cross product two vectors - |a ̅×b ̅ |=|c ̅ |=|a ̅ ||b| sinθ
coordinates of the vector с ̅={2,0,5}
magnitude of vector |c ̅ |=√(2^2+0^2+5^2 )=√29
Then, √29=3∙3 sin〖θ→θ=arcsin √29/9≈0.63 rad≈36 degree〗
angle between a ̅ and b ̅ = 36 degree
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