I have a physics problem about vectors:
Two vectors $\boldsymbol{A}$ and $\boldsymbol{B}$ have precisely equal magnitudes. For the magnitude of $\boldsymbol{A} + \boldsymbol{B}$ to be $100$ times larger than the magnitude of $\boldsymbol{A} – \boldsymbol{B}$, what must be the angle between them?
I have tried to draw a diagram and calculate the problem with geometrical methods with two simultaneous equations of the form $c^2 = a^2 + b^2 – 2ab \cos θ$:$$
|\boldsymbol{A} + \boldsymbol{B}|² = |\boldsymbol{A}|² + |\boldsymbol{B}|² – 2|\boldsymbol{A}||\boldsymbol{B}|\cos θ\\
|\boldsymbol{A} – \boldsymbol{B}|² = |\boldsymbol{A}|² + |\boldsymbol{B}|² – 2|\boldsymbol{A}||\boldsymbol{B}|\cos(π – θ)
$$
Equating these two equations in terms of $θ$ gives$$
\cos θ = -\frac{9999|\boldsymbol{A} + \boldsymbol{B}|²}{|4|\boldsymbol{A}|²|}.
$$
This is as far as i could get, any help solving the problem will be greatly appreciated
Best Answer
You haven't used the fact that $|A| = |B|$
$|A+B| = 100|A-B|\\ |A|^2 + |B|^2 + 2|A||B|\cos \theta = 100^2(|A|^2 + |B|^2 - 2|A||B|\cos \theta)$
let $a = |A| = |B|$
$a^2 + a^2 + 2a^2\cos\theta = 100^2(a^2 + a^2 - 2a^2\cos\theta)$
Isolate $\cos \theta$ and simplify
$2(100^2+ 1)a^2 \cos \theta = 2a^2(100^2-1)\\ \theta = \arccos \frac {100^2 - 1}{100^2 + 1}$