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

  • $\begingroup$ It appears that you are saying that $\mathbf{c}$ is a vector yet at the same time it is the sum of two magnitudes. But magnitudes are scalars, not vectors. Also your "equation" for $\mathbf{c}$ is incomprehensible as written and so is your diagram. $\endgroup$ – John Wayland Bales May 18 '17 at 4:50
  • $\begingroup$ Remember that the angle between two vectors is defined as the value that makes $x \cdot y = |x||y|\cos{\theta}$ true, so try showing that the dot products of (a and c) and (b and c), when scaled appropriately, are the same thing. $\endgroup$ – ConMan May 18 '17 at 4:50
  • $\begingroup$ John - I think there's just a bit of formatting issue. At a guess, I'd say the OP is trying to write $\vec{c} = |b|\vec{a} + |a|\vec{b}$, which is a perfectly fine vector expression. $\endgroup$ – ConMan May 18 '17 at 4:51
  • $\begingroup$ @ConMan OK, I believe you are correct. That makes sense. $\endgroup$ – John Wayland Bales May 18 '17 at 4:52
  • $\begingroup$ Can't you provide me an answer? $\endgroup$ – user228285 May 18 '17 at 5:01

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.

  • $\begingroup$ [+1] I like this proof using figure geometry. So sad that that there are whole books with "geometry" in their title without any figure... $\endgroup$ – Jean Marie May 18 '17 at 5:17
  • $\begingroup$ Very nice geometrical argument. $\endgroup$ – Peter Szilas May 18 '17 at 5:32

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.


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:

$\vec{c} = ||b|| \vec{a}$ + $||a|| \vec{b}$.


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