Question

Let \(\bar{a}, \bar{b}\), and \(\bar{c}\) be unit vectors. Suppose that \(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=0\) and if the angle between \(\overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\) is \(\frac{\pi}{6}\), then \(\overline{\mathrm{a}}\) is

• A. \(\pm (\overline{b}\times \overline{c})\)
• B. \(\pm \frac{1}{2}(\overset{-}{\text{ }\mathrm{b}}\times \overset{-}{\mathrm{c}})\)
• C. \(\pm 2(\overset{-}{\text{ }\mathrm{b}}\times \overset{-}{\mathrm{c}})\)
• D. \(\pm 4(\overset{-}{\text{ }\mathrm{b}}\times \overset{-}{\mathrm{c}})\)

Hint:

A vector perpendicular to both b and c is parallel to \(\bar b\times\bar c\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Since \(\bar a\cdot\bar b=\bar a\cdot\bar c=0\), \(\bar a\) is perpendicular to both, so \(\bar a=\lambda(\bar b\times\bar c)\).

Step 2: Magnitude of the Cross Product
\(|\bar b\times\bar c|=\sin\dfrac{\pi}{6}=\dfrac12\) for unit vectors.

Step 3: Use the Unit Length
\(1=|\lambda|\cdot\dfrac12\Rightarrow\lambda=\pm2\).

Step 4: Result
\(\bar a=\pm2(\bar b\times\bar c)\), which is option (C).

Final Answer:
The vector a equals plus or minus twice b cross c. \[ \boxed{\text{(C) }\pm2(\bar b\times\bar c)} \]