How do you find the angle between the vectors #u=2i-j# and #v=6i+4j#?

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Answer 1 · 2016-10-20T10:10:47

Angle between given vectors is #55.55^o#

If we have two vectors #vecu=2hati-hatj# and #vecv=6hati+4hatj#

and angle between them is #theta#, then

#costheta=(vecu*vecv)/(|u|*|v|)#

= #(2xx6+(-1)xx4)/(sqrt(2^2+(-1)^2)xxsqrt(6^2+4^2))#

= #(12-4)/(sqrt5xxsqrt40)#

= #8/sqrt200=8/sqrt(5xx5xx8)#

= #8/(5sqrt8)#

= #sqrt8/5=0.5657#

Hence #theta=55.55^o#