# How do you find a vector of length 1 in the same direction as (6,-8)?

Aug 29, 2016

The reqd. vector is $\left(\frac{6}{10} , - \frac{8}{10}\right)$.

#### Explanation:

The Reqd. Vector, say $\vec{x}$ is in the same direction as $\left(6 , - 8\right)$, we must have,

$\vec{x} = k \left(6 , - 8\right) = \left(6 k , - 8 k\right) ,$ where, $k \in {\mathbb{R}}^{+}$.

As the second requirement for $\vec{x}$ is $| | \vec{x} | | = 1$, we have,

$36 {k}^{2} + 64 {k}^{2} = 1$, i.e., $100 {k}^{2} = 1$, or, $k = \pm \frac{1}{10}$, but, as, $k > 0$,

$k = + \frac{1}{10}$.

Therefore, the reqd. vector is $\left(\frac{6}{10} , - \frac{8}{10}\right)$.

Enjoy Maths.!