If $A = < 2, 4, - 3 >$ $A = <2 ,4 ,-3 >$ , $B = < 3, 1, - 5 >$ $B = <3 ,1 ,-5 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2016-12-21T07:43:44

The angle is $=78.5º$

Let's calculate $vecC$

$vecC=vecA-vecB=〈2,4,-3〉-〈3,1,-5〉=〈-1,3.2〉$

The angle between $vecA$ and $vecC$ is obtained from the dot product definition

$vecA.vecC=∥vecA ∥* ∥vecC∥* costheta$

The dot product is $=〈-1,3.2〉.〈2,4,-3〉=(-2+12-6)=4$

The modulus of $vecA$ is $=∥vecA∥=∥〈2,4,-3〉∥=sqrt(4+16+9)=sqrt29$

The modulus of $vecC$ is $=∥vecc∥=∥〈-1,3,2〉∥=sqrt(1+9+4)=sqrt14$

Therefore,

$costheta=(vecA.vecC)/(∥vecA ∥* ∥vecC∥)=4/(sqrt29sqrt14)=0.1985$

$theta=78.5º$

If A = <2 ,4 ,-3 >A=<2,4,−3>A = <2 ,4 ,-3 >, B = <3 ,1 ,-5 >B=<3,1,−5>B = <3 ,1 ,-5 > and C=A-BC=A−BC=A-B, what is the angle between A and C?