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$ABC$ એ સમબાજુ ત્રિકોણ છે. દરેક બાજુની લંબાઈ $a$ અને તેનું પરિકેન્દ્ર $O$ છે. If $|\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{A C}|=n a$ હોય તો $n =....$
$0$
$1$
$2$
$3$
Solution
$\mathop {AB}\limits^ \to {\mkern 1mu} + {\mkern 1mu} \mathop {BC}\limits^ \to {\mkern 1mu} {\mkern 1mu} = {\mkern 1mu} {\mkern 1mu} \mathop {AC}\limits^ \to {\mkern 1mu} {\mkern 1mu} $
$ \Rightarrow {\mkern 1mu} |\mathop {AB}\limits^ \to {\mkern 1mu} + {\mkern 1mu} \mathop {BC}\limits^ \to {\mkern 1mu} {\mkern 1mu} + {\mkern 1mu} {\mkern 1mu} \mathop {AC}\limits^ \to |{\mkern 1mu} {\mkern 1mu} = |2\mathop {AC}\limits^ \to {\mkern 1mu} |{\mkern 1mu} {\mkern 1mu} = {\mkern 1mu} {\mkern 1mu} 2{\mkern 1mu} {\mkern 1mu} |\mathop {AC}\limits^ \to {\mkern 1mu} |{\mkern 1mu} = {\mkern 1mu} {\mkern 1mu} 2a{\mkern 1mu} $
$\therefore {\mkern 1mu} {\mkern 1mu} n{\mkern 1mu} {\mkern 1mu} = {\mkern 1mu} {\mkern 1mu} 2$
