What is the angle between $<-7,2,-3 >$ and $< 1,-5,1 >$ ?

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Answer 1 · 2018-02-26T12:13:23

The angle between $vecA and vecB$ is $119.26^0$

$vecA =<-7,2, -3> and vecB =<1,-5,1> ; theta$ be the

angle between them ; then we know

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

$=((-7.1)+(2* -5)+(-3*1))/(sqrt((-7)^2+2^2+(-3)^2)* (sqrt(1^2+(-5)^2+1^2))$

$= -20/(sqrt62*sqrt27)=-20/40.915 ~~ -0.489$

$:. theta=cos^-1(-0.489)~~119.26^0$

The angle between $vecA and vecB$ is $119.26^0$ [Ans]

What is the angle between <-7,2,-3 > <-7,2,-3 > and < 1,-5,1 >< 1,-5,1 >?