If a, b, c are three non-coplanar vectors such that volume of parallelopiped formed with a ,b , c as coterminous edges is equal to volume of parallelopiped formed with `a xx b, bxxc, c xxa` as coterminous edges, then :

Options

A. `[abc] = 0`
B. `[abc] =1`
C. `[abc] =-1`
D. `[abc] epsi[-1,1]`

Answer

To solve the problem, we need to analyze the given information about the volumes of the parallelepipeds formed by the vectors \( \mathbf{a}, \mathbf{b}, \mathbf{c} \) and \( \mathbf{a} \times \mathbf{b}, \mathbf{b} \times \mathbf{c}, \mathbf{c} \times \mathbf{a} \).

### Step 1: Understand the Volume of Parallelepiped
The volume \( V \) of a parallelepiped formed by three vectors \( \mathbf{u}, \mathbf{v}, \mathbf{w} \) is given by the scalar triple product:
\[ V = |\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})| \]

### Step 2: Set Up the Equation
According to the problem, we have:
\[ |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| = |\mathbf{(a \times b)} \cdot (\mathbf{(b \times c)} \times \mathbf{(c \times a)})| \]

### Step 3: Calculate the Left Side
The left side is straightforward:
\[ V_1 = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| \]

### Step 4: Calculate the Right Side
For the right side, we need to evaluate \( \mathbf{(b \times c)} \times \mathbf{(c \times a)} \). We can use the vector triple product identity:
\[ \mathbf{x} \times (\mathbf{y} \times \mathbf{z}) = (\mathbf{x} \cdot \mathbf{z}) \mathbf{y} - (\mathbf{x} \cdot \mathbf{y}) \mathbf{z} \]
Let \( \mathbf{x} = \mathbf{b} \), \( \mathbf{y} = \mathbf{c} \), and \( \mathbf{z} = \mathbf{a} \):
\[ \mathbf{(b \times c)} \times \mathbf{(c \times a)} = (\mathbf{b} \cdot \mathbf{a}) \mathbf{c} - (\mathbf{b} \cdot \mathbf{c}) \mathbf{a} \]

### Step 5: Substitute Back into the Right Side
Now substituting back into the right side:
\[ V_2 = |\mathbf{a} \times \mathbf{b} \cdot ((\mathbf{b} \cdot \mathbf{a}) \mathbf{c} - (\mathbf{b} \cdot \mathbf{c}) \mathbf{a})| \]

### Step 6: Simplify the Right Side
Using the distributive property:
\[ V_2 = |(\mathbf{a} \times \mathbf{b}) \cdot ((\mathbf{b} \cdot \mathbf{a}) \mathbf{c}) - (\mathbf{a} \times \mathbf{b}) \cdot ((\mathbf{b} \cdot \mathbf{c}) \mathbf{a})| \]
The first term simplifies to \( (\mathbf{b} \cdot \mathbf{a}) |\mathbf{a} \times \mathbf{b}| \) and the second term is zero since \( \mathbf{a} \times \mathbf{b} \) is orthogonal to \( \mathbf{a} \).

### Step 7: Equate the Two Volumes
Now we equate \( V_1 \) and \( V_2 \):
\[ |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| = |(\mathbf{b} \cdot \mathbf{a})| |\mathbf{a} \times \mathbf{b}| \]

### Step 8: Analyze the Result
From the equality, we can derive that the scalar triple product must be either \( +1 \) or \( -1 \) (since the magnitudes are equal). Thus, we conclude:
\[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \pm 1 \]

### Final Result
The answer is that the scalar triple product \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) \) must equal \( \pm 1 \).

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