A parallelogram is built on the vectors vec a=2vec m+vec n…

A parallelogram is built on the vectors $\vec a=2\vec m+\vec n$ and $\vec b=\vec m-2\vec n$, where $\vec m$ and $\vec n$ are unit vectors with $\angle(\vec m,\vec n)=60^\circ$. Find the lengths of the diagonals and the angle between them.

Explanation of Method

Diagonals of the parallelogram built on $\vec a=2\vec m+\vec n$ and $\vec b=\vec m-2\vec n$.

Step 1: Basic products

\[ \vec m^{\,2}=\vec n^{\,2}=1,\qquad \vec m\cdot\vec n=\cos60^\circ=\tfrac12. \]

Step 2: The diagonals as vectors

\[ \vec d_1=\vec a+\vec b=3\vec m-\vec n,\qquad \vec d_2=\vec a-\vec b=\vec m+3\vec n. \]

Step 3: Lengths

\[ |\vec d_1|^2=9\vec m^{\,2}-6\,\vec m\cdot\vec n+\vec n^{\,2}=9-3+1=7, \]

\[ |\vec d_2|^2=\vec m^{\,2}+6\,\vec m\cdot\vec n+9\vec n^{\,2}=1+3+9=13. \]

\[|\vec d_1|=\sqrt7\approx2.65,\qquad |\vec d_2|=\sqrt{13}\approx3.61\]

Step 4: Angle between the diagonals

\[ \vec d_1\cdot\vec d_2=(3\vec m-\vec n)\cdot(\vec m+3\vec n)=3+9\cdot\tfrac12-\tfrac12-3=4, \]

\[ \cos\varphi=\frac{4}{\sqrt7\,\sqrt{13}}=\frac{4}{\sqrt{91}}\approx0.4193. \]

\[\varphi=\arccos\frac{4}{\sqrt{91}}\approx65.2^\circ\]

$\varphi$ $O$ $\vec m$ $\vec n$ $\vec a$ $\vec b$

$\vec d_1=\vec a+\vec b,\ |\vec d_1|=\sqrt7$
$\vec d_2=\vec a-\vec b,\ |\vec d_2|=\sqrt{13}$

Drawn with $\vec m=(1,0)$, $\vec n=\left(\tfrac12,\tfrac{\sqrt3}2\right)$: $\vec a=\left(\tfrac52,\tfrac{\sqrt3}2\right)$, $\vec b=(0,-\sqrt3)$. The diagonals meet at $\varphi\approx65.2^\circ$.

Result

\[\boxed{\,|\vec d_1|=\sqrt7\approx2.65,\qquad |\vec d_2|=\sqrt{13}\approx3.61\,}\]

\[\boxed{\,\varphi=\arccos\frac{4}{\sqrt{91}}\approx65.2^\circ\,}\]

Remark

Correction. $\arccos(4/\sqrt{91})=65.21^\circ$, not $65.1^\circ$.