Two vectors of magnitudes $3$ and $4$ are perpendicular to each other. The magnitude of their resultant is:

For two perpendicular vectors, the magnitude of the resultant is $\sqrt{a^2 + b^2}$ (Pythagoras, since $\cos 90° = 0$ kills the cross-term in the general formula $\sqrt{a^2 + b^2 + 2ab\cos\theta}$). $|\vec{R}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$. Nice 3-4-5 right triangle. Same arithmetic as our Pythagoras q1002.