How do I find the magnitude of a sum of vectors from a difference of vectors?

Jun 2017
399
6
Lima, Peru
The problem is as follows:

Two vectors $\vec{r}$ and $\vec{s}$ make an angle $106^{\circ}$ and together are of the same modulus. If the magnitude of the vector $\vec{r}-\vec{s}$ is $80$. Find the magnitude of the vector $\vec{r}+\vec{s}$.

The alternatives given are:

$\begin{array}{ll}
1.&40\\
2.&50\\
3.&60\\
4.&70\\
5.&80\\
\end{array}$

What I've attempted to do here was to find the sum of the vectors as follows: Since it mentions that both vectors are in the same magnitude, then I'm only using $r$

$\left\|\vec{r}+\vec{s}\right\|^2=r^2+r^2-2r^2\cos\omega$

$\left\|\vec{r}+\vec{s}\right\|^2=2r^2(1+\cos (106)^{\circ})$

But to find these I'm given the information regarding the difference which is $80$.

This the part where I'm not sure how to use the equation regarding the vector sum.

In this specific situation how should it be used?

As this? where $\left\|r\right\|=\left\|s\right\|$

$\left\|\vec{r}-\vec{s}\right\|^2=r^2+r^2-2r(-r)\cos (180-106)^{\circ}$

$\left\|\vec{r}-\vec{s}\right\|^2=2r^2(1+\cos (74)^{\circ})$

$80^2=2r^2\left(1+\frac{7}{25}\right)$

$r=50$

Then for the sum it should be established as:

$\left\|\vec{r}+\vec{s}\right\|^2=2r^2(1-\cos (106)^{\circ})$

$\left\|\vec{r}+\vec{s}\right\|^2=2r^2(1+\cos (74)^{\circ})$

$\left\|\vec{r}+\vec{s}\right\|=\sqrt{2\left(50\right)^2\left(1+\frac{7}{25}\right)}$

To which I end up obtaining the same modulus from where I began with:

$80$.

But If I were to swap the signs I'd end up obtaining a different result.

$\left\|\vec{r}+\vec{s}\right\|^2=2r^2(1-\cos (74)^{\circ})$

$\left\|\vec{r}+\vec{s}\right\|^2=2r^2(1-\cos (74)^{\circ})$

$\left\|\vec{r}+\vec{s}\right\|=\sqrt{2\left(50\right)^2\left(1-\frac{7}{25}\right)}$

$\left\|\vec{r}+\vec{s}\right\|= 60$

So which is correct? Does it mean that the modulus of the difference in vectors is the same as the modulus of the sum? Can someone help me to clear out this doubt? Help, please!
 

skeeter

Math Team
Jul 2011
3,364
1,855
Texas
$\cos(106) = -\cos(74)$

$80^2 = 2r^2[1+\cos(74)]$ and $|r+s|^2 = 2r^2[1-\cos(74)] \implies |r+s| = \sqrt{\dfrac{80^2[1-\cos(74)]}{1+\cos(74)}}$

vector_sum_difference.jpg
 

skipjack

Forum Staff
Dec 2006
21,482
2,472
In the diagram skeeter provided, the vectors for r-s and r+s should be perpendicular. The ratio of their magnitudes is tan(θ), where θ is the appropriate angle (half of 74°). This makes obtaining the answer very easy.