1. ## Vectors

Hi, does anyone know if there is/what the general formula for the addition of vectors is?

In particular, I'm trying to work out the problem: Two vectors and have precisely equal magnitudes. In order for the magnitude of + to be larger than the magnitude of - by the factor n, what must be the angle between them?

A friend offered the formula: sum^2 = a^2 + a^2 + a^2cos(t)
diff^2 = a^2 + a^2 - a^2cos(t)

However, I've never seen these formulas before and I don't understand the premise behind them so cannot determine if they are correct...

Thank you!

2. Originally Posted by strgrl
Two vectors and have precisely equal magnitudes. In order for the magnitude of + to be larger than the magnitude of - by the factor n, what must be the angle between them?
I find this question a bit vague.
We assume that from the wording that $n >1$.
In which case in order for $\left\| {A + B} \right\|$ to be greater than $\left\| {A - B} \right\|$ the angle between $A~\&~B$ must be acute.
(The length of $A~\&~B$ is irrelevant to the problem.)

Look at what follows.
$\left\| {A + B} \right\|^2 = \left\| A \right\|^2 + 2A \cdot B + \left\| B \right\|^2$
$\left\| {A - B} \right\|^2 = \left\| A \right\|^2 - 2A \cdot B + \left\| B \right\|^2$
So in order for $\left\| {A + B} \right\| > \left\| {A - B} \right\|$ it must be the case that $A \cdot B > 0$.
That happens only if the angle between them is acute.

3. [size=3]Hello, strgrl!

Two vectors $\vec A \text{ and }\vec B$ have equal magnitudes.
In order for the magnitude of $\vec A + \vec B$ to be larger than the magnitude of $\vec A - \vec B$
by the factor $n$, what must be the angle between them?

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

However, I've never seen these formulas before
and I don't understand the premise behind them
so cannot determine if they are correct...

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, $\vec A + \vec B$, is the diagonal of their parallelogram.

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

The difference of the vectors, $\vec A - \vec B$, is the other diagonal.
Code:
            *  .  .  .  .  .  .
*  *              .
*     *  A-B      .
-B *        *        .
*           *     .
* 180°-θ       *  .
*  *  *  *  *  *  *
A
Law of Cosines: . $|A + (\text{-}B)|^2 \;=\;|A|^2 + |\text{-}B|^2 - 2\!\cdot\!|A|\!\cdot\!|\text{-}B|\cos(180^o -\theta)$

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

4. Originally Posted by Soroban

The sum of the vectors, $\vec A + \vec B$, is the diagonal of their parallelogram.
Code:
            .  .  .  .  .  .  *
.              *  *
.    A+B    *     *
.        *        * B
.     *           *
.  *            θ *
*  *  *  *  *  *  *
A
Law of Cosines: . $|\vec A + \vec B|^2 \;=\;|\vec A|^2 + |\vec B|^2 - 2\!\cdot\!|\vec A|\!\cdot\!|\vec B|\cos\theta$
If you look carefully at your own diagram, the diagonal you have described is $B-A$ and not $A+B$

5. Originally Posted by Plato
If you look carefully at your own diagram, the diagonal you have described is $B-A$ and not $A+B$
Without arrow heads I suppose there is a certain ambiguity. However, I think all the posts in this thread collectively give enough help for the OP to work out the answer to the question.