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:

For unit vectors, use the cross product to find a vector perpendicular to two given vectors, and adjust the scalar to match the unit vector condition. The magnitude of \( \vec{b} \times \vec{c} \) depends on the angle between \( \vec{b} \) and \( \vec{c} \).

Answer 1

The Correct Option is D

Step 1: Interpret the given conditions.
We have \( \vec{a}, \vec{b}, \vec{c} \) as unit vectors, so \( |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \). The conditions are:
\( \vec{a} \cdot \vec{b} = 0 \), so \( \vec{a} \) is perpendicular to \( \vec{b} \).
\( \vec{a} \cdot \vec{c} = 0 \), so \( \vec{a} \) is perpendicular to \( \vec{c} \).
The angle between \( \vec{b} \) and \( \vec{c} \) is \( \frac{\pi}{6} \), so: \[ \vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \left( \frac{\pi}{6} \right) = 1 \cdot 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}. \]
Step 2: Analyze the geometric implications.
Since \( \vec{a} \) is perpendicular to both \( \vec{b} \) and \( \vec{c} \), \( \vec{a} \) must be perpendicular to the plane spanned by \( \vec{b} \) and \( \vec{c} \). The cross product \( \vec{b} \times \vec{c} \) is a vector perpendicular to both \( \vec{b} \) and \( \vec{c} \), so \( \vec{a} \) must be parallel to \( \vec{b} \times \vec{c} \). Thus, we hypothesize: \[ \vec{a} = k (\vec{b} \times \vec{c}), \] where \( k \) is a scalar, and since \( \vec{a} \) is a unit vector, we need to determine \( k \).
Step 3: Compute the magnitude of \( \vec{b} \times \vec{c} \).
The magnitude of the cross product \( \vec{b} \times \vec{c} \) is: \[ |\vec{b} \times \vec{c}| = |\vec{b}| |\vec{c}| \sin \left( \frac{\pi}{6} \right) = 1 \cdot 1 \cdot \sin \left( \frac{\pi}{6} \right) = \sin \left( \frac{\pi}{6} \right) = \frac{1}{2}. \] If \( \vec{a} = k (\vec{b} \times \vec{c}) \), then: \[ |\vec{a}| = |k| |\vec{b} \times \vec{c}| = |k| \cdot \frac{1}{2}. \] Since \( |\vec{a}| = 1 \): \[ |k| \cdot \frac{1}{2} = 1 \implies |k| = 2 \implies k = \pm 2. \] Thus: \[ \vec{a} = \pm 2 (\vec{b} \times \vec{c}). \]
Step 4: Verify the result.
Check orthogonality: \( (\vec{b} \times \vec{c}) \cdot \vec{b} = 0 \) and \( (\vec{b} \times \vec{c}) \cdot \vec{c} = 0 \), so \( \vec{a} = \pm 2 (\vec{b} \times \vec{c}) \) satisfies \( \vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0 \).
Check the unit vector condition: The magnitude of \( \vec{a} = \pm 2 (\vec{b} \times \vec{c}) \) is 1, as computed above.
Option (D) \( \pm 2 (\vec{b} \times \vec{c}) \) matches exactly.