1. ## Vector Proof

Hi all,

Im working on a vector proof as follows:

u and v are vectors in R3.

Prove that |u|v + |v|u bisects the angle between u and v.

I have tried looking at the dot product formula cos(theta) = u.v/|v||u| but I am a bit stuck.

I figure I need to be dividing through by the magnitudes of theses vectors to give me unit vectors but not quite sure how to go about things.

Any help would be greatly appreciated, thanks

Andy

2. ## Re: Vector Proof

I don't see any need to reduce to unit vectors, although that could be done. Note that the dot product of |u|v+ |v|u with itself is |u||v|(|u|+ |v|+ 2).
Go ahead and use that formula to calculate the angle |u|v+ |v|u makes with u and the angle it makes with v and compare. You should find that a lot of things cancel.

3. ## Re: Vector Proof

Hey, sorry, I have probably been at this for too long but I feel like I am missing something.

i have only been able to get to:

|v||u| . |v||u| = |u|^2|v| + |u||v|^2 + 2|u||v|

don't know if I have done that right or where to go next.

Andy

4. ## Re: Vector Proof

Originally Posted by andy000
Hey, sorry, I have probably been at this for too long but I feel like I am missing something.

i have only been able to get to:

|v||u| . |v||u| = |u|^2|v| + |u||v|^2 + 2|u||v|

don't know if I have done that right or where to go next.

Andy
Are you adding or multiplying? |u| and |v| are scalars so all you need to do is multiply them. No fuss, no mess.

-Dan

5. ## Re: Vector Proof

There's an easier way in my opinion, but unfortunately it involves geometry rather than vectors, so perhaps isn't what the assignment is asking for. But nevertheless, here goes:

Note that |u|v and |v|u have the same magnitude. If you construct a triangle with sides (a) the origin to the tip of |u|v, (b) from there to the the tip of |u|v+|v|u, and (c) from there back to the origin it is an isosceles triangle. If the vector u makes an angle alpha with the horizontal, and the vector v angle theta, then the angle between them is theta-alpha. It's not hard to show that the angle of the vertex of the isosceles triangle is 180+alpha-theta , or 180 minus the angle between u and v; let's call this angle gamma. Since theh other two angles of the isosceles triangle must be equal, and the sum of angles of a triangle is always 180, those other two angles e must each be 1/2 of 180 minus gamma, or 1/2(theta - alpha). Hence the base of the isosceles triangle bisects the angle between u and v.

6. ## Re: Vector Proof

Originally Posted by andy000
Prove that |u|v + |v|u bisects the angle between u and v.
I have tried looking at the dot product formula cos(theta) = u.v/|v||u| but I am a bit stuck.
Let $\displaystyle \overrightarrow b = \left\| {\overrightarrow u } \right\|{\overrightarrow v } + \left\| {\overrightarrow v } \right\|{\overrightarrow u }$ then $\displaystyle \frac{{\overrightarrow b \cdot \overrightarrow v }}{{\left\| {\overrightarrow v } \right\|}} = \frac{{\overrightarrow b \cdot \overrightarrow u }}{{\left\| {\overrightarrow u } \right\|}}$
To see that note that $\displaystyle \overrightarrow b \cdot \overrightarrow v = \left\| {\overrightarrow u } \right\|\overrightarrow v \cdot \overrightarrow v + \left\| {\overrightarrow v } \right\|\overrightarrow u \cdot \overrightarrow v = \left\| {\overrightarrow u } \right\|{\left\| {\overrightarrow v } \right\|^2} + \left\| {\overrightarrow v } \right\|\overrightarrow u \cdot \overrightarrow v$

7. ## Re: Vector Proof

Hi,
Your problem is really two dimensional; work in the plane of vectors u and v. Here's a geometric solution: