# How do you find the unit vector parallel to the resultant of the vectors A= 2i - 6j -3k and B = 4i + 3j- k?

##### 1 Answer
Aug 11, 2016

The Reqd. Unit Vector $= \left(\frac{6}{\sqrt{61}} , - \frac{3}{\sqrt{61}} , - \frac{4}{\sqrt{61}}\right)$.

#### Explanation:

Let a non-null vector $\vec{x}$ be given. Then, a unit vector parallel to $\vec{x}$ is

denoted by $\hat{x}$ and is defined by,

$\hat{x} = \frac{\vec{x}}{|} | \vec{x} | |$

vecA=2hati-6hatj-3hatk=(2,-6,-3), &, vecB=(4,3,-1).

Hence, the Resultant of $\vec{A} \mathmr{and} \vec{B}$, is $\vec{A} + \vec{B}$, & is,

$\vec{A} + \vec{B} = \left(2 , - 6 , - 3\right) + \left(4 , 3 , - 1\right) = \left(2 + 4 , - 6 + 3 , - 3 - 1\right) = \left(6 , - 3 , - 4\right)$

$\therefore | | \vec{A} + \vec{B} | | = \sqrt{{6}^{2} + {\left(- 3\right)}^{2} + {\left(- 4\right)}^{2}} = \sqrt{61}$

Hence, the Reqd. Unit Vector $= \frac{\vec{A} + \vec{B}}{|} | \vec{A} + \vec{B} | |$

$= \frac{1}{\sqrt{61}} \left(6 , - 3 , - 4\right) = \left(\frac{6}{\sqrt{61}} , - \frac{3}{\sqrt{61}} , - \frac{4}{\sqrt{61}}\right)$.