If A = <5 ,2 ,-5 >A=<5,2,−5>, B = <6 ,5 ,9 >B=<6,5,9> and C=A-BC=A−B, what is the angle between A and C?

1 Answer
Narad T.
Jul 6, 2017

The angle is =56=56º

Explanation:

Let's start by calculating

vecC=vecA-vecB→C=→A−→B

vecC=〈5,2,-5〉-〈6,5,9〉=〈-1,-3,-14〉→C=⟨5,2,−5⟩−⟨6,5,9⟩=⟨−1,−3,−14⟩

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=〈5,2,-5〉.〈-1,-3,-14〉=-5-6+70=59→A.→C=⟨5,2,−5⟩.⟨−1,−3,−14⟩=−5−6+70=59

The modulus of vecA→A= ∥〈5,2,-5〉∥=sqrt(25+4+25)=sqrt54∥∥⟨5,2,−5⟩∥=√25+4+25=√54

The modulus of vecC→C= ∥〈-1,-3,-14〉∥=sqrt(1+9+196)=sqrt206∥∥⟨−1,−3,−14⟩∥=√1+9+196=√206

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=59/(sqrt54*sqrt206)=0.56cosθ=→A.→C∥∥∥→A∥⋅∥→C∥∥∥=59√54⋅√206=0.56

theta=56θ=56º

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