Question

Let \( \vec{a}, \vec{b}, \vec{c} \) be unit vectors. Suppose \( \vec{a}\cdot\vec{b} = \vec{a}\cdot\vec{c} = 0 \) and the angle between \( \vec{b} \) and \( \vec{c} \) is \( \frac{\pi}{6} \). Then \( \vec{a} \) is:

• A. \( \vec{b} \times \vec{c} \)
• B. \( \vec{c} \times \vec{b} \)
• C. \( \vec{b} + \vec{c} \)
• D. \( \pm 2(\vec{b} \times \vec{c}) \)

Hint:

If a vector is perpendicular to two vectors: \begin{itemize} \item It is parallel to their cross product. \item Normalize using magnitude. \end{itemize}

Answer 1

The Correct Option is D

Concept: Given: \[ \vec{a}\cdot\vec{b} = 0, \quad \vec{a}\cdot\vec{c} = 0 \] So \( \vec{a} \) is perpendicular to both \( \vec{b} \) and \( \vec{c} \). Thus: \[ \vec{a} \parallel (\vec{b} \times \vec{c}) \] Step 1: {\color{red}Use magnitude condition.} Since all are unit vectors, find magnitude of \( \vec{b} \times \vec{c} \): \[ |\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}|\sin\theta = \sin\frac{\pi}{6} = \frac{1}{2} \] Step 2: {\color{red}Scale to unit vector.} To make magnitude 1: \[ \vec{a} = \pm \frac{\vec{b} \times \vec{c}}{1/2} = \pm 2(\vec{b} \times \vec{c}) \] Step 3: {\color{red}Conclusion.} \[ \vec{a} = \pm 2(\vec{b} \times \vec{c}) \]