- Home
- Standard 11
- Physics
બે સદીશો $\mathop A\limits^ \to \,\, = \,\,3\hat i\,\, + \;\,\hat j\,\,$ અને $\mathop B\limits^ \to \,\, = \,\,\hat j\,\, + \,2\hat k$ આપેલા છે . આ બે સદીશો માટે $\mathop A\limits^ \to $ નો $\mathop B\limits^ \to $ ની સાપેક્ષે ઘટક સદીશના સ્વરૂપમાં શોધો.
$\frac{1}{5}\,\,\left( {\,\hat j\,\, + \;\,\hat k} \right)$
$\frac{1}{5}\,\,\left( {\,2\hat j\,\, + \;\,2\hat k} \right)$
$\frac{1}{3}\,\,\left( {\,3\hat j\,\, + \;\,2\hat k} \right)$
$\frac{1}{5}\,\,\left( {\,\hat j\,\, + \;\,2\hat k} \right)$
Solution
$\mathop A\limits^ \to $ નો $\mathop B\limits^ \to $ સાથેનો ઘટક
$\, = \,\,\left( {\frac{{\,\mathop A\limits^ \to .\,\mathop B\limits^ \to \,}}{B}} \right)\,\hat B\,\, = \,\,\left( {\frac{{\,\mathop A\limits^ \to .\,\mathop B\limits^ \to }}{B}} \right)\,\frac{{\mathop B\limits^ \to }}{B}\,$
$\, = \,\,\left[ {\frac{{\left( {3\hat i\,\, + \;\,\hat j} \right)\,\,\left( {\hat j\,\, + \;\,2\hat k} \right)}}{{\sqrt 5 }}} \right]\,\,\frac{{\left( {\hat j\,\, + \;\,2\hat k} \right)}}{{\sqrt 5 }}$
$ = \,\,\frac{1}{5}\,\,\left( {\hat j\,\, + \;\,2\hat k} \right)$
