Angle between unit vectors

Question

I need to find the angle between two unit vectors $\overrightarrow{m}$ and $\overrightarrow{n}$ if the vectors $\overrightarrow{p}=\overrightarrow{m}+2\overrightarrow{n}$ and $\overrightarrow{q}=5\overrightarrow{m}-4\overrightarrow{n}$ are perpendicular to each other.

I have no idea how to approach the problem. All I can think of from the presented information is that the angle between $\overrightarrow{p}$ and $\overrightarrow{q}$ is $\frac{\pi}{2}$ and that the angle between the two unit vectors is $cos(\overrightarrow{m}.\overrightarrow{n})$

Answer 1

Calculating the dot product we get $$\vec{p}\cdot \vec{q}=5\vec{m}^2+10\vec{m}\cdot \vec{n}-4\vec{m}\cdot \vec{n}-8\vec{n}^2=0$$ We get $$\vec{m}\cdot \vec{n}=\frac{1}{2}$$

Answer 2

Since $p, q$ are perpendicular and $m, n$ have unit length, $$ 0 = \langle p, q \rangle = \langle m + 2n, 5m - 4n \rangle = 5|m|^2-4\langle m,n\rangle + 10 \langle n, m\rangle - 8|n|^2 = -3 + 6 \langle m , n\rangle. $$ This, combined with the fact that $$ \cos \alpha = \frac{\langle m, n \rangle}{|m||n|} = \langle m, n \rangle $$ should allow you to find the required angle $\alpha$.