How to find the vector formula for the bisector of given two vectors? There are two vectors called $\vec{a}$ and $\vec{b}$. Vector $\vec{c}$ is the bisector and it can be given as
$$\vec{c} = |b|\vec{a} + |a|\vec{b}$$
How to prove that? I have used the dot product method. But there I can't find the angle between them. Then I tried to do this by using unit vectors. Then I got an answer like in the given picture. The answer which I got. Can you help me to get the correct answer
 A: Normalize $\vec{a}$ : $\frac{1}{||a||} \vec{a}$.
Normalize $\vec{b}$ : $\frac{1}{||b} \vec{b}$
Now add these two vectors to get:
$\vec{c}$ = $\frac{1}{||a||} \vec{a}$ + $\frac{1}{||b||} \vec{b}$.
Adding two unit vectors, vector addition, gives a resultant $\vec{c}$ that divides the angle between them.
$ \vec{c}$ or any $\alpha \vec{c}$ are  vectors  that have this property.
Let $\alpha = ||a||$ $||b|| $ to get: 
$\vec{c} = ||b|| \vec{a}$ + $||a|| \vec{b}$.
A: Here's a purely geometric argument.
By definition, the sum of two vectors is equal to the diagonal of the parallelogram spanned by the vectors.
Now, observe that the two vectors $|b|\vec{a}$ and $|a|\vec{b}$ have exactly the same length.  Therefore the parallelogram they span is a rhombus.  The result then follows from the fact that the diagonal of a rhombus bisects its angles.
A: Step 1 - normalise the original vectors. So define $\vec{\dot{a}} = \frac{\vec{a}}{|\vec{a}|}$ and similarly for $\vec{\dot{b}}$, then let $\vec{\dot{c}} = \vec{\dot{a}} + \vec{\dot{b}}$. It should be pretty simple to prove that the direction of $\vec{\dot{c}}$ is the same as the one of $\vec{c}$ in your post.
Step 2 - Find the angle between the new proposed bisector and the original vectors. So define $\alpha$ as the angle between $\vec{a}$ and $\vec{c}$, and then $\vec{\dot{a}} \cdot \vec{\dot{c}} = |\vec{\dot{a}}||\vec{\dot{c}}|\cos{\alpha} = |\vec{\dot{c}}|\cos{\alpha}$ since we set $|\vec{\dot{a}}| = 1$ in the first step. Similarly if $\beta$ is the angle between $\vec{b}$ and $\vec{c}$, then $\vec{\dot{b}} \cdot \vec{\dot{c}} = |\vec{\dot{c}}|\cos{\beta}$. 
But, from the way they've been defined, $\vec{\dot{a}} \cdot \vec{\dot{c}} = \vec{\dot{a}} \cdot \vec{\dot{a}} + \vec{\dot{a}} \cdot \vec{\dot{b}} = |\vec{\dot{a}}| + \vec{\dot{a}} \cdot \vec{\dot{b}} = 1 + \vec{\dot{a}} \cdot \vec{\dot{b}}$, and you can show that the other dot product has the same value. So you can conclude that $\cos{\alpha} = \cos{\beta}$, and then all you have to do is show that the angles are in the same quadrant, and hence must be equal.
