What is the angle between $< - 1, 9, 3 >$ $<-1,9,3 >$ and $< 5, - 8, 2 >$ $< 5,-8,2 >$ ?

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Answer 1 · 2018-08-02T06:41:08

$θ = {cos}^{- 1} (- \frac{71}{\sqrt{8463}})$ $theta=cos^-1(-71/(sqrt8463))$

Let , $\to a = < - 1, 9, 3 > and \to b = < 5, - 8, 2 >$ $veca=<-1,9,3> and vecb=<5,-8,2>$ be the two vectors.

$:.|veca|=sqrt((-1)^2+(9)^2+(3)^2)=sqrt(1+81+9)=sqrt91$

$and|vecb|=sqrt((5)^2+(-8)^2+(2)^2)=sqrt(25+64+4)=sqrt93$

Dot product : $a*b=(-1)xx5+9xx(-8)+3xx2$

$:.a*b=-5-72+6=-71$

Now the angle between $veca and vecb$ is:

$theta=cos^-1((a*b)/(|veca||vecb|))=cos^-1((-71)/(sqrt91sqrt93))$

$:. theta=cos^-1(-71/(sqrt8463))=$ $(140.51)^circ$

What is the angle between <-1,9,3 > <−1,9,3><-1,9,3 > and < 5,-8,2 ><5,−8,2>< 5,-8,2 >?