Find the magnitude of the vector $\mathbf{v}=3\mathbf{i}+4\mathbf{j}-12\mathbf{k}$.

Step-by-step solution

1. Compute the squares of each component: $3^2=9$, $4^2=16$, $(-12{)}^2=144$.
2. Add the squares: $9+16+144=169$.
3. Take the square root of the sum: $\sqrt{169}=13$.
4. The magnitude of $\mathbf{v}$ is $13$.

Concept tested: Magnitude of a 3‑D vector

Published 10 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked