# What is the angle between the vectors 2bb(ul hat i)+2bb(ul hat j)+bb(ul hat k) and 3bb(ul hat i)+4bb( ul hat j)?

Sep 3, 2017

$\theta = {0.37}^{\text{c}}$

#### Explanation:

An angle between two vectors $\underline{u}$ and $\underline{v}$ can be found by the following formula

$\theta = \arccos \left(\frac{\underline{u} \cdot \underline{v}}{\left\mid \underline{u} \right\mid \left\mid \underline{v} \right\mid}\right)$

Where $\underline{u} \cdot \underline{v}$ is the scalar product of the vectors and $\left\mid \underline{u} \right\mid$ is the magnitude of $\underline{u}$.

Let $\underline{u} = \left(\begin{matrix}2 \\ 2 \\ 1\end{matrix}\right)$ and $\underline{v} = \left(\begin{matrix}3 \\ 4 \\ 0\end{matrix}\right)$

$\underline{u} \cdot \underline{v} = 2 \cdot 3 + 2 \cdot 4 + 1 \cdot 0 = 14$

$\left\mid \underline{u} \right\mid \left\mid \underline{v} \right\mid = \sqrt{{2}^{2} + {2}^{2} + {1}^{1}} \sqrt{{3}^{2} + {4}^{2}} = 15$

$\therefore \theta = \arccos \left(\frac{14}{15}\right) = {0.37}^{\text{c}}$

Sep 3, 2017

${21.0}^{o}$ to 1dp

#### Explanation:

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

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

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

So for this problem, let the angle between $\boldsymbol{\underline{u}}$ and $\boldsymbol{\underline{v}}$ be $\theta$ and let:

$\boldsymbol{\underline{u}} = 2 \boldsymbol{\underline{\hat{i}}} + 2 \boldsymbol{\underline{\hat{j}}} + \boldsymbol{\underline{\hat{k}}}$
$\boldsymbol{\underline{v}} = 3 \boldsymbol{\underline{\hat{i}}} + 4 \boldsymbol{\underline{\hat{j}}}$

The vector norm is given by;

$| | \boldsymbol{\underline{u}} | | = | | 2 \boldsymbol{\underline{\hat{i}}} + 2 \boldsymbol{\underline{\hat{j}}} + \boldsymbol{\underline{\hat{k}}} | | = \sqrt{{2}^{2} + {2}^{2} + {1}^{2}} = 3$
$| | \boldsymbol{\underline{v}} | | = | | 3 \boldsymbol{\underline{\hat{i}}} + 4 \boldsymbol{\underline{\hat{j}}} | | \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \sqrt{{3}^{2} + {4}^{2}} \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 5$

And the scaler (or "dot") product is:

$\boldsymbol{\underline{u} \cdot \underline{v}} = \left(2 \boldsymbol{\underline{\hat{i}}} + 2 \boldsymbol{\underline{\hat{j}}} + \boldsymbol{\underline{\hat{k}}}\right) \cdot \left(3 \boldsymbol{\underline{\hat{i}}} + 4 \boldsymbol{\underline{\hat{j}}}\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(2\right) \left(3\right) + \left(2\right) \left(4\right) + \left(1\right) \left(0\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 6 + 8 + 0$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = 14$

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

$14 = 3 \cdot 5 \cdot \cos \theta$
$\therefore \cos \theta = \frac{14}{15}$
$\implies \theta = 21.03946 {\ldots}^{o}$

So the acute angle between the vectors is:

${\theta}_{\text{acute}} = 21.03946 {\ldots}^{o}$