How do you find the angle?

Question

#theta# lies between #i-sqrt3k# and 2#i+k-3j#

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Answer 1 · 2018-07-07T17:07:17

#:. theta=arccos[{sqrt14(2-sqrt3)}/28]#.

Let, #vecu=i-sqrt3k, and, vecv=2i+k-3j, i.e., #

#vecu=(1,0,-sqrt3), and, vecv=(2,-3,1)#.

We know that, #vecu*vecv=||vecu||*||vecv||costheta#.

#:. (1,0,-sqrt3)*(2,-3,1)=(sqrt{1^2+0^2+(-sqrt3)^2})(sqrt(4+9+1))costheta#.

#:. (1)(2)+(0)(-3)+(-sqrt3)(1)=(sqrt4)(sqrt14)costheta#.

# :. 2-sqrt3=2sqrt14costheta#.

#:. costheta=(2-sqrt3)/(2sqrt14)={sqrt14(2-sqrt3)}/28#.

#:. theta=arccos[{sqrt14(2-sqrt3)}/28]#.