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

Question

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

Answer 1

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

Answer 2

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...