A class XII student appearing for a competitive examination was asked to attempt the following questions
Let `veca,vecb` and `vecc` be three non zero vectors.
Let `veca,vecb` and `vecc` be unit vectors such that `veca.vecb=veca.vecc=0` and angle between `vecb` and `vecc` is `pi/6` then `veca=`

Options

A. `2(vecbxxvecc)`
B. `+-2(vecbxxvecc)`
C. `pm(vecbxxvecc)`
D. `2(vecbpmvecc)`

Answer

To solve the problem, we need to find the vector \(\vec{A}\) given the conditions about the vectors \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\).

### Step-by-Step Solution:

1. **Understanding the Given Information:**
We know that \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\) are unit vectors. This means:
\[ |\vec{A}| = |\vec{B}| = |\vec{C}| = 1 \]
We also know that:
\[ \vec{A} \cdot \vec{B} = 0 \quad \text{and} \quad \vec{A} \cdot \vec{C} = 0 \]
This indicates that \(\vec{A}\) is perpendicular to both \(\vec{B}\) and \(\vec{C}\).

2. **Using the Cross Product:**
Since \(\vec{A}\) is perpendicular to both \(\vec{B}\) and \(\vec{C}\), we can express \(\vec{A}\) as the cross product of \(\vec{B}\) and \(\vec{C}\):
\[ \vec{A} = k (\vec{B} \times \vec{C}) \]
where \(k\) is a scalar to ensure that \(\vec{A}\) is a unit vector.

3. **Finding the Magnitude of the Cross Product:**
The magnitude of the cross product \(\vec{B} \times \vec{C}\) is given by:
\[ |\vec{B} \times \vec{C}| = |\vec{B}| |\vec{C}| \sin \theta \]
where \(\theta\) is the angle between \(\vec{B}\) and \(\vec{C}\). Given that the angle between \(\vec{B}\) and \(\vec{C}\) is \(\frac{\pi}{6}\), we have:
\[ |\vec{B} \times \vec{C}| = 1 \cdot 1 \cdot \sin\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

4. **Setting the Scalar \(k\):**
Since \(\vec{A}\) is a unit vector, we need:
\[ |\vec{A}| = |k| \cdot |\vec{B} \times \vec{C}| = 1 \]
Substituting the magnitude of the cross product:
\[ |k| \cdot \frac{1}{2} = 1 \implies |k| = 2 \]
Thus, \(k\) can be either \(2\) or \(-2\):
\[ \vec{A} = \pm 2 (\vec{B} \times \vec{C}) \]

5. **Final Expression for \(\vec{A}\):**
Therefore, the final expression for \(\vec{A}\) is:
\[ \vec{A} = \pm 2 (\vec{B} \times \vec{C}) \]