# What is the angle between <-3,2,6 >  and <6,1,-5> ?

May 4, 2016

I got $\text{2.56 rad}$ or about ${146.6}^{\circ}$.

Let $\vec{a} = \left\langle- 3 , 2 , 6\right\rangle$ and $\vec{b} = \left\langle6 , 1 , - 5\right\rangle$. This angle between $\vec{a}$ and $\vec{b}$ is found by the following equation:

$\setminus m a t h b f \left(\vec{a} \cdot \vec{b} = | | \vec{a} | | \cdot | | \vec{b} | | \cos \theta\right)$

where:

• $\vec{a} \cdot \vec{b}$ is the dot product between $\vec{a}$ and $\vec{b}$.
• $| | \vec{a} | |$ is the norm of $\vec{a}$.
• $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

The dot product in ${\mathbb{R}}^{n}$ for some vectors $\vec{a}$ and $\vec{b}$ is defined as follows:

$\setminus m a t h b f \left(\vec{a} \cdot \vec{b}\right)$

$= \left\langle{a}_{1} , {a}_{2} , . . . , {a}_{n}\right\rangle \cdot \left\langle{b}_{1} , {b}_{2} , . . . , {b}_{n}\right\rangle$

$= \setminus m a t h b f \left({a}_{1} {b}_{1} + {a}_{2} {b}_{2} + . . . + {a}_{n} {b}_{n}\right)$

And the norm in ${\mathbb{R}}^{n}$ for some vector $\vec{u}$ is defined as:

$\setminus m a t h b f \left(| | \vec{a} | |\right)$

= sqrt(vecacdotveca

$= \sqrt{{a}_{1} {a}_{1} + {a}_{2} {a}_{2} + . . . + {a}_{n} {a}_{n}}$

$= \setminus m a t h b f \left(\sqrt{{a}_{1}^{2} + {a}_{2}^{2} + . . . + {a}_{n}^{2}}\right)$

So, we evaluate the angle as follows:

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

$\textcolor{g r e e n}{\theta = \arccos \left(\frac{\vec{a} \cdot \vec{b}}{| | \vec{a} | | \cdot | | \vec{b} | |}\right)}$

First, we should find the dot product between $\vec{a}$ and $\vec{b}$.

$\textcolor{g r e e n}{\vec{a} \cdot \vec{b}}$

$= \left\langle- 3 , 2 , 6\right\rangle \cdot \left\langle6 , 1 , - 5\right\rangle$

$= - 3 \cdot 6 + 2 \cdot 1 + 6 \cdot - 5$

$= - 18 + 2 - 30$

$= \textcolor{g r e e n}{- 46}$

Then, we should find each norm.

$\textcolor{g r e e n}{| | \vec{a} | |}$

$= \sqrt{\vec{a} \cdot \vec{a}}$

$= \sqrt{\left\langle- 3 , 2 , 6\right\rangle \cdot \left\langle- 3 , 2 , 6\right\rangle}$

$= \sqrt{- 3 \cdot - 3 + 2 \cdot 2 + 6 \cdot 6}$

$= \sqrt{9 + 4 + 36}$

$= \textcolor{g r e e n}{7}$

$\textcolor{g r e e n}{| | \vec{b} | |}$

$= \sqrt{\vec{b} \cdot \vec{b}}$

$= \sqrt{\left\langle6 , 1 , - 5\right\rangle \cdot \left\langle6 , 1 , - 5\right\rangle}$

$= \sqrt{6 \cdot 6 + 1 \cdot 1 + - 5 \cdot - 5}$

$= \sqrt{36 + 1 + 25}$

$= \textcolor{g r e e n}{\sqrt{62}}$

Finally, let's get the angle.

$\textcolor{b l u e}{\theta} = \arccos \left(\frac{\vec{a} \cdot \vec{b}}{| | \vec{a} | | \cdot | | \vec{b} | |}\right)$

$= \arccos \left(\frac{- 46}{7 \sqrt{62}}\right)$

$\approx$ $\textcolor{b l u e}{\text{2.56 rad}}$ or about $\textcolor{b l u e}{{146.6}^{\circ}}$.

And you can see that from using Wolfram Alpha on this question as well.