Question

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 \(\frac{\pi}{6}\), then

• A. \(\vec{a} = \vec{b} \times \vec{c}\)
• B. \(\vec{a} = \pm\left(\vec{b} \times \vec{c}\right)\)
• C. \(\vec{a} = 2\left(\vec{b} \times \vec{c}\right)\)
• D. \(\vec{a} = \pm 2\left(\vec{b} \times \vec{c}\right)\)

Hint:

\(\vec{a}\) is parallel to \(\vec{b}\times\vec{c}\), and \(|\vec{b}\times\vec{c}|=\sin\frac{\pi}{6}\).

Answer 1

The Correct Option is D

Step 1: Direction of a.
Since \(\vec{a} \cdot \vec{b} = 0\) and \(\vec{a} \cdot \vec{c} = 0\), \(\vec{a}\) is perpendicular to both \(\vec{b}\) and \(\vec{c}\). A vector perpendicular to both is parallel to \(\vec{b} \times \vec{c}\). So \(\vec{a} = \lambda(\vec{b} \times \vec{c})\) for some number \(\lambda\).

Step 2: Length of b x c.
\(|\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}|\sin\frac{\pi}{6} = 1 \times 1 \times \frac{1}{2} = \frac{1}{2}\).

Step 3: Find lambda.
\(\vec{a}\) is a unit vector, so \(1 = |\lambda| \times \frac{1}{2}\), which gives \(|\lambda| = 2\). So \(\lambda = \pm 2\).

Step 4: Check the options.
Option 1 and option 2 would need \(|\vec{b} \times \vec{c}| = 1\), which is false here. Option 3 misses the negative direction, because \(\vec{a}\) could point either way. Option 4 gives both directions with the right size.

Final Answer:
\(\vec{a} = \pm 2(\vec{b} \times \vec{c})\), option 4. \[ \boxed{\vec{a} = \pm 2(\vec{b} \times \vec{c})} \]