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

Concept: Scalar triple product is linear in each argument and cyclic.

Step 1: Expand using linearity. \[ [\vec{a}+\vec{b},\; \vec{b}+\vec{c},\; \vec{c}+\vec{a}] = (\vec{a}+\vec{b}) \cdot [(\vec{b}+\vec{c}) \times (\vec{c}+\vec{a})] \]

Step 2: Expand cross product. \[ (\vec{b}+\vec{c}) \times (\vec{c}+\vec{a}) = \vec{b}\times\vec{c} + \vec{b}\times\vec{a} + \vec{c}\times\vec{c} + \vec{c}\times\vec{a} \] \[ = \vec{b}\times\vec{c} + \vec{b}\times\vec{a} + 0 + \vec{c}\times\vec{a} \]

Step 3: Take dot product with \(\vec{a}+\vec{b}\). Many terms cancel due to scalar triple product properties. The result simplifies to: \[ = 2[\vec{a}\vec{b}\vec{c}] \]