Task B-4.7. Vectors $\vec{a}=2 \vec{m}+\vec{n}$ and $\vec{b}=\vec{m}-\vec{n}$ determine a parallelogram. Here, $\vec{m}$ and $\vec{n}$ are unit vectors, and the measure of the angle between $\vec{m}$ and $\vec{n}$ is $\frac{\pi}{3}$. Determine the area of the parallelogram determined by vectors $\vec{a}$ and $\vec{b}$.
## Solution.

The area of the parallelogram is $P=ab \sin \alpha$, where $a=|\vec{a}|, b=|\vec{b}|$ and $\alpha=\angle(\vec{a}, \vec{b})$.

It holds that $\vec{m}^{2}=|\vec{m}|^{2}=1, \vec{n}^{2}=|\vec{n}|^{2}=1$ and

$$
\vec{m} \cdot \vec{n}=|\vec{m}| \cdot|\vec{n}| \cos \frac{\pi}{3}=\frac{1}{2}
$$

$$
\begin{aligned}
& |\vec{a}|^{2}=\vec{a}^{2}=(2 \vec{m}+\vec{n})^{2}=4 \vec{m}^{2}+4 \vec{m} \cdot \vec{n}+\vec{n}^{2}=4+2+1=7 \\
& |\vec{b}|^{2}=\vec{b}^{2}=(\vec{m}-\vec{n})^{2}=\vec{m}^{2}-2 \vec{m} \cdot \vec{n}+\vec{n}^{2}=1-1+1=1 \\
& \vec{a} \cdot \vec{b}=(2 \vec{m}+\vec{n})(\vec{m}-\vec{n})=2 \vec{m}^{2}-\vec{m} \cdot \vec{n}-\vec{n}^{2}=2-\frac{1}{2}-1=\frac{1}{2}
\end{aligned}
$$

$$
\begin{aligned}
\cos \alpha & =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\frac{\frac{1}{2}}{\sqrt{7} \cdot 1}=\frac{1}{2 \sqrt{7}} \\
\sin \alpha & =\sqrt{1-\cos ^{2} \alpha}=\sqrt{1-\frac{1}{28}}=\frac{3 \sqrt{3}}{2 \sqrt{7}}=\frac{3 \sqrt{21}}{14} \\
P & =a b \sin \alpha=\sqrt{7} \cdot 1 \cdot \frac{3 \sqrt{3}}{2 \sqrt{7}} \\
P & =\frac{3 \sqrt{3}}{2}
\end{aligned}
$$

(2 points)