# If A = <5 ,2 ,5 >, B = <6 ,5 ,3 > and C=A-B, what is the angle between A and C?

Jul 11, 2017

The angle is =92.1º

#### Explanation:

Let's start by calculating

$\vec{C} = \vec{A} - \vec{B}$

vecC=〈5,2,5〉-〈6,5,3〉=〈-1,-3,2〉

The angle between $\vec{A}$ and $\vec{C}$ is given by the dot product definition.

vecA.vecC=∥vecA∥*∥vecC∥costheta

Where $\theta$ is the angle between $\vec{A}$ and $\vec{C}$

The dot product is

vecA.vecC=〈5,2,5〉.〈-1,-3,2〉=-5-6+10=-1

The modulus of $\vec{A}$= ∥〈5,2,5〉∥=sqrt(25+4+25)=sqrt54

The modulus of $\vec{C}$= ∥(-1,-3,2〉∥=sqrt(1+9+4)=sqrt14

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=-1/(sqrt54*sqrt14)=-0.036

$\theta = 92.1$º

Jul 11, 2017

We can start by finding $\vec{C}$. To subtract vectors, we subtract the corresponding components. For $\vec{A} - \vec{B}$, we have:

$\vec{C} = < 5 , 2 , 5 > - < 6 , 5 , 3 >$

$= < \left(5 - 6\right) , \left(2 - 5\right) , \left(5 - 3\right) >$

$= < - 1 , - 3 , 2 >$

The angle between two vectors can be found using this general formula:

$\cos \left(\theta\right) = \frac{\vec{a} \cdot \vec{b}}{| \vec{a} | \cdot | \vec{b} |}$

We can start by finding the dot product of A and C.

$\vec{A} \cdot \vec{C} = < 5 , 2 , 5 > \cdot < - 1 , - 3 , 2 >$

$= \left(5 \cdot - 1\right) + \left(2 \cdot - 3\right) + \left(5 \cdot 2\right)$

$= - 5 - 6 + 10$

$= - 1$

Now we can find the product of the magnitude of each vector.

$| \vec{A} | = | < 5 , 2 , 5 >$

$= \sqrt{{5}^{2} + {2}^{2} + {5}^{2}}$

$= \sqrt{54}$

$= 3 \sqrt{6}$

$| \vec{B} | = | < - 1 , - 3 , 2 >$

$= \sqrt{{\left(- 1\right)}^{2} + {\left(- 3\right)}^{2} + {2}^{2}}$

$= \sqrt{14}$

So the product of the magnitudes is $3 \sqrt{6} \cdot \sqrt{14} = 3 \sqrt{84} = 6 \sqrt{21}$

We now have:

$\cos \left(\theta\right) = - \frac{1}{6 \sqrt{21}} = - \frac{\sqrt{21}}{126}$

Solving for $\theta$:

$\theta = \arccos \left(- \frac{\sqrt{21}}{126}\right)$

$= {92}^{o}$