# What is the norm of <1,-3,-2 >?

Apr 1, 2016

Let $\vec{v} = \left\langle1 , - 3 , - 2\right\rangle$. The norm is written as

$\setminus m a t h b f \left(| | \vec{v} | | = \sqrt{\vec{v} \cdot \vec{v}}\right)$

$= \sqrt{\left\langle1 , - 3 , - 2\right\rangle \cdot \left\langle1 , - 3 , - 2\right\rangle}$

$= \sqrt{1 \cdot 1 + \left(- 3\right) \cdot \left(- 3\right) + \left(- 2\right) \cdot \left(- 2\right)}$

$= \sqrt{{1}^{2} + {\left(- 3\right)}^{2} + {\left(- 2\right)}^{2}}$

$= \sqrt{1 + 9 + 4}$

$= \textcolor{b l u e}{\sqrt{13}}$

This tells us that the vector has a length of $\sqrt{13} \approx 3.61$. In fact, this is really a generalization of the Pythagorean Theorem in ${\mathbb{R}}^{3}$.

Given that information, can you find the norm of $\vec{w} = \left\langle1 , 3 , - 2 , 7\right\rangle$?