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

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Answer 1 · 2016-03-06T18:04:57

#23^@#

An easy way to find the angle #theta# between any 2 vectors #A and B# in #RR^3# is from the Euclidean inner product :

#costheta=(A*B)/(||A||||B||)#

#therefore theta= cos^(-1)(((-1,-7,2)*(-3,-5,1))/(||(-1,-7,2)||||(-3,-5,1)))#

#=cos^(-1)((3+35+2)/(sqrt(1+49+4) sqrt(9+25+1)))#

#=23^@#.

(An alternative would be to use the vector cross product but this method would take longer : #sintheta=(AxxB)/(||A||||B||)#.