B2. Dana has two vectors of equal length $\frac{1}{3} \vec{a}+\vec{b}$ and $\frac{2}{3} \vec{a}+\vec{b}$, where $\vec{a}$ and $\vec{b}$ are non-zero vectors such that $|\vec{a}|=\sqrt{3}|\vec{b}|$.

(a) Calculate the magnitude of the angle between the vectors $\vec{a}$ and $\vec{b}$.

(b) Calculate the magnitude of the angle between the vectors $\frac{1}{3} \vec{a}+\vec{b}$ and $\frac{2}{3} \vec{a}+\vec{b}$.

## 62nd Mathematical Competition for High School Students in Slovenia

Selection Competition, March 15, 2018

## Problems for 3rd Year

## Time for solving: 45 minutes.
B2. (a) The vectors $\frac{1}{3} \vec{a}+\vec{b}$ and $\frac{2}{3} \vec{a}+\vec{b}$ are of equal length, so $\left|\frac{1}{3} \vec{a}+\vec{b}\right|^{2}=\left|\frac{2}{3} \vec{a}+\vec{b}\right|^{2}$. The angle between the vectors $\vec{a}$ and $\vec{b}$ is denoted by $\varphi$, and we calculate

$$
\left|\frac{1}{3} \vec{a}+\vec{b}\right|^{2}=\left(\frac{1}{3} \vec{a}+\vec{b}\right) \cdot\left(\frac{1}{3} \vec{a}+\vec{b}\right)=\frac{1}{9} \vec{a} \cdot \vec{a}+\frac{2}{3} \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{b}=\frac{1}{9}|\vec{a}|^{2}+\frac{2}{3}|\vec{a}||\vec{b}| \cos \varphi+|\vec{b}|^{2}
$$

Similarly, we get

$$
\left|\frac{2}{3} \vec{a}+\vec{b}\right|^{2}=\frac{4}{9}|\vec{a}|^{2}+\frac{4}{3}|\vec{a}||\vec{b}| \cos \varphi+|\vec{b}|^{2}
$$

Since these two expressions must be equal, it follows that

$$
\frac{1}{3}|\vec{a}|^{2}+\frac{2}{3}|\vec{a}||\vec{b}| \cos \varphi=0
$$

From this, using the relation $|\vec{a}|=\sqrt{3}|\vec{b}|$, we express $\cos \varphi=-\frac{|\vec{a}|}{2|\vec{b}|}=-\frac{\sqrt{3}}{2}$, so $\varphi=\frac{5 \pi}{6}$.

Verification of $\left|\frac{1}{3} \vec{a}+\vec{b}\right|^{2}=\left|\frac{2}{3} \vec{a}+\vec{b}\right|^{2}$

1 point

Calculation of $\left|\frac{1}{3} \vec{a}+\vec{b}\right|^{2}=\frac{1}{9}|\vec{a}|^{2}+\frac{2}{3}|\vec{a}||\vec{b}| \cos \varphi+|\vec{b}|^{2}$ 3 points

Calculation of $\left|\frac{2}{3} \vec{a}+\vec{b}\right|^{2}=\frac{4}{9}|\vec{a}|^{2}+\frac{4}{3}|\vec{a}||\vec{b}| \cos \varphi+|\vec{b}|^{2}$ 2 points

Writing the equality $\frac{1}{3}|\vec{a}|^{2}+\frac{2}{3}|\vec{a}||\vec{b}| \cos \varphi=0$ 1 point Calculation of $\cos \varphi=-\frac{\sqrt{3}}{2}$ 2 points Writing the solution $\varphi=\frac{5 \pi}{6}$ 1 point.

(b) The angle between the vectors $\frac{1}{3} \vec{a}+\vec{b}$ and $\frac{2}{3} \vec{a}+\vec{b}$ is denoted by $\alpha$. Then,

$$
\cos \alpha=\frac{\left(\frac{1}{3} \vec{a}+\vec{b}\right) \cdot\left(\frac{2}{3} \vec{a}+\vec{b}\right)}{\left|\frac{1}{3} \vec{a}+\vec{b}\right|\left|\frac{2}{3} \vec{a}+\vec{b}\right|}=\frac{\frac{2}{9}|\vec{a}|^{2}+\vec{a} \cdot \vec{b}+|\vec{b}|^{2}}{\left|\frac{2}{3} \vec{a}+\vec{b}\right|^{2}}
$$

since by assumption $\left|\frac{1}{3} \vec{a}+\vec{b}\right|=\left|\frac{2}{3} \vec{a}+\vec{b}\right|$. Considering the equality (1), the relation $|\vec{a}|=\sqrt{3}|\vec{b}|$, and the result $\varphi=\frac{5 \pi}{6}$ from part (a), we get

$$
\cos \alpha=\frac{\frac{2}{9}|\vec{a}|^{2}+|\vec{a}||\vec{b}| \cos \varphi+|\vec{b}|^{2}}{\frac{4}{9}|\vec{a}|^{2}+\frac{4}{3}|\vec{a}||\vec{b}| \cos \varphi+|\vec{b}|^{2}}=\frac{\left(\frac{2}{3}-\frac{3}{2}+1\right)|\vec{b}|^{2}}{\left(\frac{4}{3}-2+1\right)|\vec{b}|^{2}}=\frac{\frac{1}{6}}{\frac{1}{3}}=\frac{1}{2}
$$

From this, we get $\alpha=\frac{\pi}{3}$.
Writing $\cos \alpha=\frac{\frac{2}{9}|\vec{a}|^{2}+\vec{a} \cdot \vec{b}+|\vec{b}|^{2}}{\left|\frac{2}{3} \vec{a}+\vec{b}\right|^{2}}$ ..... 4 points
Calculation of $\cos \alpha=\frac{1}{2}$ ..... 5 points
Writing the solution $\alpha=\frac{\pi}{3}$ ..... 1 point

## 62nd Mathematical Competition for High School Students in Slovenia

Selection Competition, March 15, 2018

Each mathematically correct and complete solution is worth $\mathbf{2 0}$ points, even if the solving procedure is different from the official solution.

## Solutions for 3rd Year