# How do you normalize  (4 i + 4 j + 2 k)?

Apr 2, 2018

Module of vector is $\sqrt{{4}^{2} + {4}^{2} + {2}^{2}} = \sqrt{36} = 6$
So the normalized is $\frac{4}{6} \vec{i} + \frac{4}{6} \vec{j} + \frac{2}{6} \vec{k} = \frac{2}{3} \vec{i} + \frac{2}{3} \vec{j} + \frac{1}{3} \vec{k}$
Check the answer $\sqrt{{\left(\frac{2}{3}\right)}^{2} + {\left(\frac{2}{3}\right)}^{2} + {\left(\frac{1}{3}\right)}^{2}} = \sqrt{1} = 1$