If ` a xx b = c, b xx c= a` and a,b,c be moduli of the vector a, b,c respectively , then

Options

A. `a =1, b = c`
B. `c = 1, a =1`
C. `b =2, c =2a`
D. `b =1, c=a`

Answer

To solve the problem step by step, we start with the given vector equations:

1. **Given Equations**:
- \( \mathbf{a} \times \mathbf{b} = \mathbf{c} \) (Equation 1)
- \( \mathbf{b} \times \mathbf{c} = \mathbf{a} \) (Equation 2)

2. **Substituting Equation 2 into Equation 1**:
From Equation 2, we can express \( \mathbf{a} \) in terms of \( \mathbf{b} \) and \( \mathbf{c} \):
\[ \mathbf{a} = \mathbf{b} \times \mathbf{c} \]
Now, substitute this expression for \( \mathbf{a} \) into Equation 1:
\[ (\mathbf{b} \times \mathbf{c}) \times \mathbf{b} = \mathbf{c} \]

3. **Using the Vector Triple Product Identity**:
We can simplify the left-hand side using the vector triple product identity:
\[ \mathbf{x} \times (\mathbf{y} \times \mathbf{z}) = (\mathbf{x} \cdot \mathbf{z}) \mathbf{y} - (\mathbf{x} \cdot \mathbf{y}) \mathbf{z} \]
Here, let \( \mathbf{x} = \mathbf{b} \), \( \mathbf{y} = \mathbf{c} \), and \( \mathbf{z} = \mathbf{b} \):
\[ (\mathbf{b} \cdot \mathbf{b}) \mathbf{c} - (\mathbf{b} \cdot \mathbf{c}) \mathbf{b} = \mathbf{c} \]

4. **Setting Up the Equation**:
This gives us the equation:
\[ |\mathbf{b}|^2 \mathbf{c} - (\mathbf{b} \cdot \mathbf{c}) \mathbf{b} = \mathbf{c} \]
Rearranging yields:
\[ |\mathbf{b}|^2 \mathbf{c} - \mathbf{c} = (\mathbf{b} \cdot \mathbf{c}) \mathbf{b} \]
\[ (|\mathbf{b}|^2 - 1) \mathbf{c} = (\mathbf{b} \cdot \mathbf{c}) \mathbf{b} \]

5. **Analyzing the Result**:
For the equation to hold, we can analyze two cases:
- If \( |\mathbf{b}|^2 - 1 = 0 \), then \( |\mathbf{b}| = 1 \).
- If \( |\mathbf{b}|^2 - 1 \neq 0 \), then \( \mathbf{c} \) must be a scalar multiple of \( \mathbf{b} \).

6. **Finding the Magnitudes**:
From the first case, we have:
\[ |\mathbf{b}| = 1 \]
Now substituting \( |\mathbf{b}| = 1 \) back into the original equations, we can find the magnitudes of \( \mathbf{a} \) and \( \mathbf{c} \).

7. **Using the Dot Product**:
Since \( \mathbf{b} \cdot \mathbf{c} = 0 \) (from the rearranged equation), this implies that \( \mathbf{b} \) and \( \mathbf{c} \) are orthogonal. Thus, we have:
\[ |\mathbf{c}| = |\mathbf{a}| \]

8. **Conclusion**:
Therefore, we conclude:
- \( |\mathbf{b}| = 1 \)
- \( |\mathbf{c}| = |\mathbf{a}| \)

Thus, the final results are:
- \( |\mathbf{b}| = 1 \)
- \( |\mathbf{c}| = |\mathbf{a}| \)