# How do you use the definition of the scalar product, find the angles between the following pairs of vectors: A = - 3 i + j and B = 21 j - 3k?

Jan 8, 2017

 71.8 °(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:

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

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:

$\vec{u} = \left\langle3 , 1 , 0\right\rangle$ and $\vec{v} = \left\langle0 , 21 , - 3\right\rangle$

The modulus is given by;

$| | \vec{u} | | = | | \left\langle3 , 1 , 0\right\rangle | | = \sqrt{{3}^{2} + {1}^{2} + {0}^{2}} = \sqrt{9 + 1} = \sqrt{10}$
$| | \vec{v} | | = | | \left\langle0 , 21 , - 3\right\rangle | | = \sqrt{{0}^{2} + {21}^{2} + {\left(- 3\right)}^{2}} = \sqrt{441 + 9} = \sqrt{450}$

And the scaler product is:

$\vec{u} \cdot \vec{v} = \left\langle3 , 1 , 0\right\rangle \cdot \left\langle0 , 21 , - 3\right\rangle$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(3\right) \left(0\right) + \left(1\right) \left(21\right) + \left(0\right) \left(- 3\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 21$

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

$21 = \sqrt{10} \cdot \sqrt{450} \cdot \cos \theta$
$\therefore \cos \theta = \frac{21}{\sqrt{4500}}$
$\therefore \cos \theta = \frac{7 \sqrt{5}}{50}$
$\therefore \cos \theta = 0.31304 \ldots$
 :. theta = 71.7568 ... °
 :. theta = 71.8 °(3sf)