# So I understand where the formula for cos between two vectors comes from. However, if i have a paralelogram of basis vec(a) , vec(b) , and i want to find out the angle between its diagonals starting only from the fact that norm(vec(a))=3 and norm(vec(b))=2 and the angle between them pi/3 , how would i go about it?

So I understand where the formula for cos between two vectors comes from. However, if i have a paralelogram of basis $\stackrel{\to }{a}$,$\stackrel{\to }{b}$ , and i want to find out the angle between its diagonals starting only from the fact that
$‖\stackrel{\to }{a}‖=3$
and
$‖\stackrel{\to }{b}‖=2$
and the angle between them pi/3 , how would i go about it?
I know the diagonals are $\stackrel{\to }{a}-\stackrel{\to }{b}$ and $\stackrel{\to }{a}+\stackrel{\to }{b}$ respectively, but im not sure how I shoul apply the formula for sum and difference of vectors.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Kenley Rasmussen
$\stackrel{\to }{a}=||a||dir\left(a\right)=\left(||a||\mathrm{cos}\theta ,||a||\mathrm{sin}\theta \right)=\left(3\mathrm{cos}\theta ,3\mathrm{sin}\theta \right)$
$\stackrel{\to }{b}=||b||dir\left(b\right)=\left(||b||\mathrm{cos}\left(\theta +\frac{\pi }{3}\right),||a||\mathrm{sin}\left(\theta +\frac{\pi }{3}\right)\right)=\left(2\mathrm{cos}\left(\theta +\frac{\pi }{3}\right),2\mathrm{sin}\left(\theta +\frac{\pi }{3}\right)\right)$
Now, you are looking for $\alpha =\mathrm{arccos}\frac{\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)\cdot \left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)}{||\stackrel{\to }{a}+\stackrel{\to }{b}||||\stackrel{\to }{a}-\stackrel{\to }{b}||}$
You can easily calculate the parts, indeed
$\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)\cdot \left(\stackrel{\to }{a}-\stackrel{\to }{b}\right)={\stackrel{\to }{a}}^{2}-{\stackrel{\to }{b}}^{2}=9-4=5$
Besides,
$||\stackrel{\to }{a}-\stackrel{\to }{b}||=\sqrt{\left(3\mathrm{cos}\theta -2\mathrm{cos}\left(\theta +\frac{\pi }{3}\right){\right)}^{2}+\left(3\mathrm{sin}\theta -2\mathrm{sin}\left(\theta +\frac{\pi }{3}\right){\right)}^{2}}=\sqrt{9+4-12\left(\mathrm{cos}\theta \mathrm{cos}\left(\theta +\frac{\pi }{3}\right)+\mathrm{sin}\theta \mathrm{sin}\left(\theta +\frac{\pi }{3}\right)\right)}=\sqrt{7}$
$||\stackrel{\to }{a}+\stackrel{\to }{b}||=\sqrt{\left(3\mathrm{cos}\theta +2\mathrm{cos}\left(\theta +\frac{\pi }{3}\right){\right)}^{2}+\left(3\mathrm{sin}\theta +2\mathrm{sin}\left(\theta +\frac{\pi }{3}\right){\right)}^{2}}=\sqrt{9+4+12\left(\mathrm{cos}\theta \mathrm{cos}\left(\theta +\frac{\pi }{3}\right)+\mathrm{sin}\theta \mathrm{sin}\left(\theta +\frac{\pi }{3}\right)\right)}=\sqrt{19}$
Here, I've used the fact that
$\mathrm{cos}\theta \mathrm{cos}\left(\theta +\frac{\pi }{3}\right)+\mathrm{sin}\theta \mathrm{sin}\left(\theta +\frac{\pi }{3}\right)=\frac{1}{2}$
Therefore, $\alpha =\mathrm{arccos}\frac{5}{\sqrt{133}}.$
###### Did you like this example?
Kamila Frye
You have
$\begin{array}{rl}{‖\stackrel{\to }{a}-\stackrel{\to }{b}‖}^{2}& ={‖\stackrel{\to }{a}‖}^{2}-2⟨\stackrel{\to }{a},\stackrel{\to }{b}⟩+{‖\stackrel{\to }{b}‖}^{2}\\ & =9-\overline{)2}×3×2×\overline{)\mathrm{cos}\left(\frac{\pi }{3}\right)}+4\\ & =7.\end{array}$
Can you deal with $‖\stackrel{\to }{a}+\stackrel{\to }{b}‖$ now?