If A = <8 ,1 ,-5 >A=<8,1,−5>, B = <6 ,-2 ,4 >B=<6,−2,4>, and C=A-BC=A−B, what is the angle between A and C?

1 Answer
Narad T.
Jan 5, 2017

The answer is =45.9=45.9º

Explanation:

Let's start by calculating

vecC=vecA-vecB→C=→A−→B

vecC=〈8,1,-5〉-〈6,-2,4〉=〈2,3,-9〉→C=⟨8,1,−5⟩−⟨6,−2,4⟩=⟨2,3,−9⟩

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=〈8,1,-5〉.〈2,3,-9〉=16+3+45=64→A.→C=⟨8,1,−5⟩.⟨2,3,−9⟩=16+3+45=64

The modulus of vecA→A= ∥〈8,1,-5〉∥=sqrt(64+1+25)=sqrt90∥∥⟨8,1,−5⟩∥=√64+1+25=√90

The modulus of vecC→C= ∥〈2,3,-9〉∥=sqrt(4+9+81)=sqrt94∥∥⟨2,3,−9⟩∥=√4+9+81=√94

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=64/(sqrt90*sqrt94)=0.696cosθ=→A.→C∥∥∥→A∥⋅∥→C∥∥∥=64√90⋅√94=0.696

theta=45.9θ=45.9º

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