What is the angle between #<0,-7,6> # and #<-1,6,-6>#?

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Answer 1 · 2018-08-02T09:25:29

# theta=cos^-1(-78/(sqrt6205))#

Let , #veca=<0,-7,6> and vecb=<-1,6,-6># be the two vectors.

#:.|veca|=sqrt((0)^2+(-7)^2+(6)^2)=sqrt(0+49+36)=sqrt85#

#and|vecb|=sqrt((-1)^2+(6)^2+(-6)^2)=sqrt(1+36+36)#=#sqrt73#

Dot product : #a*b=<0,-7,6>*<-1,6,-6>#

#:.a*b=(0)xx(-1)+(-7)xx(6)+(6)xx(-6)#

#:.a*b=0-42-36=-78#

Now the angle between #veca and vecb# is:

#theta=cos^-1((a*b)/(|veca||vecb|))=cos^-1((-78)/(sqrt85sqrt73))#

#:. theta=cos^-1(-78/(sqrt6205))=##(171.97)^circ#