# What is the unit vector of this vector v = 2i - j + k?

Dec 16, 2016

$u = < \frac{2}{\sqrt{6}} , - \frac{1}{\sqrt{6}} , \frac{1}{\sqrt{6}} >$.

#### Explanation:

To determine the unit vector, divide the given vector by its magnitude.

The magnitude of the vector is given by $\sqrt{{\left(i\right)}^{2} + {\left(j\right)}^{2} + {\left(k\right)}^{2}}$, where $i , j$, and $k$ are those components of the vector.

For $v = 2 i - j + k$, equivalent to $v = < 2 , - 1 , 1 >$, the magnitude is given by

$\sqrt{{\left(2\right)}^{2} + {\left(- 1\right)}^{2} + {\left(1\right)}^{2}} = \sqrt{6}$.

Thus, the unit vector is found by

$\frac{< 2 , - 1 , 1 >}{\sqrt{6}}$

Equivalently, $u = < \frac{2}{\sqrt{6}} , - \frac{1}{\sqrt{6}} , \frac{1}{\sqrt{6}} >$.