# Homework Help: Vector Proof

1. Sep 14, 2011

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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.

2. Sep 14, 2011

### lanedance

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 $\phi = 2\theta$

3. Sep 15, 2011

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.

4. Sep 15, 2011

### HallsofIvy

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.

5. Sep 15, 2011

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.

6. Sep 15, 2011

Can anyone help me?

7. Sep 16, 2011

### lanedance

try doing what Halls suggested and let us see what you get

8. Sep 16, 2011

### vela

Staff Emeritus
Is this actually true?

9. Sep 16, 2011

### vela

Staff Emeritus
You're not supposed to assume it's twice the angle. That's what you're trying to prove!

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 $\vec{c}$.