If $A = < 6, - 7, 5 >$ $A = <6 ,-7 ,5 >$ , $B = < 1, - 3, - 8 >$ $B = <1 ,-3 ,-8 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

Question

1352 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-08T15:49:06

The angle is $=36º$

Let's start by calculating

$vecC=vecA-vecB$

$vecC=〈6,-7,5〉-〈1,-3,-8〉=〈5,-4,13〉$

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

$vecA.vecC=〈6,-7,5〉.〈5,-4,13〉=30+28+65=123$

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

The modulus of $vecC$ = $∥〈5,-4,13〉∥=sqrt(25+16+169)=sqrt210$

So,

$costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=123/(sqrt110*sqrt210)=0.81$

$theta=36$ º

If A = <6 ,-7 ,5 >A=<6,−7,5>A = <6 ,-7 ,5 >, B = <1 ,-3 ,-8 >B=<1,−3,−8>B = <1 ,-3 ,-8 > and C=A-BC=A−BC=A-B, what is the angle between A and C?