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

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Answer 1 · 2018-08-02T06:41:08

#theta=cos^-1(-71/(sqrt8463))#

Let , #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#