# How do you find two unit vectors that make an angle of 60° with v = ‹3, 4›?

Jul 31, 2016

Tthe reqd. unit vectors are, $\left(\frac{3}{10} - \frac{2}{5} \sqrt{3} , \frac{2}{5} + 3 \frac{\sqrt{3}}{10}\right)$, or,

$\left(\frac{3}{10} + 2 \frac{\sqrt{3}}{5} , \frac{2}{5} - 3 \frac{\sqrt{3}}{10}\right)$.

#### Explanation:

Let $\vec{u} = \left(x , y\right)$ be the reqd. unit vector.

$\therefore | | \vec{u} | | = 1 \Rightarrow {x}^{2} + {y}^{2} = 1. \ldots \ldots \ldots \ldots \ldots . \left(1\right)$.

Given that, Angle btwn. $\vec{u} \mathmr{and} \vec{v}$ is $\frac{\pi}{3}$, we take these vectors' Dot Product, to get,

$\vec{u} \cdot \vec{v} = | | u | | | | v | | \cos \left(\hat{\vec{u} , \vec{v}}\right)$

$\therefore \left(x , y\right) \cdot \left(3 , 4\right) = 1 \left(\sqrt{{3}^{2} + {4}^{2}}\right) \cos \left(\frac{\pi}{3}\right)$

$\therefore 3 x + 4 y = 1 \cdot 5 \cdot \frac{1}{2} = \frac{5}{2} \Rightarrow 3 x = \frac{5}{2} - 4 y$

$\Rightarrow x = \frac{1}{3} \left(\frac{5}{2} - 4 y\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(2\right)$.

Using $\left(2\right)$ in $\left(1\right)$, we get,

$\frac{1}{9} {\left(\frac{5}{2} - 4 y\right)}^{2} + {y}^{2} = 1 \Rightarrow \frac{25}{4} - 20 y + 16 {y}^{2} + 9 {y}^{2} = 9$

$\Rightarrow 25 {y}^{2} - 20 y = 9 - \frac{25}{4}$.

To make the $L . H . S .$ complete square, we add $4$ on both sides.

$\therefore 25 {y}^{2} - 20 y + 4 = 9 - \frac{25}{4} + 4$.

$\therefore {\left(5 y - 2\right)}^{2} = \frac{27}{4}$

$\therefore 5 y - 2 = \pm 3 \frac{\sqrt{3}}{2} , i . e . , 5 y = 2 \pm 3 \frac{\sqrt{3}}{2} , s o , y = \frac{2}{5} \pm 3 \frac{\sqrt{3}}{10}$

By $\left(2\right)$, then, $x = \frac{1}{3} \left\{\frac{5}{2} - 4 \left(\frac{2}{5} \pm 3 \frac{\sqrt{3}}{10}\right)\right\}$.

Thus, the reqd. unit vectors are, $\left(\frac{3}{10} - \frac{2}{5} \sqrt{3} , \frac{2}{5} + 3 \frac{\sqrt{3}}{10}\right)$, or,

$\left(\frac{3}{10} + 2 \frac{\sqrt{3}}{5} , \frac{2}{5} - 3 \frac{\sqrt{3}}{10}\right)$.

An Alternative Method to solve this problem, is, instead of starting

with $\vec{u} = \left(x , y\right)$, we may suppose that,

$\vec{u} = \left(\cos \theta , \sin \theta\right)$, where, we may, preferably restrict

$\theta \in \left[0 , \frac{\pi}{2}\right]$.