| 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. |
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) \[\frac{2i+k}{3},\,\,\frac{-8i+3j-7k}{3},\,\,\frac{-7i+3j-5k}{3}\]
Correct Answer: D
Solution :
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 \[c'=\frac{-7i+3j-5k}{3}\]
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