The angle between the vectors imath-jmath and jmath-k is

The correct answer is Option (4) → \(\displaystyle\frac{2\pi}{3}\)

Let \(\vec{a}=\hat{\imath}-\hat{\jmath}\) and \(\vec{b}=\hat{\jmath}-\hat{k}\). The scalar product is

    \(\vec{a}\cdot\vec{b} = (1)(0)+(-1)(1)+(0)(-1) = -1\)

The magnitudes are

    \(|\vec{a}| = \sqrt{2}, \quad |\vec{b}| = \sqrt{2}\)

So the cosine of the angle is

    \( \displaystyle \cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|} = \frac{-1}{2} \)

The cosine is negative, so the angle is obtuse:

    \( \displaystyle \theta = \frac{2\pi}{3} \)

Why the other options are incorrect

• Option (1) \(\displaystyle\frac{\pi}{3}\) — Incorrect. \(\cos\frac{\pi}{3} = \frac{1}{2}\); it drops the minus sign of \(\vec{a}\cdot\vec{b} = -1\).

• Option (2) \(\displaystyle\frac{\pi}{6}\) — Incorrect. \(\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\), not \(-\frac{1}{2}\).

• Option (3) \(\displaystyle\frac{5\pi}{6}\) — Incorrect. \(\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}\), not \(-\frac{1}{2}\).