If \(\vec a,\vec b,\vec c\) are unit vectors such that \(\vec a \cdot \vec b =\vec a \cdot \vec c = 0\), and the angle between \(\vec b\) and \(\vec c\) is \(\displaystyle\frac{\pi}{6}\), then
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\vec a = \pm 2(\vec b\times \vec c)\)
Since \(\vec a \cdot \vec b = 0\) and \(\vec a \cdot \vec c = 0\), \(\vec a\) is perpendicular to both \(\vec b\) and \(\vec c\). The vectors \(\vec b\) and \(\vec c\) are not parallel (the angle between them is \(\frac{\pi}{6}\)), so \(\vec a\) is parallel to \(\vec b \times \vec c\).
The magnitude of the cross product:
\( \displaystyle |\vec b \times \vec c| = |\vec b|\,|\vec c|\sin\frac{\pi}{6} \)
\( \displaystyle = 1 \times 1 \times \frac{1}{2} = \frac{1}{2} \)
\(\vec a\) is a unit vector along \(\pm(\vec b \times \vec c)\):
\( \displaystyle \vec a = \pm\frac{\vec b \times \vec c}{|\vec b \times \vec c|} = \pm\frac{\vec b \times \vec c}{1/2} \)
\(\vec a = \pm 2(\vec b \times \vec c)\)
Why the other options are incorrect
Option (1) \(\vec b = \pm(\vec a\times \vec c)\) — Incorrect. \(\vec a \times \vec c\) is perpendicular to \(\vec c\), but \(\vec b \cdot \vec c = \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2} \ne 0\).
Option (2) \(\vec a = \pm(\vec b\times \vec c)\) — Incorrect. \(|\vec b \times \vec c| = \frac{1}{2}\), so this vector has magnitude \(\frac{1}{2}\), not 1. It would hold only if \(\vec b\) and \(\vec c\) were perpendicular.
Option (4) \(\vec c = \pm 2(\vec a\times \vec b)\) — Incorrect. \(\vec a \times \vec b\) is perpendicular to \(\vec b\), but \(\vec c \cdot \vec b = \frac{\sqrt{3}}{2} \ne 0\).