Length of Parallelogram Diagonals

Question

I have a problem with vectors. It is said that there are $2$ vectors $\overrightarrow{m}$ and $\overrightarrow{n}$ with length $1$ and $\angle (m,n)=60^\circ$. $\overrightarrow{a}$ and $\overrightarrow{b}$ are $2$ vectors such that $\overrightarrow{a}=2\overrightarrow{m}+\overrightarrow{n}$ and $\overrightarrow{b}=\overrightarrow{m}-2\overrightarrow{n}$ and represent the sides of parallelogram. I have to calculate the length of the diagonals of this parallelogram. Thank you very much!

Answer 1

The diagonal vectors will be addition and subtraction of these two vectors. I am calculating one, you can do other one with similar procedure.

You have to calculate length of diagonal i.e. $| \overrightarrow{a}+ \overrightarrow{b}|$

$$| \overrightarrow{a}+ \overrightarrow{b}|= |(2\overrightarrow{m}+\overrightarrow{n}) + (\overrightarrow{m}-2\overrightarrow{n})|=|3\overrightarrow{m}-\overrightarrow{n}|=\sqrt{|3\overrightarrow{m}|^2+|\overrightarrow{n}|^2+2|\overrightarrow{3m}||\overrightarrow{n}| \cos \angle(m,n)}=\sqrt{9+1+6\times \cos 60^\circ}=\sqrt{13}$$

Do the same for other one. (subtract vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ in that case)

Answer 2

A parallelogram with vector "sides" a and b has diagonals $a+b$ and $a-b$.

In this case it means $(2m+n)+(m-2n)=3m-n$ and $2m+n-(m-2n)=m+3n.$

The square of their lengths is the dot product of these vectors with themselves:

$(3m-n).(3m-n)=9m^2+n^2-6 m.n=9+1-6 \cos(60°)=7$ and

$(m+3n).(m+3n)=m^2+9n^2+6 m.n=1+9+6 \cos(60°)=13.$

It suffices now to take the square roots of these values.