Given that.
Vector A, \(A = i + 2j - k\)
Vector B, \(B = -i + j - 2k\)
Let θ is the angle between A and B. It can be calculated as:
\(A.B = |A||B| \cos\theta\)
\(|A| = \sqrt{1^2+ 2^2 + (-1)^2} = \sqrt 6\)
\(|B| = \sqrt{(-1)^2 + 1^2 + (-2)^2} = \sqrt 6\)
\(A.B = -1 + 2 + 2 =3\)
So,
\(\cos\theta =\frac 3{(\sqrt6)^2}\)
\(\cos\theta = \frac 3{(\sqrt 6)^2}\)
\(\cos\theta = \frac 12\)
\(\theta = 60°\)
So, the angle between vector A and B is 60 degrees. Hence, this is the required solution.
