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 \( 2[\vec{a}\vec{b}\vec{c}] \).