How to find angle between two vectors given only the magnitudes and the difference in magnitudes This is a physics first year problem, the same problem has been posted on the Math Stack Exchange website, but the answer given there is beyond the scope of what is taught in my course! For now, all my teacher has taught me are the cosine law and just the laws of vectors in general.
Here is the problem,

Vector $\vec B$ is 5.0 cm long and vector $\vec A$ is 4.0 cm long. Find the angle, in degrees, between these two vectors when |$\vec A$ + $\vec B$| = 3.0 cm and |$\vec A$ - $\vec B$| = 3.0 cm.

I really don't know where to start with this one, and have been stuck for a hot minute now... any help would be appreciated, and it would really help me get the concept of vectors for this course!
 A: $$|\vec{A}+\vec{B}|=|(x+x')i + (y+y')j|=\sqrt{(x+x')^2+(y+y')^2}=3\implies(x+x')^2+(y+y')^2=9$$
$$|\vec{A}-\vec{B}|=|(x-x')i + (y-y')j|=\sqrt{(x-x')^2+(y-y')^2}=3\implies(x-x')^2+(y-y')^2=9$$
$$|\vec{A}|=\sqrt{x^2+y^2}=4 \implies x^2+y^2=16$$
$$|\vec{B}|=\sqrt{x'^2+y'^2}=4 \implies x'^2+y'^2=25$$
You have lots of equations. Use them to get $x,x',y,y'$ and you can use the dot product formula to get the angle.

If you want the solution:
$$(x+x')^2+(y+y')^2=9 \implies x^2+2xx'+x'^2+y^2+2yy'+y'^2=9 \implies 2xx'+2yy'+16+25=9\implies 2(xx'+yy')=-32\implies xx'+yy'=-16$$
This is exactly the dot product formula, so:
$$\vec{A}\cdot\vec{B}=-16=|A||B|\cos(\theta)=20\cos(\theta)\implies \cos(\theta)=-4/5$$
so
$$\theta = 2.498 \text{  or  } \theta = -2.498$$
Note that as Jensen Paull said, the problem is impossible with the condition:
$$|\vec{A}-\vec{B}|=3$$
Without it you would get to my solution.
A: $|\vec A+\vec B|^2=(\vec A+\vec B)\cdot (\vec A+\vec B)=\vec A\cdot \vec A + 2(\vec A\cdot \vec B)+\vec B\cdot \vec B=|\vec A|^2+2(\vec A\cdot \vec B)+|\vec B|^2$
$|\vec A-\vec B|^2=(\vec A-\vec B)\cdot (\vec A-\vec B)=\vec A\cdot \vec A - 2(\vec A\cdot \vec B)+\vec B\cdot \vec B=|\vec A|^2-2(\vec A\cdot \vec B)+|\vec B|^2$
Subtract the equations to get:
$|\vec A+\vec B|^2-|\vec A-\vec B|^2=4(\vec A\cdot \vec B)$
So you can work out what the dot product of $\vec A$ and $\vec B$ is, which should help.
As an aside, the second equation is a vector version of the cosine rule for triangles. If vectors $\vec a$ and $\vec b$ are two sides of a triangle, the third side $\vec c=\vec a-\vec b$. The cosine rule for this triangle says: $c^2=a^2+b^2-2ab\,\mathrm{cos}\,C$, Where $a$, $b$, and $c$ are the magnitudes of the vectors, and the last term on the right is just two times the dot product of $\vec a$ and $\vec b$. If the dot product is zero, then you have a right-angled triangle, and this reduces to Pythagoras' Theorem.
A: Just calculate the dot product of each one (the difference and sumed terms) with itself (basically, square them) and then you can find the cosine of the angle between them.
A: Follow up, given differences and addition of vectors, find angle inbetween
This problem is impossible since
$$\cos(\theta) = u$$
$$\cos(\theta) = - u$$
Solutions are:
$$u=0, \cos(\theta) = 0$$
Which is not valid on this problem.
