# Vector Proof

## Homework Statement

If
c=|a|b+|b|a where a,b, and c are all non zero vectors, show that c bisects the angle between a and b

## The Attempt at a Solution

I'm taking the approach to prove that the angle between b and c= the angle between c and a

I have written the fact that a$\bullet$b=|a||b|cos $\theta$
However, I'm not sure how I can work the equation for c to be able to apply the identity.

lanedance
Homework Helper
as you know they are in the same plane and c will be between a and b
$$a \bullet b = |a||b|cos\theta$$
$$c \bullet b = |c||b|cos\phi$$

and the show $\phi = 2\theta$

I'm not quite sure what you are suggesting.

I can draw the relationship

|a||b|cos $\theta$ = |c||b| cos 2$\theta$

I can eliminate the |b| from both sides, but I don't know where to go from there, since |c| doesn't seem to help when substituting.

HallsofIvy
Homework Helper
Then get rid of "c". You are given that c= |a|b+ |b|a so that c.b= (|a|b+ |b|a).b= |a||b|^2+ |b|a.b.

I can replace the c$\bullet$b side with what you've suggested but then how am I supposed to include that it is twice the angle. I loose this ability without the trigonometric function.

Can anyone help me?

lanedance
Homework Helper
try doing what Halls suggested and let us see what you get

vela
Staff Emeritus
Homework Helper
I can draw the relationship

|a||b|cos $\theta$ = |c||b| cos 2$\theta$
Is this actually true?

vela
Staff Emeritus
I can replace the c$\bullet$b side with what you've suggested but then how am I supposed to include that it is twice the angle. I [strike]loose[/strike] lose this ability without the trigonometric function.
Note that HallsofIvy's suggestion $\vec{c}\cdot\vec{b} = ab^2+b(\vec{a}\cdot\vec{b})$ has the dot product of $\vec{a}$ and $\vec{b}$ in it. That will introduce $\cos\theta$ into the equation.
You can get rid of $c = \sqrt{\vec{c}\cdot\vec{c}}$ from the righthand side by again using the definition of $\vec{c}$.