# What is the norm of < -5 , -2, 4 >?

$\sqrt{45}$
We're basically calculating the norm of the vector $O A = \left(- 5 , - 2 , 4\right)$. So we apply the formula for any $M \left(x , y , z\right) \in {\mathbb{R}}^{3} : | | O M | | = \sqrt{{x}^{2} + {y}^{2} + {z}^{2}}$.
Here, $| | O A | | = \sqrt{{\left(- 5\right)}^{2} + {\left(- 2\right)}^{2} + {4}^{2}} = \sqrt{25 + 4 + 16} = \sqrt{45}$