If A = <3 ,-1 ,-8 >, B = <5 ,6 ,-9 > and C=A-B, what is the angle between A and C?

1 Answer
Dec 24, 2017

The angle is $= {96.3}^{\circ}$

Explanation:

Let's start by calculating

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

vecC=〈3,-1,-8〉-〈5,6,-9〉=〈-2,-7,1〉

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=〈3,-1,-8〉.〈-2-7,1〉=-6+7-8=-7

The modulus of $\vec{A}$= ∥〈3,-1,-8〉∥=sqrt(9+1+64)=sqrt74

The modulus of $\vec{C}$= ∥〈-2,-7,1〉∥=sqrt(4+49+1)=sqrt54

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=-7/(sqrt74*sqrt54)=-0.11

$\theta = \arccos \left(- 0.11\right) = {96.3}^{\circ}$