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

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Answer 1 · 2018-03-24T12:10:21

The angle between $vecA and vecC$ is $63.23^0$

$vec C=vecA- vecB = (< 2,4,-3 >) - (< 3,4,1 >)$

$=(< (2-3),(4-4),(-3-1) >) =< -1,0,-4 >$

$vecA =<2,4,-3> and vecC =<-1,0,-4>$ Let $theta$ be the

angle between them ; then we know

$cos theta= (vecA*vecC)/(||vecA||*||vecC||)$

$=(2* (-1)+(4*0)+(-3*(-4)))/(sqrt(2^2+4^2+(-3)^2)* (sqrt((-1)^2+0^2+(-4)^2))$

$= 10/(sqrt29*sqrt17)=10/22.2 ~~ 0.4504$

$:. theta=cos^-1(0.4504)~~63.23^0$

The angle between $vecA and vecC$ is $63.23^0$ [Ans]

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