If $A = < 2, - 8, 5 >$ $A = <2 ,-8 ,5 >$ , $B = < 5, - 7, - 5 >$ $B = <5 ,-7 ,-5 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2017-06-19T05:47:59

The angle is $=59.1º$

Let's start by calculating

$vecC=vecA-vecB$

$vecC=〈2,-8,5〉-〈5,-7,-5〉=〈-3,-1,10〉$

The angle between $vecA$ and $vecC$ is given by the dot product definition.

$vecA.vecC=∥vecA∥*∥vecC∥costheta$

Where $theta$ is the angle between $vecA$ and $vecC$

The dot product is

$vecA.vecC=〈2,-8,5〉.〈-3,-1,10〉=-6+8+50=52$

The modulus of $vecA$ = $∥〈2,-8,5〉∥=sqrt(4+64+25)=sqrt93$

The modulus of $vecC$ = $∥〈-3,-1,10〉∥=sqrt(9+1+100)=sqrt110$

So,

$costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=52/(sqrt93*sqrt110)=0.51$

$theta=59.1$ º

If A = <2 ,-8 ,5 >A=<2,−8,5>A = <2 ,-8 ,5 >, B = <5 ,-7 ,-5 >B=<5,−7,−5>B = <5 ,-7 ,-5 > and C=A-BC=A−BC=A-B, what is the angle between A and C?