# Determine the unit vector

1. Aug 16, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
The vector -i+j+k bisects the angle between the vector c and 3i+4j. Determine a unit vector along c.

2. Relevant equations

3. The attempt at a solution

Taking the dot product of the two vectors (other than c) gives me the cosine of the angle = 1/5√3.
This is also equal to the angle between the angle bisector and c. But now I can't take the dot product anymore as I don't know anything about c.

2. Aug 16, 2013

### haruspex

If you have two vectors, u and v, can you write down in terms of u and v a vector that bisects the angle between them?

3. Aug 16, 2013

### utkarshakash

lu+mv where l and m are constants.

4. Aug 16, 2013

### haruspex

That can give you any angle in the plane formed by the vectors. To get the bisector there will be a relationship between l, m, |u| and |v|.

Last edited: Aug 16, 2013
5. Aug 17, 2013

### utkarshakash

OK So here's what I did:
I assumed c=xi+yj+zk

$-i+j+k= \lambda \left( \dfrac{3i+4j}{5} + \dfrac{ \vec{c}}{|\vec{c}|} \right)$

Equating the respective components of both sides I get three equations

$\frac{3}{5} + \frac{x}{|\vec{c}|} = \frac{-1}{\lambda} \\ \frac{4}{5} + \frac{y}{|\vec{c}|} = \frac{1}{\lambda} \\ \lambda \frac{z}{|\vec{c}|} = 1$

Now taking the dot product of angle bisector with c gives me the value of |c|=5(-x+y+z). Using this in the above three equations gives me x=y=z=0 !:surprised

6. Aug 17, 2013

### haruspex

I don't see how it gives you that. How about taking the dot product of c with itself?