# What is the angle between <5,1,4 > and < 3,1,9 >?

Dec 16, 2016

 32.7 °(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 $\vec{B}$ 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:

$\vec{u} = \left\langle5 , 1 , 4\right\rangle$ and $\vec{v} = \left\langle3 , 1 , 9\right\rangle$

The modulus is given by;

$| \vec{u} | = | \left\langle5 , 1 , 4\right\rangle | = \sqrt{{5}^{2} + {1}^{2} + {4}^{2}} = \sqrt{25 + 1 + 16} = \sqrt{42}$
$| \vec{v} | = | \left\langle3 , 1 , 9\right\rangle | = \sqrt{{3}^{2} + {1}^{2} + {9}^{2}} = \sqrt{9 + 1 + 81} = \sqrt{91}$

And the scaler product is:

$\vec{u} \cdot \vec{v} = \left\langle5 , 1 , 4\right\rangle \cdot \left\langle3 , 1 , 9\right\rangle$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(5\right) \left(3\right) + \left(1\right) \left(1\right) + \left(4\right) \left(9\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 15 + 1 + 36$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 52$

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

$52 = \sqrt{42} \cdot \sqrt{91} \cdot \cos \theta$
$\therefore \cos \theta = \frac{52}{\sqrt{42} \cdot \sqrt{91}}$
$\therefore \cos \theta = 0.841120 \ldots$
 :. theta = 32.741412 °
 :. theta = 32.7 °(3sf)