Question

\([\vec{a} + \vec{b},\; \vec{b} + \vec{c},\; \vec{c} + \vec{a}]\) is equal to

• A. \([\vec{a}\vec{b}\vec{c}]\)
• B. \(\Sigma(\vec{a}\cdot\vec{b})\vec{c}\)
• C. \(2[\vec{a}\vec{b}\vec{c}]\)
• D. \(|\vec{a}||\vec{b}||\vec{c}|\)

Hint:

Use cyclic symmetry and cancellation in triple products. Remember: \([\vec{a}\vec{b}\vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c})\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The box product \( [\vec{u} \quad \vec{v} \quad \vec{w}] \) represents the scalar triple product \( \vec{u} \cdot (\vec{v} \times \vec{w}) \). We use the linearity and cyclic properties of the scalar triple product to simplify the expression.
Step 2: Detailed Explanation:
Let \( E = [\vec{a} + \vec{b} \quad \vec{b} + \vec{c} \quad \vec{c} + \vec{a}] \). By definition: \[ E = (\vec{a} + \vec{b}) \cdot [(\vec{b} + \vec{c}) \times (\vec{c} + \vec{a})] \] Expanding the cross product: \[ = (\vec{a} + \vec{b}) \cdot [\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{c} + \vec{c} \times \vec{a}] \] Since \( \vec{c} \times \vec{c} = \vec{0} \): \[ = (\vec{a} + \vec{b}) \cdot [\vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{a}] \] Perform the dot product: \[ = \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{a} \cdot (\vec{b} \times \vec{a}) + \vec{a} \cdot (\vec{c} \times \vec{a}) + \vec{b} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{b} \times \vec{a}) + \vec{b} \cdot (\vec{c} \times \vec{a}) \] In a box product, if any two vectors are identical, the result is zero: - \( \vec{a} \cdot (\vec{b} \times \vec{a}) = [\vec{a}\vec{b}\vec{a}] = 0 \) - \( \vec{a} \cdot (\vec{c} \times \vec{a}) = 0 \) - \( \vec{b} \cdot (\vec{b} \times \vec{c}) = 0 \) - \( \vec{b} \cdot (\vec{b} \times \vec{a}) = 0 \) Remaining terms: \[ E = [\vec{a}\vec{b}\vec{c}] + [\vec{b}\vec{c}\vec{a}] \] Since \( [\vec{b}\vec{c}\vec{a}] = [\vec{a}\vec{b}\vec{c}] \) (cyclic property): \[ E = 2[\vec{a}\vec{b}\vec{c}] \].
Step 3: Final Answer:
The result is \( 2[\vec{a}\vec{b}\vec{c}] \).