# How do you find the two unit vectors in R^2 parallel to the line y=3x+4?

Feb 24, 2017

Reqd. vectors are $\left(\pm \frac{1}{\sqrt{10}} , \pm \frac{3}{\sqrt{10}}\right) .$

#### Explanation:

Note that the slope the given line is $3.$ So, it is not vertical.

Suppose that, this line makes an angle of $\theta$ with the $+ v e$

direction of the $X -$Axis, where, $\theta \in \left(0 , \pi\right) - \left\{\frac{\pi}{2}\right\} .$

Clearly, then, the Unit vector $\vec{u}$ parallel to the line is given

by, $\vec{u} = \left(\cos \theta , \sin \theta\right) .$

Now, by the Defn. of Slope, we have,

$\tan \theta = 3 , \theta \in \left(0 , \pi\right) - \left\{\frac{\pi}{2}\right\} .$

$\text{But, } \tan \theta > 0 \Rightarrow 0 < \theta < \frac{\pi}{2.}$

${\sec}^{2} \theta = 1 + {\tan}^{2} \theta = 1 + 9 = 10 \Rightarrow \sec \theta = \pm \sqrt{10.}$

$\theta \in \left(0 , \frac{\pi}{2}\right) \Rightarrow \cos \theta = \frac{1}{\sec} \theta = + \frac{1}{\sqrt{10.}}$

Also, $\sin \theta = \tan \theta \sec \theta = + \frac{3}{\sqrt{10.}}$

Hence, $\vec{u} = \left(\frac{1}{\sqrt{10}} , \frac{3}{\sqrt{10}}\right) .$

The other vector parallel to the line is $- \vec{u} .$

Enjoy Maths.!