Proving angle bisectors given two different non-parallel vectors $\vec a$ and $\vec b$

Question

Given two different non-parallel vectors, $\vec{a}$ and $\vec{b}$, prove that $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are the bisector vectors of the angle formed by $\vec{a}$ and $\vec{b}$.

This question is in my book of vector geometry, but it is kind of weird for me. I do not know if I am approaching it the right way.

I know that the formula for the angle bisector vector is given like this:

Let $u$ and $v$ be vectors of non-zero length. Let $\|u\|$ and $\|v\|$ be their respective lengths. Then $\|u\|v+\|v\|u$ is the angle bisector of $u$ and $v$. So the only way $\vec{a}+\vec{b}$ is the bisector vector is that both $\vec{a}$ and $\vec{b}$ have a length of one unit, but the question does not talk about this particular case, it seems phrased for all cases. Also I do not see how $\vec{a}$-$\vec{b}$ could satisfy this formula.

Any ideas?

Thanks in advance.

Answer 1

You are rigth, if you don't have equality hipothesis ∥u∥=∥v∥ the conclusion is false in general, you can draw any two vectors with diferent length and see difente angles.

Answer 2

A bit of geometry:

In a Cartesian coordinate system

Consider $\vec a = \vec {AB}$, $\vec b =\vec {BC}.$

Complete to form a

parallelogram $ABCD.$

$\vec a + \vec b = \vec {AC}$, a diagonal in the parallelogram .

1) $\angle CAB = \angle DCB := \alpha$, since $AB || DC$.

2) $\angle CAD = \angle ACB := \beta$, since $AD || BC$.

$\triangle ABC$ is isosceles , i.e. $\alpha =\beta$

$ \iff$

$||\vec {AB} || = ||\vec {BC}||$, i.e. they have the same length.

Answer 3

Only if the two vectors are of Equal Length, the statement is correct. Else (the book) is wrong. In this special case of rhombus ( not parallelogram) the sum and difference vectors are perpendicular.