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: Understand the given conditions.
The vectors \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors, meaning \( |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \). The provided 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 \( \frac{\pi}{6} \), therefore:\n\[\n\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \left( \frac{\pi}{6} \right) = 1 \cdot 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}.\n\]
Step 2: Analyze the geometric implications.
Since \( \vec{a} \) is perpendicular to both \( \vec{b} \) and \( \vec{c} \), \( \vec{a} \) is perpendicular to the plane defined by \( \vec{b} \) and \( \vec{c} \). The cross product \( \vec{b} \times \vec{c} \) is also perpendicular to both \( \vec{b} \) and \( \vec{c} \), implying \( \vec{a} \) is parallel to \( \vec{b} \times \vec{c} \). We hypothesize:\n\[\n\vec{a} = k (\vec{b} \times \vec{c}),\n\] where \( k \) is a scalar. We need to determine \( k \) because \( \vec{a} \) is a unit vector.
Step 3: Calculate the magnitude of \( \vec{b} \times \vec{c} \).
The magnitude of the cross product \( \vec{b} \times \vec{c} \) is:\n\[\n|\vec{b} \times \vec{c}| = |\vec{b}| |\vec{c}| \sin \left( \frac{\pi}{6} \right) = 1 \cdot 1 \cdot \sin \left( \frac{\pi}{6} \right) = \sin \left( \frac{\pi}{6} \right) = \frac{1}{2}.\n\] If \( \vec{a} = k (\vec{b} \times \vec{c}) \), then:\n\[\n|\vec{a}| = |k| |\vec{b} \times \vec{c}| = |k| \cdot \frac{1}{2}.\n\] Since \( |\vec{a}| = 1 \):\n\[\n|k| \cdot \frac{1}{2} = 1 \implies |k| = 2 \implies k = \pm 2.\n\] Thus:\n\[\n\vec{a} = \pm 2 (\vec{b} \times \vec{c}).\n\]
Step 4: Verify the result.
Check orthogonality: \( (\vec{b} \times \vec{c}) \cdot \vec{b} = 0 \) and \( (\vec{b} \times \vec{c}) \cdot \vec{c} = 0 \), hence \( \vec{a} = \pm 2 (\vec{b} \times \vec{c}) \) 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 calculated previously.
Option (D) \( \pm 2 (\vec{b} \times \vec{c}) \) is the correct answer.