If A = <2 ,4 ,-9 >A=<2,4,−9>, B = <-1 ,8 ,-5 >B=<−1,8,−5> and C=A-BC=A−B, what is the angle between A and C?

1 Answer
Narad T.
Mar 7, 2017

The angle is =66.2=66.2º

Explanation:

Let's start by calculating

vecC=vecA-vecB→C=→A−→B

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

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=〈2,4,-9〉.〈3,-4,-4〉=6-16+36=26→A.→C=⟨2,4,−9⟩.⟨3,−4,−4⟩=6−16+36=26

The modulus of vecA→A= ∥〈2,4,-9〉∥=sqrt(4+16+81)=sqrt101∥∥⟨2,4,−9⟩∥=√4+16+81=√101

The modulus of vecC→C= ∥〈3,-4,-4〉∥=sqrt(9+16+16)=sqrt41∥∥⟨3,−4,−4⟩∥=√9+16+16=√41

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=26/(sqrt101*sqrt41)=0.404cosθ=→A.→C∥∥∥→A∥⋅∥→C∥∥∥=26√101⋅√41=0.404

theta=66.2θ=66.2º

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