કારણ $(R) : \,|\vec p \,\, - \,\,\vec q |\,\, = \,\,|\vec p \,\, + \,\,\vec q| $
$ \Rightarrow \,\,|2\vec a \,\, - \,\,\vec b {|^2}\,\, = \,\,{5^2}\,\, $
$\Rightarrow \,\,4{a^2}\,\, + \;\,{b^2}\,\, - \,\,4\vec a .\,\vec b \,\, = \,\,25$
$ \Rightarrow \,\,16\,\, + \,\,9\,\, - \,\,4\,\vec a .\,\vec b \,\, = \,\,25\,\,\,$
$\therefore \,\,\vec a .\,\vec b \,\, = \,\,0$
$\therefore \,\,\,|2\vec a \,\, + \,\,\vec b |\,\,\, $
$= \,\,\sqrt {|2\vec a \,\, + \,\,\vec b {|^2}\,} \,\, $
$= \,\,\sqrt {\left( {4{a^2}\,\, + \;\,{b^2}\,\, + \;\,4\,\,\left( {\vec a .\,\vec b } \right)} \right)} $
$ = \,\,\sqrt {\left( {16\,\, + \;\,9\,\, + \;\,0} \right)} \,\, $
$= \,\,5\,\,\,\,$
$\therefore \,\,|\vec p - \,\vec q |\,\, = \,\,|\vec p + \,\vec q |\,\,\,$
જ્યારે $\vec p \bot \,\vec q $ માત્ર ત્યારે જ શક્ય છે
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$ \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(2+\lambda) \hat{\mathrm{i}}+(1-3 \lambda) \hat{\mathrm{j}}+(3+4 \lambda) \hat{\mathrm{k}}, \lambda \in \mathbb{R} $
$ \mathrm{L}_2: \overrightarrow{\mathrm{r}}=2(1+\mu) \hat{\mathrm{i}}+3(1+\mu) \hat{\mathrm{j}}+(5+\mu) \hat{k}, \mu \in \mathbb{R}$
વચ્ચેનું ન્યૂનતમ અંતર $\frac{m}{\sqrt{n}}$ હોય, જ્યાં $\operatorname{gcd}(m, n)=1$, તો $m+n$ નું મૂલ્ય ........... છે.