Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $(2\vec a + \vec b)$ and $(\overrightarrow a - 3\vec b)$, externally in the ratio $1:2$ . Also, show that $P$ is the mid point of the line segment $RQ$.
Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $(2\vec a + \vec b)$ and $(\overrightarrow a - 3\vec b)$, externally in the ratio $1:2$ . Also, show that $P$ is the mid point of the line segment $RQ$.
Official Solution
VVidaara Team
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The position vector of $R$ is given by
$\vec r = \cfrac{{1(\vec a - 3\vec b) - 2(2\vec a + \vec b)}}{{1 - 2}} = \cfrac{{ - 3\vec a - 5\vec b}}{{ - 1}} = 3\vec a + 5\vec b$
Mid point of $RQ\,\,{\rm{is}}\,\, = \cfrac{{(3\vec a + 5\vec b) + (\vec a - 3\vec b)}}{2} = 2\vec a + \vec b$, which is $P$.
Hence, $P$ is the middle point of $RQ$
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