# Angle of Parallelogram diagonals

• January 22nd 2009, 01:58 PM
faktor-g
Angle of Parallelogram diagonals
Hello. I need help on this problem

Calculate the angle between parallelogram diagonals builded onto vectors $a = 2p+q$ and $b=p-3q$, if $|p|=5, |q|=2$, and the angle $(p, q)=\frac{\pi}{3}$
• January 22nd 2009, 02:59 PM
Plato
Quote:

Originally Posted by faktor-g
Calculate the angle between parallelogram diagonals builded onto vectors $a = 2p+q$ and $b=p-3q$, if $|p|=5, |q|=2$, and the angle $(p, q)=\frac{\pi}{3}$

This is a nightmare of a problem in so far as calculation goes.
I will not do it for you. But I gladly give you these hints.
The diagonals of this parallelogram are $a + b\;\& \,a - b$.
$\left( {a + b} \right) \cdot \left( {a - b} \right) = a \cdot a - b \cdot b = \left\| a \right\|^2 - \left\| b \right\|^2$.
$\frac{1}{2} = \cos \left( {\frac{\pi }{3}} \right) = \frac{{p \cdot q}}{{\left\| p \right\|\left\| q \right\|}}$.

Now you can put this all together. Maybe someone here who does other people’s work can understand this question and give you a complete solution. But if you really want to understand you will do it for yourself.
• January 23rd 2009, 06:38 AM
earboth
Quote:

Originally Posted by faktor-g
Hello. I need help on this problem

Calculate the angle between parallelogram diagonals builded onto vectors $a = 2p+q$ and $b=p-3q$, if $|p|=5, |q|=2$, and the angle $(p, q)=\frac{\pi}{3}$

1. I've made an exact drawing of the parallelogram. So you have the possibility to control your results.

2. Maybe you can use the cosine rule to calculate the angle in question:

Let $\mu$ denote the angle included by the diagonals of the parallelogram. Then

$(|a|)^2 = (\frac12 |a+b|)^2 + (\frac12|a-b|)^2 - 2 \cdot \frac12 |a+b| \cdot \frac12 |a-b| \cdot \cos(\mu)$

Solve for $\cos(\mu)$ and afterwards determine the value of $\mu$.