- #1
TheMadCapBeta
Hello everyone.
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| ]
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
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| ]
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