Statement-1: Let $\vec a, \vec b, \vec c$ be three coterminous edges of a parallelopiped of volume 2 cubic units and r is any vector in space, then $|(\vec r.\vec a) (\vec b×\vec c)+(\vec r.\vec b) (\vec c×\vec a)+(\vec r. \vec c) (\vec a×\vec b)=2|\vec r|$
Statement-2: Any vector in space can be written as a linear combination of three non-coplanar vectors.
Answer & explanation
Correct answer: option 1
Clearly, statement-2 is true.
We have, $\left|[\vec a\,\,\vec b\,\,\vec c]\right|=2$
$\begin{bmatrix}\vec a×\vec b&\vec b×\vec c&\vec c×\vec a\end{bmatrix}=[\vec a\,\,\vec b\,\,\vec c]^2 = 4≠0$
$⇒\vec a×\vec b, \vec b×\vec c, \vec c×\vec a$ are non-coplanar.
Using statement-2, we have
$\vec r = x(\vec a×\vec b) + y (\vec b×\vec c) +z (\vec c×\vec a)$ ...(i)
Taking dot products with $\vec a, \vec b$ and $\vec c$ respectively, we get
$\vec r.\vec a=y[\vec a\,\,\vec b\,\,\vec c], \vec r.\vec b =z[\vec a\,\,\vec b\,\,\vec c]$ and $\vec r.\vec c =[\vec a\,\,\vec b\,\,\vec c]$
Substituting the values of x, y, z in (i), we get
$\vec r[\vec a\,\,\vec b\,\,\vec c]=(\vec r.\vec a) (\vec b×\vec c) + (\vec r.\vec b) (\vec c×\vec a) + (\vec r.\vec c) (\vec a×\vec b)$
$⇒\left|(\vec r.\vec a) (\vec b×\vec c) + (\vec r.\vec b) (\vec c×\vec a) + (\vec r.\vec c) (\vec a×\vec b)\right|$
$⇒|\vec r||\vec a\,\,\vec b\,\,\vec c|$
$⇒\left|(\vec r.\vec a) (\vec b×\vec c) + (\vec r.\vec b) (\vec c×\vec a) + (\vec r.\vec c) (\vec a×\vec b)\right|=2|\vec r|$