# How do you normalize  (2i + 3j – 7k) ?

Feb 27, 2016

#### Answer:

$\hat{v}$ = $\left(\frac{2}{\sqrt{62}}\right) \hat{i} + \left(\frac{3}{\sqrt{62}}\right) \hat{j} - \left(\frac{7}{\sqrt{62}}\right) \hat{k}$

#### Explanation:

Normalizing vector v=(2i+3j–7k) means finding a unit vector in the direction given by vector $v$.

For a vector $u = a \hat{i} + b \hat{j} + c \hat{k}$, this is represented by $\hat{u}$, and is equal to $\frac{u}{|} u |$,

where $| u |$ is absolute value of $u$ and is given by $\sqrt{{a}^{2} + {b}^{2} + {c}^{2}}$.

Hence in te given case

hat v=(2i+3j–7k)/sqrt(2^2+3^3+(-7)^2) or

(2i+3j–7k)/sqrt62 i.e.

$\left(\frac{2}{\sqrt{62}}\right) \hat{i} + \left(\frac{3}{\sqrt{62}}\right) \hat{j} - \left(\frac{7}{\sqrt{62}}\right) \hat{k}$