If a unit vector a makes an angle (π)/3 with imath , (π)/4 with jmath…

The correct answer is Option (3) → \( \displaystyle\frac{\pi}{3} \) and \( \displaystyle\frac{\hat{\imath}}{2}+\frac{\hat{\jmath}}{\sqrt{2}}+\frac{\hat{k}}{2} \)

The components of a unit vector are its direction cosines, and their squares add up to 1:

    \( \displaystyle \cos^2\frac{\pi}{3}+\cos^2\frac{\pi}{4}+\cos^2\theta=1 \)

    \( \displaystyle \frac{1}{4}+\frac{1}{2}+\cos^2\theta=1 \)

    \( \displaystyle \ \Rightarrow\ \cos^2\theta=\frac{1}{4} \)

θ is acute, so \( \cos\theta=\frac{1}{2} \) and \( \theta=\frac{\pi}{3} \).

    \( \displaystyle \vec{a}=\cos\frac{\pi}{3}\,\hat{\imath}+\cos\frac{\pi}{4}\,\hat{\jmath}+\cos\frac{\pi}{3}\,\hat{k} \)

    \( \displaystyle =\frac{\hat{\imath}}{2}+\frac{\hat{\jmath}}{\sqrt{2}}+\frac{\hat{k}}{2} \)

Why the other options are incorrect

• Option (1) \( \displaystyle\frac{\pi}{3} \) and \( \displaystyle\frac{\sqrt{3}}{2}\hat{\imath}+\frac{\hat{\jmath}}{\sqrt{2}}+\frac{\sqrt{3}}{2}\hat{k} \) — Incorrect. The \( \hat{\imath} \) component must be \( \cos\frac{\pi}{3}=\frac{1}{2} \), not \( \frac{\sqrt{3}}{2} \) (which is the sine); this vector is not even a unit vector.

• Option (2) \( \displaystyle\frac{\pi}{6} \) and \( \displaystyle\frac{\sqrt{3}}{2}\hat{\imath}+\frac{\hat{\jmath}}{\sqrt{2}}+\frac{\sqrt{3}}{2}\hat{k} \) — Incorrect. The same wrong components as Option (1), and \( \cos\frac{\pi}{6}=\frac{\sqrt{3}}{2} \) would make the squares add to more than 1.

• Option (4) \( \displaystyle\frac{\pi}{6} \) and \( \displaystyle\frac{\hat{\imath}}{2}+\frac{\hat{\jmath}}{\sqrt{2}}+\frac{\hat{k}}{2} \) — Incorrect. The vector is right, but its \( \hat{k} \) component \( \frac{1}{2} \) corresponds to \( \theta=\frac{\pi}{3} \), not \( \frac{\pi}{6} \).

NCERT (2026-27 edition):
Vector (or Cross) Product of Two Vectors