$|\vec a + \,\vec b {|^2}\,\, = \,\,3$
$|\,\,\vec a {|^2}\,\, + \;\,|\vec b {|^2}\,\, + \;\,2\,\,\left( {\vec a .\,\,\vec b } \right)\,\, = \,\,3$
$1\,\, + \;\,1\,\, + \;\,2\,\left( {\vec a .\,\,\vec b } \right)\,\, = \,\,3\,\,\,$
$\therefore \,\vec a .\,\,\vec b \,\, = \,\,1/2\,\,\,\,.........\left( 1 \right)$
હવે $\,\vec c \, - \,\,\vec a \,\, - \,\,2\,\vec b \,\, = \,\,3\,\left( {\vec a \, \times \,\,\vec b } \right)\,$
$\left( {\vec c \, - \,\,\vec a \,\, - \,\,2\,\vec b } \right)\,\,.\,\vec b \,\, = \,\,3\,\left\{ {\left( {\vec a \, \times \,\,\vec b } \right)\,\,.\,\vec b } \right\}$
$\vec c \,.\,\,\vec b \, - \,\,\,\vec a .\,\,\vec b \,\, - \,\,2\,\,\left( {\vec b \,.\,\,\vec b } \right)\,\, = \,\,3\,\,\left[ {\vec a \,\vec b \,\,\vec b } \right]$
$\vec c \,.\,\,\vec b \,\, - \,\,\frac{1}{2}\,\,\, - \,\,2\,\,\left( 1 \right)\,\, = \,\,0$ $\left\{ {\because \,\,\,\left[ {\vec a \,\vec b \,\,\vec b \,\, = \,\,0} \right]} \right\}$ ${\left( 1 \right)}$ નો ઉપયોગ કરતો
$ \,\vec c \,.\,\,\vec b \,\, = \,\,\frac{5}{2}\,$
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$f ( n )=\left[\begin{array}{ll}2 n , \,\,\, \,\,\,\,\,\,n =2,4,6,8, \ldots . \\ n -1,\,\,\, n =3,7,11,15, \ldots . \\ \frac{ n +1}{2}, \,\,\, \,\,\,n =1,5,9,13, \ldots \ldots\end{array}\right.$
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