\(\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}\) are non-zero non-coplanar vectors and \(\overrightarrow{p}=\frac{\overrightarrow{b}\times\overrightarrow{c}}{[bca]},\overrightarrow{q}=\frac{\overrightarrow{c}\times\overrightarrow{a}}{[cab]},\overrightarrow{r}=\frac{\overrightarrow{a}\times\overrightarrow{b}}{[abc]}\), then [abc] =

Concept:

If \(\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}\) are three vectors, then \(\overrightarrow{a}\cdot(\overrightarrow{b}\times \overrightarrow{c})\) is called the scalar triple product or box product, denoted by [abc]

• [abc] = [bca] = [cab] (It follows cyclic order)

For four vectors, \(\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c},\overrightarrow{d}\) quadruple product is given by:

\((\overrightarrow{a}\times\overrightarrow{b})\times (\overrightarrow{c}\times\overrightarrow{d})\) = [abd]c - [abc]d

Calculation:

We have, \(\overrightarrow{p}=\frac{\overrightarrow{b}\times\overrightarrow{c}}{[bca]},\overrightarrow{q}=\frac{\overrightarrow{c}\times\overrightarrow{a}}{[cab]},\overrightarrow{r}=\frac{\overrightarrow{a}\times\overrightarrow{b}}{[abc]}\)

[pqr] = \([\frac{\overrightarrow{b}\times\overrightarrow{c}}{[bca]}\frac{\overrightarrow{c}\times\overrightarrow{a}}{[cab]}\frac{\overrightarrow{a}\times\overrightarrow{b}}{[abc]}]\)

\(\rm[\vec p\ \vec q\ \vec r]=\frac{1}{[\vec a\ \vec b\ \vec c]}\left[(\vec b\times \vec c)\ (\vec c \times \vec a)\ (\vec a \times \vec b)\right]\)

\(\Rightarrow \rm[\vec p \vec q\vec r]=\frac{1}{[\vec a\vec b\vec c]}(\vec b\times \vec c)\left((\vec c \times \vec a)\times(\vec a \times \vec b)\right)\)      ...........(1)

∵ \(\rm (\vec c \times \vec a)\times(\vec a \times \vec b)=\left((\vec c\times \vec a).\vec b\right)\vec a-\left((\vec c \times \vec a).\vec a\right)\vec b\)

\(\Rightarrow \rm (\vec c \times \vec a)\times(\vec a \times \vec b)=[\vec c\ \vec a\ \vec b]\vec a-0\)

\(\Rightarrow \rm (\vec c \times \vec a)\times(\vec a \times \vec b)=[\vec a\ \vec b\ \vec c]\vec a\)      .......(2)

put this in (1), 

\(\Rightarrow \rm[\vec p \vec q\vec r]=\frac{1}{[\vec a\vec b\vec c]}(\vec b\times \vec c).[\vec a\ \vec b\ \vec c]\vec a\)

\(\rm=\frac{[\vec a\ \vec b \ \vec c]}{[\vec a\ \vec b\ \vec c]}(\vec b\times \vec c).\vec a\)

\(\rm =1[\vec b\ \vec c\ \vec a]=[\vec a\ \vec b\ \vec c]\)

The correct answer is Option 1