What is the angle between #<6 , 3 , 1 > # and # < 3 , -9 , 8 > #?

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Answer 1 · 2016-12-18T07:11:48

The answer is #=90.008º#

The angle is given by the dot product definition

#vecu.vecv=∥vecu∥*∥vecv∥*costheta#

Here, #vecu=〈6,3,1〉#

and #vecv=〈3,-9,8〉#

The dot product is

#vecu.vecv=〈6,3,1〉.〈3,-9,8〉=(18-27+8)=-1#

The modulus of #vecu=∥〈6,3,1〉∥=sqrt(36+9+1)=sqrt46#

The modulus of #vecv=∥〈3,-9,8〉∥=sqrt(9+81+64)=sqrt154#

Then,

#costheta=(vecu.vecv)/(∥vecu∥*∥vecv∥)=-1/(sqrt46sqrt154)=-0.00014#

and #theta=90.008º#