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Question

Question: \[a \times (b \times c) + b \times (c \times a) + c \times (a \times b) =\]...

a×(b×c)+b×(c×a)+c×(a×b)=a \times (b \times c) + b \times (c \times a) + c \times (a \times b) =

A

0

B

2[a b c]

C

a + b + c

D

3[a b c]

Answer

0

Explanation

Solution

∵a×(b×c)=(b.c)a−(a.b)c\because\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{b}.\mathbf{c})\mathbf{a} - (\mathbf{a}.\mathbf{b})\mathbf{c}

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

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

=0= 0, {∵a.b=b.aetc.}\{\because\mathbf{a}.\mathbf{b} = \mathbf{b}.\mathbf{a}etc.\}.