Question

If \( \hat{a}, \hat{b} \) and \( \hat{c} \) are unit vectors such that \[ \hat{a} \cdot \hat{b} = \hat{a} \cdot \hat{c} = 0 \] and the angle between \( \hat{b} \) and \( \hat{c} \) is \( \frac{\pi}{6} \), then prove that: \[ \hat{a} = \pm 2 (\hat{b} \times \hat{c}) \]

Hint:

If a unit vector is perpendicular to two other vectors, it's parallel to their cross product. Use the magnitude identity \[ |\vec{a} \times \vec{b}| = |\vec{a}| \cdot |\vec{b}| \cdot \sin\theta \] to determine the scalar multiple.

Answer 1

We are given: \begin{itemize} \item \( \hat{a}, \hat{b}, \hat{c} \) are unit vectors. \item \( \hat{a} \cdot \hat{b} = 0 \) \item \( \hat{a} \cdot \hat{c} = 0 \) \item Angle between \( \hat{b} \) and \( \hat{c} \) is \( \frac{\pi}{6} \) \end{itemize} Step 1: Geometrical Meaning Since \( \hat{a} \) is perpendicular to both \( \hat{b} \) and \( \hat{c} \), it must be perpendicular to the plane containing \( \hat{b} \) and \( \hat{c} \). This implies: \[ \hat{a} \parallel (\hat{b} \times \hat{c}) \Rightarrow \hat{a} = \lambda (\hat{b} \times \hat{c}) \quad \text{for some scalar } \lambda \] Step 2: Use Magnitudes Take magnitude on both sides: \[ |\hat{a}| = |\lambda| \cdot |\hat{b} \times \hat{c}| \] Given \( \hat{a} \) is a unit vector: \[ 1 = |\lambda| \cdot |\hat{b}| \cdot |\hat{c}| \cdot \sin\theta \] Since \( \hat{b} \) and \( \hat{c} \) are unit vectors and \( \theta = \frac{\pi}{6} \), we have: \[ 1 = |\lambda| \cdot 1 \cdot 1 \cdot \sin\left(\frac{\pi}{6}\right) = |\lambda| \cdot \frac{1}{2} \Rightarrow |\lambda| = 2 \] Step 3: Final Answer \[ \lambda = \pm 2 \quad \Rightarrow \hat{a} = \pm 2 (\hat{b} \times \hat{c}) \] Hence Proved. \quad \(\boxed{\hat{a} = \pm 2 (\hat{b} \times \hat{c})}\)