Angle between 2 vectors.Sorry for asking so many "am i right" question, in a corespondance course needthehelp

Have no other way of checking my work, i have no teacher and appriciate your guys/gals help so much.





3 points





A = (0,0,1) B = (1,0,0) C = (1,1,1)





between the two vectors BA and BC





I have 60 degrees or 1/3 pi

Angle between 2 vectors.Sorry for asking so many "am i right" question, in a corespondance course needthehelp
you need to take the dot product between the vectors and then divide by their norms


cos theta = BA . BC /|BA| |BC|


=(-1,0,1)(0,1,1) /sqrt(2)sqrt(2)


= 1/2


cos(theta) = 1/2


theta = pi/3
Reply:Use dot product:


BA*BC = |AB|*|BC|*cos(theta)


(-1,0,1)*(0,1,1) = sqrt(2)*sqrt(2)*cos(theta)


0+0+1 = ..


1 = 2*cos(theta)


cos(theta) = 1/2


theta = 60 deg or pi/3





I agree with your answer...