# What is  || < -3, -3, 0 > || ?

$3 \sqrt{2}$

#### Explanation:

Anything enclosed in $< >$ denote a vector which has direction componential magnitudes (i.e. -3 units on the x axis, -3 units on the y-axis, and 0 units at the z-axis.)

Furthermore, anything enclosed in $\left\lVert < > \right\rVert$ denotes the norm or the 'overall' magnitude of the vector brought about by its respective components.

For example, in 2D, $| | < 2 , 3 > | |$ is the magnitude brought about the magnitude in both the x and y axes. In this case, it follows the Pythagorean Theorem where the magnitude of the vector is $\setminus \sqrt{4 + 9}$ and is consequently the norm. Its direction is $\arctan \left(\frac{y}{x}\right)$ in degrees.

In your case, a 3D vector, finding the norm is just analogous to the Pythagorean Theorem and that is, $m a g n i t u \mathrm{de} = \sqrt{\left({x}^{2} + {y}^{2} + {z}^{2}\right)}$ or also $\sqrt{\left({\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2} + {\left({z}_{2} - {z}_{1}\right)}^{2}\right)}$ for two given vectors of the same order.

So,
$\left\lVert < - 3 , - 3 , 0 > \right\rVert = \setminus \sqrt{{\left(- 3\right)}^{2} + {\left(- 3\right)}^{2} + {\left(0\right)}^{2}} = 3 \setminus \sqrt{2}$