Prove c Bisects Angle Between a & b | Dot Product Proof

[ProPatto16]

Homework Statement

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.

Homework Equations

cos$$\theta$$ = (a.b)/(|a||b|)

The Attempt at a Solution

no idea. i struggle with proofs :(
please help!

 

[HallsofIvy]

So, as you say, the angle between a and b satisfies $cos(\theta)= (a\cdot b)/(|a||b|)$

Further, the angle between a and c satisfies $cos(\phi)= (a\cdot c)/(|a||c|)$

You are told that c= |a|b+ |b|a so $a\cdot c= |a|(a\cdot b)+ |a|^2|b|= |a|(a\cdot b+ |a||b|)$ and
$$|c|= \sqrt{2|a|^2|b|^2+ 2|a||b|a\cdot b}$$.

Now use the "double angle formula", $cos(2\phi)= 2cos^2(\phi)- 1$ and show that is equal to $cos(\theta)$.

 

[ProPatto16]

cos$$\theta$$= (a.b)/(|a||b|)

cos$$\phi$$= (a.c)/(|a||c|)

a.c = |a|(a.b + |a||b|) and |c|= that sq root you have.

subbing in:

cos$$\phi$$= [|a|(a.b + |a||b|)] / [(|a| multiplied by sqrt you have)]

|a| in numerator and denominator cancel, then square both sides gives:

cos²$$\phi$$= (a.b + |a||b|)² / (2|a|²|b|²+2|a||b|a.b)

then take out the 2 in the denominator so:

2cos²$$\phi$$= (a.b + |a||b|)² / (|a|²|b|²+|a||b|a.b)

now I am looking at that result with cos$$\theta$$ and i see obviously when the numerator is evaluated it will cancel out terms to gives a +1 which will move over to other side to become -1 and then leave 2cos²$$\phi$$-1= (a.b)/(|a||b|)

2 things.

number 1. I am unsure of where you got that first root equation of |c| from. i know how to use it obviously but just unsure where it came from and trying to understand every part of the question.

number 2.
im unsure how to evaluate the numerator.

(a.b + |a||b|)²

if i simply say it equals (a.b)² + |a|²|b|² it becomes:

(a.b)²+|a|²|b|² / |a|²|b|² + |a||b|a.b

then cancel (a.b) from bottom and take off square on top. then |a||b|² on top and bottom becomes +1 leaving(a.b)/(|a||b|) with the +1 becomes -1 on the other side

the only thing I am not sure about is the cancellation of the a.b on the bottom. is that allowed? if so is that correct?

 

[ProPatto16]

if i take out the +1 first that leaves only a multiplication on the denominator so then yes that cancellation of a.b would be legitimate yes?

 

[ProPatto16]

it all works out. i just can't work out where the original |c| root equation comes from.

 

[HallsofIvy]

If you mean
$$|c|= \sqrt{2|a|^2|b|^2+ 2|a||b|a\cdot b}$$
It is simply from
$$|c|= \sqrt{c\cdot c}$$

You are told that c= |a|b+ |b|a so that
$$c\cdot c= (|a|b+ |b|a)\cdot (|a|b+ |b|a)$$
$$= |a|^2b\cdot b+ |a||b|b\cdot a+ |b||a|a\cdot b+ |b|^2 a\cdot a$$
$$= 2|a|^2|b|^2+ 2|a||b|a\cdot b$$

 

[ProPatto16]

ah yes. thank you. i had tunnel vision. was trying to get |c| out of cos(theta) equation haha. the method I've shown above seem complete?