If \( \vec{a},\vec{b} \) and \( \vec{c} \) are three unit vectors such that \( \vec{a}\cdot\vec{b}=\vec{a}\cdot\vec{c}=0 \) and angle between \( \vec{b} \) and \( \vec{c} \) is \( \displaystyle\frac{\pi}{6} \), then
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \vec{a}=\pm 2(\vec{b}\times\vec{c}) \)
\( \vec{a} \) is perpendicular to both \( \vec{b} \) and \( \vec{c} \), so it is parallel to \( \vec{b}\times\vec{c} \):
\( \displaystyle \vec{a}=k(\vec{b}\times\vec{c}) \)
Taking magnitudes, \( |\vec{a}|=1 \) and
\( \displaystyle |\vec{b}\times\vec{c}|=|\vec{b}||\vec{c}|\sin\frac{\pi}{6} \)
\( \displaystyle =1\times 1\times\frac{1}{2} \)
\( \displaystyle =\frac{1}{2} \)
\( \displaystyle 1=|k|\cdot\frac{1}{2} \)
\( \displaystyle |k|=2 \)
\( \displaystyle k=\pm 2 \)
So \( \vec{a}=\pm 2(\vec{b}\times\vec{c}) \).
Why the other options are incorrect
Option (1) \( \vec{a}=\vec{b}\times\vec{c} \) — Incorrect. \( \vec{b}\times\vec{c} \) has magnitude \( \frac{1}{2} \), not 1, so it cannot equal the unit vector \( \vec{a} \); the sign ± is also missing.
Option (2) \( \vec{a}=\pm(\vec{b}\times\vec{c}) \) — Incorrect. The factor 2 is missing: \( \pm(\vec{b}\times\vec{c}) \) has magnitude \( \frac{1}{2} \).
Option (3) \( \vec{a}=2(\vec{b}\times\vec{c}) \) — Incorrect. \( \vec{a} \) may point either way along \( \vec{b}\times\vec{c} \); only perpendicularity is given, so both signs are possible.