Animated Solution for Physics - Kinematics: Two vectors $\mathbf{A}$ and $\mathbf{B}$ have equal magnitudes. The magnitude of $(\mathbf{A}+\mathbf{B})$ is '$n$' times the magnitude of $(\mathbf{A}-\mathbf{B})$. The angle between $\mathbf{A}$ and $\mathbf{B}$ is

Imagine you are standing at the origin of a coordinate system. You have two vectors, $\mathbf{A}$ and $\mathbf{B}$, pointing outwards.

The problem gives us a beautiful symmetry: both vectors have the exact same length. Let's call this common magnitude $A$.

So, we can write $|\mathbf{A}|=|\mathbf{B}|=A$. Let the angle between them be $\theta$. This angle $\theta$ is exactly what we need to find!

To relate the sum and difference of these vectors, we need to use the parallelogram law of vector addition.

The magnitude of the sum of two vectors is given by the formula $|\mathbf{A}+\mathbf{B}|=\sqrt{A^2+B^2+2AB\cos \theta }$.

Since $A=B$, we can substitute this into our formula. This gives us $|\mathbf{A}+\mathbf{B}{|}^2=A^2+A^2+2A^2\cos \theta$.

Factoring out $2A^2$, we get a neat expression: $|\mathbf{A}+\mathbf{B}{|}^2=2A^2(1+\cos \theta )$.

Similarly, the magnitude of the difference is $|\mathbf{A}-\mathbf{B}|=\sqrt{A^2+B^2-2AB\cos \theta }$.

Following the exact same logic, we find that $|\mathbf{A}-\mathbf{B}{|}^2=2A^2(1-\cos \theta )$.

Now, let's look at the core condition given in the problem. We are told that the magnitude of the sum is $n$ times the magnitude of the difference.

Mathematically, this translates to $|\mathbf{A}+\mathbf{B}|=n|\mathbf{A}-\mathbf{B}|$.

Square roots can be messy to deal with, so let's square both sides of this equation. This gives us $|\mathbf{A}+\mathbf{B}{|}^2=n^2|\mathbf{A}-\mathbf{B}{|}^2$.

Now, we substitute the expressions we derived earlier into our squared condition.

This yields $2A^2(1+\cos \theta )=n^2\cdot 2A^2(1-\cos \theta )$.

Notice how the $2A^2$ terms are present on both sides? They beautifully cancel each other out! We are left with $1+\cos \theta =n^2(1-\cos \theta )$.

We can rearrange this into a ratio: $\frac{n^2}{1}=\frac{1+\cos \theta }{1-\cos \theta }$.

To isolate $\cos \theta$ quickly, we can use a brilliant algebraic trick called Componendo and Dividendo.

Applying this rule, we get $\frac{n^2-1}{n^2+1}=\frac{(1+\cos \theta )-(1-\cos \theta )}{(1+\cos \theta )+(1-\cos \theta )}$.

The right side simplifies perfectly to $\frac{2\cos \theta }{2}$, which is just $\cos \theta$.

We have successfully isolated our trigonometric term!

We found that $\cos \theta =\frac{n^2-1}{n^2+1}$.

To find the angle $\theta$ itself, we simply take the inverse cosine of both sides.

This gives us our final, elegant answer: $\theta ={\cos }^{-1}\left(\frac{n^2-1}{n^2+1}\right)$.

As a fun thought experiment, imagine if $n=1$. This would mean $\cos \theta =0$, which implies $\theta =90^{\circ }$. This perfectly aligns with the geometric fact that the diagonals of a rhombus are equal only when it is a square!