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Question: \(\lbrack\mathbf{b} \times \mathbf{cc} \times \mathbf{aa} \times \mathbf{b}\rbrack\) is equal to...

[b×cc×aa×b]\lbrack\mathbf{b} \times \mathbf{cc} \times \mathbf{aa} \times \mathbf{b}\rbrack is equal to

A

a×(b×c)\mathbf{a \times (b \times c)}

B

2[abc]2\lbrack\mathbf{abc}\rbrack

C

[abc]2\lbrack\mathbf{abc}\rbrack^{2}

D

[abc]\lbrack\mathbf{abc}\rbrack

Answer

[abc]2\lbrack\mathbf{abc}\rbrack^{2}

Explanation

Solution

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

Let a×b=d\mathbf{a} \times \mathbf{b} = \mathbf{d}

so , (b×c)[(c×a)×d]=(b×c)[(d.a)c−(d.c).a](\mathbf{b} \times \mathbf{c})\lbrack(\mathbf{c} \times \mathbf{a}) \times \mathbf{d}\rbrack = (\mathbf{b} \times \mathbf{c})\lbrack(\mathbf{d}.\mathbf{a})\mathbf{c} - (\mathbf{d}.\mathbf{c}).\mathbf{a}\rbrack

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

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

=[abc][abc]=[abc]2= \lbrack\mathbf{abc}\rbrack\lbrack\mathbf{abc}\rbrack = \lbrack\mathbf{abc}\rbrack^{2}.