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If the position vector $\vec{a}$ of the point $(5, \mathrm{n})$ is such that $|\overrightarrow{\mathrm{a}}|=13$, then the value/values of $n$ an be
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The correct answer is:
$\pm 12$
$\sqrt{(5-0)^{2}+(n-0)^{2}}=13$
$25+n^{2}=169$
$n^{2}=144$
$n=\pm 12$
$25+n^{2}=169$
$n^{2}=144$
$n=\pm 12$
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