$d\,\, = \,\,\left| {\frac{{\left( {_{{a_2}}^ \to \,\, - \,\,_{{a_1}}^ \to } \right).\,\,\left( {_{{b_1}}^ \to \,\, \times \,\,_{{b_2}}^ \to } \right)}}{{|_{{b_1}}^ \to \,\, \times \,\,_{{b_2}}^ \to |}}} \right|$
આપેલ સમીકરણો ને અનુક્રમે સમીકરણ $_r^ \to \,\, = \,\,_{{a_1}}^ \to \, + \,\,\lambda \,_{{b_1}}^ \to \,$ અને $_r^ \to \,\, = \,\,_{{a_2}}^ \to \,\, + \,\,\mu _{{b_2}}^ \to $ સાથે સરખાવતા
$_{{a_1}}^ \to \,\, = \,\,4\hat i\,\,\, - \,\,\hat j,\,\,_{{a_2}}^ \to \,\, = \,\,\hat i\,\, - \,\,\hat j\,\, + \;\,2\hat k$
$_{{b_1}}^ \to \,\, = \,\,\,\hat i\,\,\, + \;\,2\hat j\,\, - \,\,3\hat k\,$ અને $\,_{{b_2}}^ \to \,\, = \,\,2\hat i\,\, + \;\,4\hat j\,\, - \,\,5\hat k$
$\therefore \,\,_{{a_2}}^ \to \,\, - \,\,_{{a_1}}^ \to \, = \,\, - \,\,3\hat i\,\, + \,\,0\hat j\,\, + \;\,2\hat k$
અને, $_{{b_1}}^ \to \,\, \times \,\,_{{b_2}}^ \to \,\,\, = \,\,\left| {\begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k} \\
1&2&{ - 3} \\
2&4&{ - 5}
\end{array}} \right|\,\, = \,\,2\hat i\,\,\, - \,\,\hat j\,\, + \;\,0\hat k$
$ \Rightarrow \,\,\left( {_{{a_2}}^ \to \,\, - \,\,_{{a_1}}^ \to } \right)\,\,.\,\,\left( {_{{b_1}}^ \to \,\, \times \,\,_{{b_2}}^ \to } \right)\,\, = \,\,\left( { - 3\hat i\,\, + \;\,0\hat j\,\, + \;\,2\hat k} \right)\,\,.\,\,\left( {2\hat i\,\, - \,\,\hat j\,\, + \;\,0\hat k} \right)\,\, = \,\, - 6$
અને, $|_{{b_1}}^ \to \,\, \times \,\,_{{b_2}}^ \to |\,\, = \,\,\sqrt {4\,\, + \;\,1\,\, + \;\,0} \,\, = \,\,\sqrt 5 $
$\therefore \,\,d\,\, = \,\,\left| {\frac{{\left( {_{{a_2}}^ \to \,\, - \,\,_{{a_1}}^ \to } \right).\,\,\left( {_{{b_1}}^ \to \,\, \times \,\,_{{b_2}}^ \to } \right)}}{{|_{{b_1}}^ \to \,\, \times \,\,_{{b_2}}^ \to |}}} \right|\,\, = \,\,\left| {\frac{{ - 6}}{{\sqrt 5 }}} \right|\,\, = \,\,\frac{6}{{\sqrt 5 }}$
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$f(x)=\left\{\begin{array}{cc}2 \sin \left(-\frac{\pi x}{2}\right), & \text { if } x<-1 \\ \left|a x^{2}+x+b\right|, & \text { if }-1 \leq x \leq 1 \\ \sin (\pi x), & \text { if } x>1\end{array}\right.$
વડે વ્યાખ્યાયીત છે. જો $f(x)$ એ $R$ પર સતત હોય, તો $a+b $ ..... .
વિધાન $1$:=$S=\{x:f(x)=f^{-1}(x)\}=$$\left\{ {1,2} \right\}$
વિધાન $2$: $f $ એ એક-એક અને વ્યાપત છે અને ${f^{ - 1}}\left( x \right) = 1 + \sqrt {x - 1} \;,x \ge 1$