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Question

Find the cosine of the angle between the vectors a=i+j+k and b=2i3j+2k

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Solution

We have
ab=abcosθ
cosθ=(abab)
ab=axbx+ayby+acbc
=(1)(2)+(1)(3)+(1)(2)=1
a=12+12+12=3
b=22+(3)2+22=17
Therefore, we have
cosθ=(1317)=0.14

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