# What is the norm of < 7 , 5, 1 >?

Feb 14, 2016

The 'norm' of a vector is a unit vector in the same direction. To find it we 'normalize' the vector. In this case, the vector is $< \frac{7}{\sqrt{75}} , \frac{5}{\sqrt{75}} , \frac{1}{\sqrt{75}} >$ or $< \frac{7}{8.66} , \frac{5}{8.66} , \frac{1}{8.66} >$.

#### Explanation:

To normalize a vector, we divide each of its elements by the length of the vector. To find the length:

$l = \sqrt{{7}^{2} + {5}^{2} + {1}^{2}} = \sqrt{49 + 25 + 1} = \sqrt{75} = 8.66$

The norm can be expressed two ways, depending on your preference (and the preferences of the person who will mark your work):

$< \frac{7}{\sqrt{75}} , \frac{5}{\sqrt{75}} , \frac{1}{\sqrt{75}} >$ or $< \frac{7}{8.66} , \frac{5}{8.66} , \frac{1}{8.66} >$