Question

Find the value of: \[ \vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times (\vec{c} + \vec{a}) + \vec{c} \times (\vec{a} + \vec{b}) \]

• A. 1
• B. 0
• C. -1
• D. 3

Hint:

The cross product is anti-commutative, meaning \( \vec{u} \times \vec{v} = -\vec{v} \times \vec{u} \). Use this property to simplify expressions involving cross products.

Answer 1

The Correct Option is B

We will use the distributive property of the cross product: \[ \vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} \] \[ \vec{b} \times (\vec{c} + \vec{a}) = \vec{b} \times \vec{c} + \vec{b} \times \vec{a} \] \[ \vec{c} \times (\vec{a} + \vec{b}) = \vec{c} \times \vec{a} + \vec{c} \times \vec{b} \] Now substitute these into the original expression: \[ \vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times (\vec{c} + \vec{a}) + \vec{c} \times (\vec{a} + \vec{b}) \] \[ = (\vec{a} \times \vec{b} + \vec{a} \times \vec{c}) + (\vec{b} \times \vec{c} + \vec{b} \times \vec{a}) + (\vec{c} \times \vec{a} + \vec{c} \times \vec{b}) \] Rearranging terms: \[ = (\vec{a} \times \vec{b} + \vec{b} \times \vec{a}) + (\vec{a} \times \vec{c} + \vec{c} \times \vec{a}) + (\vec{b} \times \vec{c} + \vec{c} \times \vec{b}) \] Now, using the property that \( \vec{u} \times \vec{v} = -\vec{v} \times \vec{u} \), we have: \[ \vec{a} \times \vec{b} = - \vec{b} \times \vec{a}, \quad \vec{a} \times \vec{c} = - \vec{c} \times \vec{a}, \quad \vec{b} \times \vec{c} = - \vec{c} \times \vec{b} \] Thus, the terms cancel each other out: \[ \vec{a} \times \vec{b} + \vec{b} \times \vec{a} = 0, \quad \vec{a} \times \vec{c} + \vec{c} \times \vec{a} = 0, \quad \vec{b} \times \vec{c} + \vec{c} \times \vec{b} = 0 \] So, the entire expression equals zero: \[ 0 \] Thus, the correct answer is: \[ \boxed{0} \]