot \underline { a } -\underline { b } \cdot \underline { b } +2\underline { a } \cdot \underline { b } \\ \\ \therefore \quad 2\left| \underline { a } \right| \left| \underline { b } \right| \cdot cos\theta =2\underline { a } \cdot \underline { b } \\ \\ \therefore \quad \underline { a } \cdot \underline { b } =\quad \left| \underline { a } \right| \left| \underline { b } \right| \cdot cos\theta $
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April 24th, 2014, 10:07 AM   #2
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Attachment URL....

https://drive.google.com/file/d/0B-J...it?usp=sharing

Last edited by perfect_world; April 24th, 2014 at 10:10 AM.
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