Which of the following is a unit vector for all real values of \(\theta\) and \(\phi\)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\sin\theta\cos\phi\,\hat{\imath}+\sin\theta\sin\phi\,\hat{\jmath}+\cos\theta\,\hat{k}\)
A unit vector has magnitude 1, so the squares of its components must add up to 1 for every \(\theta\) and \(\phi\):
\( \sin^2\theta\cos^2\phi+\sin^2\theta\sin^2\phi+\cos^2\theta \)
\( = \sin^2\theta\left(\cos^2\phi+\sin^2\phi\right)+\cos^2\theta \)
\( = \sin^2\theta+\cos^2\theta = 1 \)
Why the other options are incorrect
Option (2) \(\cos\theta\,\hat{\imath}+\sin\theta\,\hat{\jmath}+\hat{k}\) — Incorrect. The squares add up to \(\cos^2\theta+\sin^2\theta+1 = 2\), so its magnitude is \(\sqrt{2}\).
Option (3) \(\sin\theta\,\hat{\imath}+\cos\theta\,\hat{\jmath}+\sin\theta\,\hat{k}\) — Incorrect. The squares add up to \(1+\sin^2\theta\), which is 2 at \(\theta = \frac{\pi}{2}\).
Option (4) \( \displaystyle \frac{1}{3}\left(\hat{\imath}+\hat{\jmath}+\hat{k}\right) \) — Incorrect. The squares add up to \(\frac{3}{9} = \frac{1}{3}\), so its magnitude is \(\frac{1}{\sqrt{3}}\).