If `vec( a) , vec( b) , vec( c ) ` are non coplanar non-zero vectors such that `vec( b) xxvec( c) = vec( a ), vec( a ) xx vec( b) =vec( c ) ` and `vec( c ) xx vec( a ) = vec ( b ) ` , then which of the following is not true

Options

A. `| vec( a) | =1`
B. `[ vec( a ) vec( b ) vec( c ) ] =1`
C. `| vec( a) | + |vec( b)| +| vec( c ) | =3`
D. `| vec( a) | cancel(=) |vec( b)| cancel(=)| vec( c ) | `

Answer

To solve the problem, we need to analyze the given vector equations and derive the necessary conclusions step by step.

### Step 1: Understand the given conditions
We have three non-coplanar non-zero vectors \( \vec{a}, \vec{b}, \vec{c} \) such that:
1. \( \vec{b} \times \vec{c} = \vec{a} \)
2. \( \vec{a} \times \vec{b} = \vec{c} \)
3. \( \vec{c} \times \vec{a} = \vec{b} \)

### Step 2: Calculate the scalar triple product
The scalar triple product \( \text{box}(\vec{a}, \vec{b}, \vec{c}) \) is defined as:
\[ \text{box}(\vec{a}, \vec{b}, \vec{c}) = \vec{a} \cdot (\vec{b} \times \vec{c}) \]
From the first condition, we know that \( \vec{b} \times \vec{c} = \vec{a} \). Thus:
\[ \text{box}(\vec{a}, \vec{b}, \vec{c}) = \vec{a} \cdot \vec{a} = |\vec{a}|^2 \]

### Step 3: Calculate other scalar triple products
Using the other conditions:
- For \( \text{box}(\vec{b}, \vec{c}, \vec{a}) \):
\[ \text{box}(\vec{b}, \vec{c}, \vec{a}) = \vec{b} \cdot (\vec{c} \times \vec{a}) = \vec{b} \cdot \vec{b} = |\vec{b}|^2 \]

- For \( \text{box}(\vec{c}, \vec{a}, \vec{b}) \):
\[ \text{box}(\vec{c}, \vec{a}, \vec{b}) = \vec{c} \cdot (\vec{a} \times \vec{b}) = \vec{c} \cdot \vec{c} = |\vec{c}|^2 \]

### Step 4: Establish equality of the scalar triple products
From the properties of scalar triple products, we have:
\[ \text{box}(\vec{a}, \vec{b}, \vec{c}) = \text{box}(\vec{b}, \vec{c}, \vec{a}) = \text{box}(\vec{c}, \vec{a}, \vec{b}) \]
This implies:
\[ |\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 \]

### Step 5: Set a common magnitude
Let \( |\vec{a}| = |\vec{b}| = |\vec{c}| = k \). Then:
\[ |\vec{a}|^2 = k^2 \]

### Step 6: Use the identity for the scalar triple product
There is an important identity:
\[ \text{box}(\vec{b} \times \vec{c}, \vec{c} \times \vec{a}, \vec{a} \times \vec{b}) = \text{box}(\vec{a}, \vec{b}, \vec{c})^2 \]
Substituting the conditions:
\[ \text{box}(\vec{a}, \vec{b}, \vec{c}) = \text{box}(\vec{a}, \vec{b}, \vec{c})^2 \]
Let \( x = \text{box}(\vec{a}, \vec{b}, \vec{c}) \), then:
\[ x = x^2 \]
This gives us:
\[ x(x - 1) = 0 \]
Thus, \( x = 0 \) or \( x = 1 \).

### Step 7: Analyze the implications
Since \( \vec{a}, \vec{b}, \vec{c} \) are non-coplanar, \( x \) cannot be 0. Therefore, we conclude:
\[ \text{box}(\vec{a}, \vec{b}, \vec{c}) = 1 \]

### Step 8: Conclusion about magnitudes
Since \( |\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1 \), we have:
\[ |\vec{a}| = |\vec{b}| = |\vec{c}| = 1 \]

### Step 9: Check the options
Now we need to check which of the statements is not true based on our findings. The options generally relate to the magnitudes and relationships of the vectors.