# What is the norm or <-7 ,6,-1 >?

Jun 11, 2016

$\sqrt{86}$

#### Explanation:

Let $x = \left({x}_{1} , {x}_{2} , \ldots . , {x}_{n}\right)$ be any vector in an n-dimensional vector space X.
Then the norm of $x$ is given by

$| | x | | = \sqrt{{x}_{1}^{2} + {x}_{2}^{2} + \ldots . . + {x}_{n}^{2}}$.

So in this particular case we work in a 3-dimesnional vector space and get that

||(-7;6;1)||=sqrt((-7)^2+6^2+(-1)^2

$= \sqrt{86}$