Question

$(\mathbf{a} + \mathbf{b}).(\mathbf{b} + \mathbf{c}) \times (\mathbf{a} + \mathbf{b} + \mathbf{c}) =$

• A. – [a b c]
• B. [a b c]
• C. 0
• D. 2[a b c]

Answer 1

Answer: (B)

**$a b c**$

Answer 2

$(\mathbf{a} + \mathbf{b}).(\mathbf{b} + \mathbf{c}) \times (\mathbf{a} + \mathbf{b} + \mathbf{c})$

$= (\mathbf{a} + \mathbf{b}).\{ - \mathbf{a} \times \mathbf{b} + \mathbf{c} \times \mathbf{a} + \mathbf{b} \times \mathbf{b} + \mathbf{c} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{c}\}$

$= (\mathbf{a} + \mathbf{b}).\{ - \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a} + \mathbf{c} \times \mathbf{b}\}$

$\lbrack\because\mathbf{b} \times \mathbf{b} = 0\text{and}\mathbf{c} \times \mathbf{c} = 0\rbrack$

$= (\mathbf{a} + \mathbf{b}).( - \mathbf{a} \times \mathbf{b} + \mathbf{c} \times \mathbf{a})$

$= - \lbrack\mathbf{aab}\rbrack + \lbrack\mathbf{aca}\rbrack - \lbrack\mathbf{bab}\rbrack + \lbrack\mathbf{bca}\rbrack$

$= 0 + 0 - 0 + \lbrack\mathbf{bca}\rbrack = \lbrack\mathbf{abc}\rbrack$.