How do you find the angel between u=27i+14j, v=i-7j?

Dec 13, 2016

71°(3sf)

Explanation:

The angle $\theta$ between two vectors $\vec{A}$ and $\vec{B}$ is related to the modulus (or magnitude) and scaler (or dot) product of $\vec{A}$ and $\vec{A}$ by the relationship:

$\vec{A} \cdot \vec{B} = | A | | B | \cos \theta$

By convention when we refer to the angle between vectors we choose the acute angle.

So for this problem, let the angle betwen $\vec{u}$ and $\vec{v}$ be $\theta$ then:

$\vec{u} = 27 \underline{\hat{i}} + 14 \underline{\hat{j}} = \left(\begin{matrix}27 \\ 14\end{matrix}\right)$ and $\vec{v} = 1 \underline{\hat{i}} - 7 \underline{\hat{j}} = \left(\begin{matrix}1 \\ - 7\end{matrix}\right)$

The modulus is given by;

$| \vec{u} | = | \left(\begin{matrix}27 \\ 14\end{matrix}\right) | \setminus \setminus = \sqrt{{27}^{2} + {14}^{2}} = \sqrt{729 + 196} = \sqrt{925}$
$| \vec{v} | = | \left(\begin{matrix}1 \\ - 7\end{matrix}\right) | = \sqrt{{1}^{2} + - {7}^{2}} = \sqrt{1 + 49} = \sqrt{50}$

And the scaler product is:

$\vec{u} \cdot \vec{v} = \left(\begin{matrix}27 \\ 14\end{matrix}\right) \cdot \left(\begin{matrix}1 \\ - 7\end{matrix}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(27\right) \left(1\right) + \left(14\right) \left(- 7\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 27 - 98$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = - 71$

And so using $\vec{A} \cdot \vec{B} = | A | | B | \cos \theta$ we have:

$- 71 = \sqrt{925} \cdot \sqrt{50} \cdot \cos \theta$
$\therefore \cos \theta = \frac{- 71}{\sqrt{46250}}$
$\therefore \cos \theta = - 0.33014 \ldots$
 :. theta = 109.277 °
 :. theta = 109 °(3sf)

So the acute angle between the vectors is 71°(3sf)