# Angle between two vectors given magnitudes only

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

• Sorry didn't complete the question... been edited Jun 27, 2018 at 2:18

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

$2(100^2+ 1)a^2 \cos \theta = 2a^2(100^2-1)\\ \theta = \arccos \frac {100^2 - 1}{100^2 + 1}$ • The solution is correct thankyou, although I dont understand why | A + B | has an angle of π-θ. I got it that between A and -B the angle = π-θ Jun 27, 2018 at 2:29
• Your solution gives an answer of 1.14... should the angle between the vectors not be almost 180 degrees? Jun 27, 2018 at 2:39
• They are almost parallel. To add vectors place them head to tail (forming a parallelogram.) The difference between the two vectors is the opposite diagonal (from head to head) Jun 27, 2018 at 3:08
• The diagram clears it up thanks Jun 27, 2018 at 3:22

Intuitively they need to be almost in the same direction. The sum will be (almost) twice the length of one vector and the difference will be the (small) sideways component between them. We can divide by the length and work with unit vectors, then choose our coordinates so that $A=(1,0),B=(\cos \theta, \sin \theta)$. The angle between them is then $\pi-\theta$
The length of the sum is then $\sqrt{(1+\cos \theta)^2+\sin^2 \theta}=\sqrt{2+2\cos \theta}$.
The length of the difference is $\sqrt{(1-\cos \theta)^2+\sin^2 \theta}=\sqrt{2-2 \cos \theta}$
Let $\cos \theta=c$ to save typing and we want $$\frac {\sqrt{2+2c}}{\sqrt{2-2c}}=100\\\frac {2+2c}{2-2c}=10000\\ 2+2c=20000-20000c\\c=\frac{19998}{20002}\\ \theta=\arccos \frac {19998}{20002} \approx 0.0199999$$