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Simplify (vec(b)+vec(c )).{(vec(c )+vec(...

Simplify `(vec(b)+vec(c )).{(vec(c )+vec(a))xx (vec(a)+vec(b))}`

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To simplify the expression \((\vec{b} + \vec{c}) \cdot ((\vec{c} + \vec{a}) \times (\vec{a} + \vec{b}))\), we can follow these steps: ### Step 1: Identify the vectors Let: - \(\vec{x} = \vec{b} + \vec{c}\) - \(\vec{y} = \vec{c} + \vec{a}\) - \(\vec{z} = \vec{a} + \vec{b}\) Thus, we need to simplify \(\vec{x} \cdot (\vec{y} \times \vec{z})\). ### Step 2: Use the scalar triple product The expression \(\vec{x} \cdot (\vec{y} \times \vec{z})\) can be interpreted as the scalar triple product, which can be expressed as the determinant of a matrix formed by the vectors: \[ \vec{x} \cdot (\vec{y} \times \vec{z}) = \text{det}(\vec{x}, \vec{y}, \vec{z}) \] ### Step 3: Substitute the vectors Substituting the vectors back into the determinant: \[ \text{det}(\vec{b} + \vec{c}, \vec{c} + \vec{a}, \vec{a} + \vec{b}) \] ### Step 4: Expand the determinant Using the properties of determinants, we can expand this determinant: \[ \text{det}(\vec{b} + \vec{c}, \vec{c} + \vec{a}, \vec{a} + \vec{b}) = \text{det}(\vec{b}, \vec{c}, \vec{a}) + \text{det}(\vec{c}, \vec{a}, \vec{b}) + \text{det}(\vec{a}, \vec{b}, \vec{c}) \] This can be rewritten as: \[ = \text{det}(\vec{b}, \vec{c}, \vec{a}) + \text{det}(\vec{c}, \vec{a}, \vec{b}) + \text{det}(\vec{a}, \vec{b}, \vec{c}) \] ### Step 5: Combine the determinants Notice that the determinants can be rearranged: \[ = \text{det}(\vec{b}, \vec{c}, \vec{a}) + \text{det}(\vec{b}, \vec{a}, \vec{c}) + \text{det}(\vec{a}, \vec{b}, \vec{c}) \] Using the property of determinants that states that swapping two rows changes the sign of the determinant, we can see that: \[ \text{det}(\vec{c}, \vec{a}, \vec{b}) = -\text{det}(\vec{b}, \vec{c}, \vec{a}) \] Thus, the total simplifies to: \[ = 2 \cdot \text{det}(\vec{b}, \vec{c}, \vec{a}) \] ### Final Result The simplified expression is: \[ 2 \cdot \text{det}(\vec{b}, \vec{c}, \vec{a}) \]
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