# If A = <6 ,4 ,9 >, B = <7 ,-1 ,5 > and C=A-B, what is the angle between A and C?

May 17, 2018

The angle is $= {42.9}^{\circ}$

#### Explanation:

Start by calculating

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

vecC=〈6,4,9〉-〈7,-1,5〉=〈-1,5,4〉

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=〈6,4,9〉.〈-1,5,4〉=-6+20+36=50

The modulus of $\vec{A}$= ∥〈6,4,9〉∥=sqrt(36+16+81)=sqrt133

The modulus of $\vec{C}$= ∥〈-1,5,4〉∥=sqrt(1+25+9)=sqrt35

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=50/(sqrt133*sqrt35)=0.7328

$\theta = \arccos \left(0.7328\right) = {42.9}^{\circ}$