Let A,B and C be unit vectors. Suppose `A*B=A*C=0` and the angle betweenn B and C is `(pi)/(4)`. Then,

Options

A. `A=+-2(BxxC)`
B. `A=+-sqrt(2)(BxxC)`
C. `A=+-3(B+C)`
D. `A=+-sqrt(3)(BxxC)`.

Answer

To solve the problem step-by-step, we start with the information given:

1. **Understanding the Given Information**:
- A, B, and C are unit vectors, which means:
\[ |A| = |B| = |C| = 1 \]
- The dot products are given as:
\[ A \cdot B = 0 \quad \text{and} \quad A \cdot C = 0 \]
- The angle between vectors B and C is:
\[ \theta = \frac{\pi}{4} \]

2. **Interpreting the Dot Products**:
- Since \(A \cdot B = 0\), vector A is perpendicular to vector B.
- Since \(A \cdot C = 0\), vector A is also perpendicular to vector C.
- Therefore, we can conclude that vector A is mutually perpendicular to both B and C.

3. **Using the Cross Product**:
- If a vector A is perpendicular to two other vectors (B and C), we can express A in terms of the cross product of B and C:
\[ A = k (B \times C) \]
where \(k\) is a scalar.

4. **Finding the Magnitude of the Cross Product**:
- The magnitude of the cross product \(B \times C\) can be calculated using the formula:
\[ |B \times C| = |B| |C| \sin(\theta) \]
- Since both B and C are unit vectors, their magnitudes are 1:
\[ |B \times C| = 1 \cdot 1 \cdot \sin\left(\frac{\pi}{4}\right) \]
- The sine of \(\frac{\pi}{4}\) is \(\frac{1}{\sqrt{2}}\):
\[ |B \times C| = \frac{1}{\sqrt{2}} \]

5. **Substituting Back to Find A**:
- Now substituting the magnitude of the cross product back into the equation for A:
\[ A = k (B \times C) \]
- Since A is a unit vector, we set the magnitude of A to 1:
\[ |A| = |k| \cdot |B \times C| = 1 \]
- Therefore, we have:
\[ |k| \cdot \frac{1}{\sqrt{2}} = 1 \implies |k| = \sqrt{2} \]
- Thus, we can express A as:
\[ A = \pm \sqrt{2} (B \times C) \]

6. **Final Result**:
- The final expression for vector A is:
\[ A = \pm \sqrt{2} (B \times C) \]