• # question_answer Direction: Let a, b and c are three non-coplanar vectors, i.e., $[a\,b\,c]\,\ne 0$. The three new vectors $a',\,\,b'$ and c' defined by the equation $a'=\frac{b\times c}{[a\,\,b\,\,c]},\,\,b'=\frac{c\times a}{[a\,\,b\,\,c]}$and$c'=\frac{a\times b}{[a\,\,b\,\,c]}$ are called reciprocal system to the vectors a, b and c. The reciprocal set of the vectors $2i+3j-k,$$i-j-2k$ and $-i+2j+2k$ are A)  $\frac{2i+k}{3},\,\,\frac{-8i+3j-7k}{3},\,\,\frac{7i+3j+5k}{3}$ B)  $\frac{2i+k}{3},\,\,\frac{-8i+3j+7k}{3},\,\,\frac{7i+3j+5k}{3}$ C) $\frac{2i+k}{3},\,\,\frac{-8i+3j+7k}{3},\,\,\frac{7i-3j+5k}{3}$ D)  $\frac{2i+k}{3},\,\,\frac{-8i+3j-7k}{3},\,\,\frac{-7i+3j-5k}{3}$

Let $a=2i+3j-k,$ $b=i-j-2k$ and      $c=-i+2j+2k$ Here,   $[a\,b\,c]=3$ $\left[ \because \,\left| \begin{matrix} 2 & 3 & -1 \\ 1 & -1 & -2 \\ -1 & 2 & 2 \\ \end{matrix} \right|=2(2)-3(0)-1(1)=3 \right]$and   $b\times c\,\left| \begin{matrix} i & j & k \\ 1 & -1 & -2 \\ -1 & 2 & 2 \\ \end{matrix} \right|=2i+k$ $a'\,=\frac{b\times c}{[a\,b\,c]}=\frac{2i+k}{3}$ Similarly, $b'\,=\frac{-8i+3j-7k}{3}$and     $c'=\frac{-7i+3j-5k}{3}$