How do you find a unit vector in the direction of the vector v=(-3,-4)?

Aug 6, 2016

$\left(- \frac{3}{5} , - \frac{4}{5}\right)$.

Explanation:

The reqd. vector is in the direction of $\vec{v} = \left(- 3 , - 4\right)$.

Therefore, it must be of the form

$k \left(- 3 , - 4\right) = \left(- 3 k , - 4 k\right) , w h e r e , k > 0$

As this is a unit vector, $| | \left(\left(- 3 k , - 4 k\right)\right) | | = 1$.

$\therefore \sqrt{{\left(- 3 k\right)}^{2} + {\left(- 4 k\right)}^{2}} = 1$.

$\therefore 5 k = 1 \Rightarrow k = \pm \frac{1}{5} , b u t , k > 0 \Rightarrow k \ne - \frac{1}{5.} \therefore k = \frac{1}{5}$.

Hence, the unit vector$= \left(- \frac{3}{5} , - \frac{4}{5}\right)$.