Follow up, given differences and addition of vectors, find angle inbetween Following on from this question.
How to find angle between two vectors given only the magnitudes and the difference in magnitudes
I am very confused about how this is possible
We are given the relations:
$$|\vec{a}|=4$$
$$|\vec{b}|= 5$$
$$|\vec{A}+\vec{B}| = 3$$
$$|\vec{A}-\vec{B}| = 3$$
Square both sides
$$|\vec{A}+\vec{B}|^2 = 9$$
$$|\vec{A}-\vec{B}|^2 = 9$$
Turn the square into a dot product relation
$$(\vec{A}+\vec{B})\cdot(\vec{A}+\vec{B})  = 9$$
$$(\vec{A}-\vec{B})\cdot (\vec{A}-\vec{B}) = 9$$
Expand the dot product
$$\vec{a} \cdot \vec{a} +2\vec{a}\cdot \vec{b}+ \vec{b} \cdot \vec{b}  = 9$$
$$\vec{a} \cdot \vec{a} -2\vec{a}\cdot \vec{b}+ \vec{b} \cdot \vec{b}  = 9$$
$\vec{k} \cdot \vec{k} = |\vec{k}|^2$
$\vec{a}\cdot \vec{b} = |\vec{a}||\vec{b}| \cos(\theta)$
$$|\vec{a}|^2  + 2|\vec{a}||\vec{b}| \cos(\theta) +|\vec{b}|^2 = 9$$
$$|\vec{a}|^2  - 2|\vec{a}||\vec{b}| \cos(\theta) +|\vec{b}|^2 = 9$$
Rearrange for $\cos(\theta)$ for each
Naively, one would use each equation and then solve for theta, but theta independantly satisfies both equations  and thus the solution should also satisfy both.
It's in the form
$$\cos(\theta) = u$$
$$\cos(\theta) = - u$$
$\theta$ has to satisfy  both, Drawing a graph, the only way that this is possible  is if
$$u=0$$
But it's not, so I don't see how there is a solution?
 A: I'm not sure I could post this as an answer, but I don't see any problem with your reasoning. You simply proved that two vectors can have a sum and a difference with the exact same length only if they're perpendicular.
I wasn't particulary aware of this result, but it's reasonable I think.
A: There are several forms of the polarization identity

*

*$\vec A \cdot \vec B=\displaystyle\frac{1}{4}\left( |\vec A +\vec B |^2 - |\vec A -\vec B |^2 \right)$

*$\vec A \cdot \vec B=\displaystyle\frac{1}{2}\left( |\vec A +\vec B |^2 - |\vec A |^2 - |\vec B |^2 \right)$

*$\vec A \cdot \vec B=\displaystyle\frac{1}{2}\left( |\vec A|^2 +|\vec B |^2 - |\vec A -\vec B |^2 \right)$
With the given

*

*$|\vec A|=4$

*$|\vec B|=5$

*$|\vec A + \vec B|=3$

*$|\vec A - \vec B|=3$
one obtains

*

*$\vec A \cdot \vec B=\displaystyle\frac{1}{4}\left( 3^2 - 3^2 \right) =0$

*$\vec A \cdot \vec B=\displaystyle\frac{1}{2}\left( 3^2 - 4^2 - 5^2 \right)=-16$

*$\vec A \cdot \vec B=\displaystyle\frac{1}{2}\left( 4^2 + 5^2 - 3^2 \right)=16$
which is clearly inconsistent, which is in accord with @Frobenius 's comment that this is an impossible problem.

side comments:

*

*If instead the given were
$|\vec A|=3$,  $|\vec B|=4$,  $|\vec A + \vec B|=5$, $|\vec A - \vec B|=5$,
then we have a consistent problem.

*If instead the given were
somehow $|\vec A|^2=-4^2$,  $|\vec B|^2=5^2$,  $|\vec A + \vec B|=3$, $|\vec A - \vec B|=3$,
then we have a consistent problem.

