- #1

TheMadCapBeta

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

<Let

**bold**lower case letter represent a vector, and |

**a**| represent the length of vector.>

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**.

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

**a**and

**b**.

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|^2**and

**a · 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| ]

[(|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