How can c be shown to bisect the angle between a and b?

[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

 

[Hurkyl]

Well, your main problem is that just about the very first thing you did was to assume what you're trying to prove! You can't use the fact that the two angles are equal because you're trying to prove it.

 

[TheMadCapBeta]

What do you suggest?

 

[theunknot]

dude,

divide by the magnitudes,

c/(|a||b|)=a(unit vector)+ b(unit vector)

now there's the parallelogram method for adding vectors, where a and b form two sides of a parallelogram, and since they are UNIT vectors, the parallelogram will have equal sides. Therefor, each half of the rhombus is an isosceles triangle. c will be cut across the diagonal of this rhombus.

 

[Hurkyl]

Your idea of computing angles with dot products is good; but you need to use that to prove the angles are the same.

 

[TheMadCapBeta]

Originally posted by theunknot
dude,

divide by the magnitudes,

c/(|a||b|)=a(unit vector)+ b(unit vector)

now there's the parallelogram method for adding vectors, where a and b form two sides of a parallelogram, and since they are UNIT vectors, the parallelogram will have equal sides. Therefor, each half of the rhombus is an isosceles triangle. c will be cut across the diagonal of this rhombus.

c= |a||b|( a/|a| + b/|b|)

c/|a||b| = a/|a| + b/|b|

Here's what I got out of your reply. c/|a||b| will be the resultant vector of the two unit vectors on the right side. And yes, since they are UNIT vectors , the parallelogram will be an equilateral quadtrilateral. The c vector will be the hypotenuse, bisecting the angle...

theunknot. Good job pointing this out, I really didn't see it.
But yes, Hurkyl is right. The angle must be proven symmetrical.

I'm going to try and work this out more. Any more advice would be welcome. Thanks