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Prove c bisects angle between a and b | Vector Algebra
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[QUOTE="Calpalned, post: 4995461, member: 525107"] [h2]Homework Statement [/h2] If c = |a|b + |b|a, where a, b, and c are all nonzero vectors, show that c bisects the angle between a and b. [h2]Homework Equations[/h2] Angle between a & b is cos[SUP]-1[/SUP](a dot b)/(|a||b|) Angle between a & c is cos[SUP]-1[/SUP](a dot c)/(|a||c|) Angle AC is half of angle AB [h2]The Attempt at a Solution[/h2] Given c = |a|b + |b|a, I plug that into the equation for the angle between a and c. I eventually get (|a||b|)(|a|b + |b|a[SUP]2[/SUP] = (2|a||c|)(a dot b). Is this right? I would also like to confirm: - if (a dot a) is always equal to |a|^2 - how to differentiate absolute value and magnitude as they use the same symbol - When I got (2|a||c|)(a dot b), do I do regular multiplication or use the dot product? [/QUOTE]
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Prove c bisects angle between a and b | Vector Algebra
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