If A = <2 ,-5 ,9 >A=<2,−5,9>, B = <-9 ,1 ,-5 >B=<−9,1,−5> and C=A-BC=A−B, what is the angle between A and C?

1 Answer
Narad T.
Dec 13, 2017

The angle is =25.4^@=25.4∘

Explanation:

Let's start by calculating

vecC=vecA-vecB→C=→A−→B

vecC=〈2,-5,9〉-〈-9,1,-5> = <11,-6,14>→C=⟨2,−5,9⟩−⟨−9,1,−5>=<11,−6,14>
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=〈2,-5,9〉.〈11,-6,14〉=22+30+126=178→A.→C=⟨2,−5,9⟩.⟨11,−6,14⟩=22+30+126=178

The modulus of vecA→A= ∥〈2,-5,9〉∥=sqrt(4+25+81)=sqrt110∥∥⟨2,−5,9⟩∥=√4+25+81=√110

The modulus of vecC→C= ∥〈11,-6,14〉 ∥=sqrt(121+36+196)∥∥⟨11,−6,14⟩∥=√121+36+196

=sqrt353=√353

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=178/(sqrt110*sqrt353)=0.903cosθ=→A.→C∥∥∥→A∥⋅∥→C∥∥∥=178√110⋅√353=0.903

theta=25.4^@θ=25.4∘

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