If $A = <6 ,-7 ,5 >$ , $B = <1 ,-5 ,-8 >$ and $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2017-08-02T04:23:50

The angle is $=42.4^@$

Let's start by calculating

$vecC=vecA-vecB$

$vecC=〈6,-7,5〉-〈1,-5,-8〉=〈5,-2,13〉$

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

$vecA.vecC=〈6,-7,5〉.〈5,-2,13〉=30+14+65=109$

The modulus of $vecA$ = $∥〈6,-7,5〉∥=sqrt(36+49+25)=sqrt110$

The modulus of $vecC$ = $∥〈5,-2,13〉∥=sqrt(25+4+169)=sqrt198$

So,

$costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=109/(sqrt110*sqrt198)=0.74$

$theta=42.4^@$

If A = <6 ,-7 ,5 >A = <6 ,-7 ,5 >, B = <1 ,-5 ,-8 >B = <1 ,-5 ,-8 > and C=A-BC=A-B, what is the angle between A and C?