The angle between the vectors \(\hat{i}-\hat{j}\)and \(\hat{j}-\hat{k}\) is:

\(\frac{2π}{3}\)

Let \(\vec{a}=\hat{i}-\hat{j}\) and \(\vec{b}=\hat{j}-\hat{k}\)

\(\vec{a}.\vec{b}=|\vec{a}||\vec{b}|cosθ⇒(\hat{i}-\hat{j}).(\hat{j}-\hat{k})=\sqrt{1+1}\sqrt{1+1}cosθ\)

$cosθ=\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}=\frac{-1}{\sqrt{2}\sqrt{2}}=\frac{-1}{2}⇒θ=\frac{2π}{3}$

Option 3 is correct.