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Question: Let $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ be three non-coplanar vectors and $...

Let a,b,c\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} be three non-coplanar vectors and r\overrightarrow{r} be any arbitrary vector, then the expression (a×b)×(r×c)+(b×c)×(r×a)+(c×a)×(r×b)(\overrightarrow{a} \times \overrightarrow{b}) \times (\overrightarrow{r} \times \overrightarrow{c}) + (\overrightarrow{b} \times \overrightarrow{c}) \times (\overrightarrow{r} \times \overrightarrow{a}) + (\overrightarrow{c} \times \overrightarrow{a}) \times (\overrightarrow{r} \times \overrightarrow{b}) is always equal to:

A

[abc]r[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]\overrightarrow{r}

B

2[abc]r2[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]\overrightarrow{r}

C

4[abc]r4[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]\overrightarrow{r}

D

0\overrightarrow{0}

Answer

2[a b c]r

Explanation

Solution

The given expression is E=(a×b)×(r×c)+(b×c)×(r×a)+(c×a)×(r×b)E = (\overrightarrow{a} \times \overrightarrow{b}) \times (\overrightarrow{r} \times \overrightarrow{c}) + (\overrightarrow{b} \times \overrightarrow{c}) \times (\overrightarrow{r} \times \overrightarrow{a}) + (\overrightarrow{c} \times \overrightarrow{a}) \times (\overrightarrow{r} \times \overrightarrow{b}).

Using the vector triple product identity P×(Q×R)=(PR)Q(PQ)R\overrightarrow{P} \times (\overrightarrow{Q} \times \overrightarrow{R}) = (\overrightarrow{P} \cdot \overrightarrow{R})\overrightarrow{Q} - (\overrightarrow{P} \cdot \overrightarrow{Q})\overrightarrow{R}:

Term 1: (a×b)×(r×c)=((a×b)c)r((a×b)r)c=[abc]r[abr]c(\overrightarrow{a} \times \overrightarrow{b}) \times (\overrightarrow{r} \times \overrightarrow{c}) = ((\overrightarrow{a} \times \overrightarrow{b}) \cdot \overrightarrow{c})\overrightarrow{r} - ((\overrightarrow{a} \times \overrightarrow{b}) \cdot \overrightarrow{r})\overrightarrow{c} = [\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]\overrightarrow{r} - [\overrightarrow{a}\overrightarrow{b}\overrightarrow{r}]\overrightarrow{c}.

Term 2: (b×c)×(r×a)=((b×c)a)r((b×c)r)a=[abc]r[bcr]a(\overrightarrow{b} \times \overrightarrow{c}) \times (\overrightarrow{r} \times \overrightarrow{a}) = ((\overrightarrow{b} \times \overrightarrow{c}) \cdot \overrightarrow{a})\overrightarrow{r} - ((\overrightarrow{b} \times \overrightarrow{c}) \cdot \overrightarrow{r})\overrightarrow{a} = [\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]\overrightarrow{r} - [\overrightarrow{b}\overrightarrow{c}\overrightarrow{r}]\overrightarrow{a}.

Term 3: (c×a)×(r×b)=((c×a)b)r((c×a)r)b=[abc]r[car]b(\overrightarrow{c} \times \overrightarrow{a}) \times (\overrightarrow{r} \times \overrightarrow{b}) = ((\overrightarrow{c} \times \overrightarrow{a}) \cdot \overrightarrow{b})\overrightarrow{r} - ((\overrightarrow{c} \times \overrightarrow{a}) \cdot \overrightarrow{r})\overrightarrow{b} = [\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]\overrightarrow{r} - [\overrightarrow{c}\overrightarrow{a}\overrightarrow{r}]\overrightarrow{b}.

Adding the three terms: E=3[abc]r([abr]c+[bcr]a+[car]b)E = 3[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]\overrightarrow{r} - ([\overrightarrow{a}\overrightarrow{b}\overrightarrow{r}]\overrightarrow{c} + [\overrightarrow{b}\overrightarrow{c}\overrightarrow{r}]\overrightarrow{a} + [\overrightarrow{c}\overrightarrow{a}\overrightarrow{r}]\overrightarrow{b}).

Using the identity [abc]r=[rbc]a+[rca]b+[rab]c=[bcr]a+[car]b+[abr]c[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]\overrightarrow{r} = [\overrightarrow{r}\overrightarrow{b}\overrightarrow{c}]\overrightarrow{a} + [\overrightarrow{r}\overrightarrow{c}\overrightarrow{a}]\overrightarrow{b} + [\overrightarrow{r}\overrightarrow{a}\overrightarrow{b}]\overrightarrow{c} = [\overrightarrow{b}\overrightarrow{c}\overrightarrow{r}]\overrightarrow{a} + [\overrightarrow{c}\overrightarrow{a}\overrightarrow{r}]\overrightarrow{b} + [\overrightarrow{a}\overrightarrow{b}\overrightarrow{r}]\overrightarrow{c}, the term in parenthesis is equal to [abc]r[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]\overrightarrow{r}.

Therefore, E=3[abc]r[abc]r=2[abc]rE = 3[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]\overrightarrow{r} - [\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]\overrightarrow{r} = 2[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]\overrightarrow{r}.