[size=3]Hello, strgrl!

Two vectors $\displaystyle \vec A \text{ and }\vec B$ have equal magnitudes.

In order for the magnitude of $\displaystyle \vec A + \vec B$ to be larger than the magnitude of $\displaystyle \vec A - \vec B$

by the factor $\displaystyle n$, what must be the angle between them?

A friend offered the formula: . $\displaystyle \begin{array}{ccc}\text{sum}^2 &=& a^2 + a^2 + a^2\cos\theta \\\text{diff}^2 &=& a^2 + a^2 - a^2\cos\theta \end{array}$

However, I've never seen these formulas before

and I don't understand the premise behind them

so cannot determine if they are correct...

Can someone please help explain these formulas?

Or offer another method of solving this problem? I believe your friend is using the Law of Cosines ... but incorrectly.

The sum of the vectors, $\displaystyle \vec A + \vec B$, is the diagonal of their parallelogram.

Code:

. . . . . . *
. * *
. A+B * *
. * * B
. * *
. * θ *
* * * * * * *
A

Law of Cosines: .$\displaystyle |\vec A + \vec B|^2 \;=\;|\vec A|^2 + |\vec B|^2 - 2\!\cdot\!|\vec A|\!\cdot\!|\vec B|\cos\theta $

The difference of the vectors, $\displaystyle \vec A - \vec B$, is the *other *diagonal. Code:

* . . . . . .
* * .
* * A-B .
-B * * .
* * .
* 180°-θ * .
* * * * * * *
A

Law of Cosines: .$\displaystyle |A + (\text{-}B)|^2 \;=\;|A|^2 + |\text{-}B|^2 - 2\!\cdot\!|A|\!\cdot\!|\text{-}B|\cos(180^o -\theta)$

. . which simplifies to: .$\displaystyle |A - B|^2\;=\;|A|^2 + |B|^2 + 2\!\cdot\!|A|\!\cdot\!|B|\cos\theta $

Your turn . . .