Hello. I need help on this problem
Calculate the angle between parallelogram diagonals builded onto vectors $\displaystyle a = 2p+q$ and $\displaystyle b=p-3q$, if $\displaystyle |p|=5, |q|=2$, and the angle $\displaystyle (p, q)=\frac{\pi}{3}$
Hello. I need help on this problem
Calculate the angle between parallelogram diagonals builded onto vectors $\displaystyle a = 2p+q$ and $\displaystyle b=p-3q$, if $\displaystyle |p|=5, |q|=2$, and the angle $\displaystyle (p, q)=\frac{\pi}{3}$
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 $\displaystyle a + b\;\& \,a - b$.
$\displaystyle \left( {a + b} \right) \cdot \left( {a - b} \right) = a \cdot a - b \cdot b = \left\| a \right\|^2 - \left\| b \right\|^2 $.
$\displaystyle \frac{1}{2} = \cos \left( {\frac{\pi }{3}} \right) = \frac{{p \cdot q}}{{\left\| p \right\|\left\| q \right\|}}$.
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.
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 $\displaystyle \mu$ denote the angle included by the diagonals of the parallelogram. Then
$\displaystyle (|a|)^2 = (\frac12 |a+b|)^2 + (\frac12|a-b|)^2 - 2 \cdot \frac12 |a+b| \cdot \frac12 |a-b| \cdot \cos(\mu)$
Solve for $\displaystyle \cos(\mu)$ and afterwards determine the value of $\displaystyle \mu$.