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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

John Wayland Bales
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user228285
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  • 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. – John Wayland Bales May 18 '17 at 04:50
  • 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. – ConMan May 18 '17 at 04:50
  • 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. – ConMan May 18 '17 at 04:51
  • @ConMan OK, I believe you are correct. That makes sense. – John Wayland Bales May 18 '17 at 04:52
  • Can't you provide me an answer? – user228285 May 18 '17 at 05:01
  • But I have to show that expression. Is it incorrect? – user228285 May 18 '17 at 05:02
  • It's not wrong. I've provided a mostly-complete answer where you just have to prove a few assumptions to make it rigorous. – ConMan May 18 '17 at 05:06

3 Answers3

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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.

mweiss
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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:

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

Peter Szilas
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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.

ConMan
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