If ` a xx b = c, b xx c = a` and a, b,c be moduli of the vectors 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 equations and analyze them systematically.

### Step 1: Write down the given equations
We have two equations:
1. \( \mathbf{a} \times \mathbf{b} = \mathbf{c} \) (Equation 1)
2. \( \mathbf{b} \times \mathbf{c} = \mathbf{a} \) (Equation 2)

### Step 2: Substitute 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, we substitute this expression for \( \mathbf{a} \) into Equation 1:
\[ (\mathbf{b} \times \mathbf{c}) \times \mathbf{b} = \mathbf{c} \]

### Step 3: Use the vector triple product identity
We can use the vector triple product identity, which states:
\[ \mathbf{x} \times (\mathbf{y} \times \mathbf{z}) = (\mathbf{x} \cdot \mathbf{z}) \mathbf{y} - (\mathbf{x} \cdot \mathbf{y}) \mathbf{z} \]
Applying this to our equation:
\[ (\mathbf{b} \times \mathbf{c}) \times \mathbf{b} = (\mathbf{b} \cdot \mathbf{b}) \mathbf{c} - (\mathbf{b} \cdot \mathbf{c}) \mathbf{b} \]
Thus, we rewrite our equation:
\[ (\mathbf{b} \cdot \mathbf{b}) \mathbf{c} - (\mathbf{b} \cdot \mathbf{c}) \mathbf{b} = \mathbf{c} \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ (\mathbf{b} \cdot \mathbf{b}) \mathbf{c} - \mathbf{c} = (\mathbf{b} \cdot \mathbf{c}) \mathbf{b} \]
Factoring out \( \mathbf{c} \):
\[ (\mathbf{b} \cdot \mathbf{b} - 1) \mathbf{c} = (\mathbf{b} \cdot \mathbf{c}) \mathbf{b} \]

### Step 5: Analyze the coefficients
For the above equation to hold true for all vectors, we can compare coefficients. This leads us to two conditions:
1. \( \mathbf{b} \cdot \mathbf{b} - 1 = 0 \) implies \( \mathbf{b} \cdot \mathbf{b} = 1 \) (which means the magnitude of \( \mathbf{b} \) is 1).
2. \( \mathbf{b} \cdot \mathbf{c} = 0 \) (which means \( \mathbf{b} \) and \( \mathbf{c} \) are orthogonal).

### Step 6: Conclusion about magnitudes
Since \( \mathbf{b} \) is a unit vector, we have:
\[ |\mathbf{b}| = 1 \]
Now, we need to consider the implications for \( \mathbf{a} \) and \( \mathbf{c} \). From Equation 1, since \( \mathbf{a} \times \mathbf{b} = \mathbf{c} \), and knowing that \( \mathbf{b} \) is a unit vector, we can conclude that for the cross product to yield a vector \( \mathbf{c} \), the magnitudes of \( \mathbf{a} \) and \( \mathbf{c} \) must be equal:
\[ |\mathbf{a}| = |\mathbf{c}| \]

### Final Result
Thus, we conclude:
- The magnitude of \( \mathbf{b} \) is \( 1 \).
- The magnitudes of \( \mathbf{a} \) and \( \mathbf{c} \) are equal.