If A = <8 ,0 ,4 >A=<8,0,4>, B = <6 ,5 ,4 >B=<6,5,4> and C=A-BC=A−B, what is the angle between A and C?

1 Answer
Narad T.
Feb 11, 2017

The angle is =70.6=70.6º

Explanation:

Let's start by calculating

vecC=vecA-vecB→C=→A−→B

vecC=〈8,0,4〉-〈6,5,4〉=〈2,-5,0〉→C=⟨8,0,4⟩−⟨6,5,4⟩=⟨2,−5,0⟩

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=〈8,0,4〉.〈2,-5,0〉=16+0+0=16→A.→C=⟨8,0,4⟩.⟨2,−5,0⟩=16+0+0=16

The modulus of vecA→A= ∥〈8,0,4〉∥=sqrt(64+0+16)=sqrt80∥∥⟨8,0,4⟩∥=√64+0+16=√80

The modulus of vecC→C= ∥〈2,-5,0〉∥=sqrt(4+25+0)=sqrt29∥∥⟨2,−5,0⟩∥=√4+25+0=√29

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=16/(sqrt80*sqrt29)=0.332cosθ=→A.→C∥∥∥→A∥⋅∥→C∥∥∥=16√80⋅√29=0.332

theta=70.6θ=70.6º

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