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

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Answer 1 · 2016-10-23T21:38:44

#theta ~~ 0.76 radians#

Let #barA = <1, 5, 9>#
Let #barB = <-2,3,2>#

The #barA*barB# is:

#barA*barB = (1)(-2) + (5)(3) + (9)(2) = 31#

#|barA| = sqrt(1^2 + 5^2 + 9^2) = sqrt(107)#

#|barB| = sqrt((-2)^2 + 3^2 + 2^2) = sqrt(17)#

Use #barA*barB = |barA||barB|cos(theta)#

Solve for #theta#

#theta = cos^-1((barA*barB)/(|barA||barB|))#

#theta = cos^-1((31)/(sqrt(107)sqrt(17)))#

#theta ~~ 0.76 radians#