# Find |u| ? U=<2,-3>

Apr 11, 2018

$| \setminus u |$ asks for the length of the vector.

#### Explanation:

In other words, you just need to use pythagoras theorem.

$| \setminus u | = \sqrt{{\left(2\right)}^{2} + {\left(- 3\right)}^{2}}$

$| \setminus u | = \sqrt{13}$

Apr 11, 2018

$| u | = \sqrt{13}$

#### Explanation:

$\text{given "ulu=< x,y>" then the magnitude is}$

$| \underline{u} | = \sqrt{{x}^{2} + {y}^{2}}$

$\text{here "x=2" and } y = - 3$

$\Rightarrow | \underline{u} | = \sqrt{{2}^{2} + {\left(- 3\right)}^{2}} = \sqrt{13}$