# What is the angle between <-3,5,6>  and <-1,9,-3>?

Mar 5, 2016

$\cos \theta = \frac{< - 3 , 5 , 6 > \cdot < - 1 , 9 , - 3 >}{\left(\sqrt{{\left(- 3\right)}^{2} + {5}^{2} + {6}^{2}}\right) \sqrt{{\left(- 1\right)}^{2} + {9}^{2} + {\left(- 3\right)}^{2}}}$
$\cos \theta = \frac{\left(- 3\right) \left(- 1\right) + \left(5\right) \left(9\right) + \left(6\right) \left(- 3\right)}{\left(\sqrt{70}\right) \left(\sqrt{91}\right)} = \frac{30}{\sqrt{6370}} \to \theta = {\cos}^{-} 1 \left(\frac{30}{\sqrt{6370}}\right) = {67.92}^{\circ}$

#### Explanation:

Find the dot product of the vectors then divide by the magnitude of each. Finally take the cosine inverse to find $\theta$

Mar 5, 2016

Use the definition of a dot product to find that:

$\theta = {67.1}^{o}$ or $1.17$ radians

#### Explanation:

The dot product is a very convenient tool that allows you to find the angle between any two vectors. It is given by:

$\vec{a} \cdot \vec{b} = | \setminus \vec{a} | \cdot | \setminus \vec{b} | \cdot \cos \theta$

Where $\vec{a} \cdot \vec{b}$ represents the dot product between $\vec{a}$ and $\vec{b}$, given by:

$\vec{a} \cdot \vec{b} = \left({a}_{1} \cdot {b}_{1}\right) + \left({a}_{2} \cdot {b}_{2}\right) + \left({a}_{3} \cdot {b}_{3}\right) + \ldots$

Note: The dot product will always result in a scalar value.

$| \vec{a} |$ and $| \vec{b} |$ are the magnitudes of $\vec{a}$ and $\vec{b}$, respectively. They are also scalar values, calculated by using:

$| \vec{a} | = {\left({a}_{1}^{2} + {a}_{2}^{2} + {a}_{3}^{2} + \ldots\right)}^{\frac{1}{2}}$

With this information, the angle between the vectors <−3,5,6> and <−1,9,−3> can be found:

Step 1: Calculate the dot product
$\vec{a} \cdot \vec{b} = \left({a}_{1} \cdot {b}_{1}\right) + \left({a}_{2} \cdot {b}_{2}\right) + \left({a}_{3} \cdot {b}_{3}\right) + \ldots$
$\vec{a} \cdot \vec{b} = \left(- 3 \cdot - 1\right) + \left(5 \cdot 9\right) + \left(6 \cdot - 3\right)$
$\vec{a} \cdot \vec{b} = 3 + 45 + \left(- 18\right)$
$\vec{a} \cdot \vec{b} = 30$

Step 2: Calculate the magnitude of each vector
$| \vec{a} | = {\left({a}_{1}^{2} + {a}_{2}^{2} + {a}_{3}^{2} + \ldots\right)}^{\frac{1}{2}}$
$| \vec{a} | = {\left({\left(- 3\right)}^{2} + {\left(5\right)}^{2} + {\left(6\right)}^{2}\right)}^{\frac{1}{2}}$
$| \vec{a} | = {\left(9 + 25 + 36\right)}^{\frac{1}{2}}$
$| \vec{a} | = {\left(70\right)}^{\frac{1}{2}}$
$| \vec{a} | = 8.367$

Following the same procedure for $| \vec{b} |$ gives:
$| \vec{b} | = 9.220$

$| \vec{a} | \cdot | \vec{b} |$ then is: $8.367 \cdot 9.220 = 77.143$

Step 3: Solve for $\theta$:

$\vec{a} \cdot \vec{b} = | \setminus \vec{a} | \cdot | \setminus \vec{b} | \cdot \cos \theta$
$30 = 77.143 \cdot \cos \theta$
$\cos \theta = \frac{30}{77.143}$
$\cos \theta = 0.389$
$\theta = {\cos}^{-} 1 \left(0.389\right)$
$\theta = {67.1}^{o}$ or $1.17$ radians