If a⃗ is a unit vector perpendicular to b⃗ and (a⃗ + 2b⃗) · (3a⃗ - b⃗) = -5, find |b⃗|.

As per question $|\vec{a}| = 1$, $\vec{a} \cdot \vec{b} = 0$ $\therefore (\vec{a} + 2\vec{b}) \cdot (3\vec{a} - \vec{b}) = -5$ $\Rightarrow 3|\vec{a}|^2 + 5(\vec{a} \cdot \vec{b}) - 2|\vec{b}|^2 = -5$ $\Rightarrow 2|\vec{b}|^2 = 8 \Rightarrow |\vec{b}| = 2$