Statement-1: Any vector in space can be uniquely written as the linear combination of three non-coplanar vectors.
Statement-2: If $\vec a,\vec b,\vec c$ are three non-coplanar vectors and r is any vector in space, then
$[\vec a\,\,\vec b\,\,\vec r]\vec c+[\vec b\,\,\vec c\,\,\vec r]\vec a+[\vec c\,\,\vec a\,\,\vec r]\vec b=[\vec a\,\,\vec b\,\,\vec c]\vec r$
Answer & explanation
Correct answer: option 2
Clearly, statement-1 is true.
We have,
$\vec r = x\vec a+y\vec b +z\vec c$ ...(i)
Taking product successively with $\vec b ×\vec c, \vec c×\vec a$ and $\vec a×\vec b$, we obtain
$x=\frac{[\vec b\,\,\vec c\,\,\vec r]}{[\vec a\,\,\vec b\,\,\vec c]},y=\frac{[\vec c\,\,\vec a\,\,\vec r]}{[\vec a\,\,\vec b\,\,\vec c]}$ and $z=\frac{\vec a\,\,\vec b\,\,[\vec r]}{[\vec a\,\,\vec b\,\,\vec c]}$
Substituting the values of x, y, z in (i), we get
$[\vec a\,\,\vec b\,\,\vec r]\vec c+[\vec b\,\,\vec c\,\,\vec r]\vec a+[\vec c\,\,\vec a\,\,\vec r]\vec b=[\vec a\,\,\vec b\,\,\vec c]\vec r$
So, statement-2 is true. But, statement-2 is not a correct explanation for statement-1.