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I have to solve this proof, and I'm having a little trouble. Let me explain.

<Letboldlower case letter represent a vector, and |a| represent the length of vector.>

Ifc=|a|b+|b|a, wherea, b,andcare all nonzero vectors, show thatcbisects the angle betweenaandb.

I'm trying to prove this by Corollay, since both angles will be equal (and half of the whole betweenaandb.

So:

cos(x) =(b · c) / |b||c|

cos(x) =(a · c) / |a||c|

[(b · c)/|b||c| ] = [ (a · c) / |a||c| ]

[(b · (|a|b + |b|a))/|b||c| ] = [(a · (|a|b + |b|a))/|a||c| ]

Then distributing the numerator on each side and by dot product

b · b = |b|^2anda · b = |a||b|cos(x)

So:

[(|a||b|^2 + |b||a||b|cos(x))/|b||c| ] =

[(|a||a||b|cos(x) + |b||a|^2)/|a||c| ]

And this is basically as far as I got. I saw some oppurtunites to factor out some components but it didn't really come to much.

Any help would be greatly appreciated since I have to hand it in by wednesday.

Thanks again.

-Doug

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# Vector proof: need some help

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