Let vectors a =(2,1,−4), b=(−3,0,1), and c=(−1,−1,2)
what is the angle θab between vector a and b?

The to find the angle between the two vectors, the following method can be used: first, computing the scalar product between the two vectors A and B, after, their length. the next is to use the main formula cos T = A*B / //A// //B//, where A*B is scalar product T= teta Practice: A*B = (2, 1,-4) * (-3, 0,1) =(2x-3)+(1x0)+(-4x1)=-6-4= -10 //A// = sqrt( 2² + 1² +4²) = sqrt(4+1+16) = sqrt(21)=4.58 //B// = sqrt(10)=3.16 so, cosT = -10 / 3.16*4.58 = -10/14.48 = -0.69, cosT = -0.69, therefore, T= arccos(-0.69)=46,33°

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