If A = <5 ,-1 ,3 >A=<5,−1,3>, B = <2 ,-2 ,5 >B=<2,−2,5> and C=A-BC=A−B, what is the angle between A and C?

1 Answer
Narad T.
Feb 14, 2017

The angle is =68.8=68.8º

Explanation:

Let's start by calculating

vecC=vecA-vecB→C=→A−→B

vecC=〈5,-1,3〉-〈2,-2,5〉=〈3,1,-2〉→C=⟨5,−1,3⟩−⟨2,−2,5⟩=⟨3,1,−2⟩

The angle between vecA→A and vecC→C is given by the dot product definition.

vecA.vecC=∥vecA∥*∥vecC∥costheta→A.→C=∥→A∥⋅∥→C∥cosθ

Where thetaθ is the angle between vecA→A and vecC→C

The dot product is

vecA.vecC=〈5,-1,3〉.〈3,1,-2〉=15-1-6=8→A.→C=⟨5,−1,3⟩.⟨3,1,−2⟩=15−1−6=8

The modulus of vecA→A= ∥〈5,-1,3〉∥=sqrt(25+1+9)=sqrt35∥∥⟨5,−1,3⟩∥=√25+1+9=√35

The modulus of vecC→C= ∥〈3,1,-2〉∥=sqrt(9+1+4)=sqrt14∥∥⟨3,1,−2⟩∥=√9+1+4=√14

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=8/(sqrt35*sqrt14)=0.36cosθ=→A.→C∥∥∥→A∥⋅∥→C∥∥∥=8√35⋅√14=0.36

theta=68.8θ=68.8º

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