## Task B-3.3.

Given vectors $\vec{u}=\vec{m}+3 \vec{n}, \vec{v}=7 \vec{m}-5 \vec{n}, \vec{w}=\vec{m}-4 \vec{n} \text{ and } \vec{z}=7 \vec{m}-2 \vec{n}$, where $\vec{m}, \vec{n} \neq \overrightarrow{0}$. If vectors $\vec{u} \text{ and } \vec{v}$, and $\vec{w} \text{ and } \vec{z}$ are perpendicular, determine the angle between vectors $\vec{m} \text{ and } \vec{n}$.
## Solution.

Since the scalar product of orthogonal vectors is equal to zero, according to the conditions of the problem, we have:

$$
\begin{aligned}
& 0=\vec{u} \cdot \vec{v}=(\vec{m}+3 \vec{n}) \cdot(7 \vec{m}-5 \vec{n}) \\
& 0=\vec{w} \cdot \vec{z}=(\vec{m}-4 \vec{n}) \cdot(7 \vec{m}-2 \vec{n})
\end{aligned}
$$

By applying the properties of the scalar product, we get:

$$
\begin{aligned}
& 0=\vec{u} \cdot \vec{v}=(\vec{m}+3 \vec{n}) \cdot(7 \vec{m}-5 \vec{n})=7|\vec{m}|^{2}+16 \vec{m} \cdot \vec{n}-15|\vec{n}|^{2} \\
& 0=\vec{w} \cdot \vec{z}=(\vec{m}-4 \vec{n}) \cdot(7 \vec{m}-2 \vec{n})=7|\vec{m}|^{2}-30 \vec{m} \cdot \vec{n}+8|\vec{n}|^{2}
\end{aligned}
$$

From equation $(*)$, we have

$$
\vec{m} \cdot \vec{n}=\frac{15|\vec{n}|^{2}-7|\vec{m}|^{2}}{16}
$$

and from equation $(* *)$, we get

$$
\vec{m} \cdot \vec{n}=\frac{8|\vec{n}|^{2}+7|\vec{m}|^{2}}{30}
$$

By equating the previous two equations and simplifying the resulting expression, we obtain

$$
30\left(15|\vec{n}|^{2}-7|\vec{m}|^{2}\right)=16\left(8|\vec{n}|^{2}+7|\vec{m}|^{2}\right)
$$

thus we conclude that $|\vec{m}|=|\vec{n}|$.

Substituting $|\vec{m}|=|\vec{n}|$ into one of the equations $(*)$ or $(* *)$, we get $2 \vec{m} \cdot \vec{n}=|\vec{m}|^{2}$, from which, by applying the definition of the scalar product, we conclude that $\cos \varangle(\vec{m}, \vec{n})=\frac{1}{2}$.

From $\cos \varangle(\vec{m}, \vec{n})=\frac{1}{2}$, we conclude that $\varangle(\vec{m}, \vec{n})=\frac{\pi}{3}$, or $\varangle(\vec{m}, \vec{n})=60^{\circ}$.