We have
\(\vec A.\vec B\)= | A | | B | cos θ
Here | A | = \(\sqrt{1^2+1^2+(-2)^2}=\sqrt6\)
| B | =\(\sqrt{1^2+1^2+(-1)^2}=\sqrt3\)
\(\vec A.\vec B=(\hat i+\hat j-\hat k).(\hat i+\hat j-\hat k)\)
= 1 × 1 + 1 × 1 + (-2)(-1)
= 1 + 1 + 2 = 4
∴ cos θ = \(\frac{\vec A.\vec B}{|A||B|}=\frac{4}{\sqrt6\sqrt3}\)
cos θ = \(\frac{4}{\sqrt{18}}\)