How do you find the angle between the vectors #u=-6i-3j# and #v=-8k+4j#?

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Answer 1 · 2016-09-03T17:32:07

#approx 101.537^@#

Let #theta=/_(vec u, vec v)#.

#" Then, "vec u*vec v=||vec u||||vec v||cos theta, ...............(star)#.

Here, #vec u =-6i-3j=(-6,-3,0), and, vec v=(0,4,-8)#

#:. vec u*vec v=0-12+0=-12#, and,

#||vec u||=sqrt(36+9)=sqrt45=3sqrt5, and, ||vec v||=4sqrt5#.

Hence, by #(star)#, we have,

#cos theta=-12/((3sqrt5)(4sqrt5))=-1/5#.

#:. theta=arc (cos (-0.2)=pi-arc cos (0.2)~~(180-78.463)^@#

#:. theta~~101.537^@#