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Question: \(a.\lbrack(b + c) \times (a + b + c)\rbrack\) is equal to...

a.[(b+c)×(a+b+c)]a.\lbrack(b + c) \times (a + b + c)\rbrack is equal to

A

[a b c]

B

2[a b c]

C

3[a b c]

D

0

Answer

0

Explanation

Solution

a.[(b+c)×(a+b+c)]\mathbf{a}.\lbrack(\mathbf{b} + \mathbf{c}) \times (\mathbf{a} + \mathbf{b} + \mathbf{c})\rbrack

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

=[aba]+[abb]+[a b c]+[aca]+[a c b]+[a c c]= \lbrack\mathbf{aba}\rbrack + \lbrack\mathbf{abb}\rbrack + \lbrack\mathbf{a}\ \mathbf{b}\ \mathbf{c}\rbrack + \lbrack\mathbf{aca}\rbrack + \lbrack\mathbf{a}\ \mathbf{c}\ \mathbf{b}\rbrack + \lbrack\mathbf{a}\ \mathbf{c}\ \mathbf{c}\rbrack

0+0+[abc]+0−[abc]+0=0.0 + 0 + \lbrack\mathbf{abc}\rbrack + 0 - \lbrack\mathbf{abc}\rbrack + 0 = 0.