The position vector of the point which divides the join of points $2\overrightarrow {\rm{a}} - 3\overrightarrow {\rm{b}}$ and $\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}}$ in the ratio 3: 1, is
The position vector of the point which divides the join of points $2\overrightarrow {\rm{a}} - 3\overrightarrow {\rm{b}}$ and $\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}}$ in the ratio 3: 1, is
Official Solution
Let the position vector of the point $R$ divides the join of points
$2\overrightarrow {\rm{a}} - 3\overrightarrow {\rm{b}}$ and $\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}}$.
$\therefore$ Position vector $R = \frac{{3(\overrightarrow {\rm{a}} + \overrightarrow {\rm{b}} ) + 1(2\overrightarrow {\rm{a}} - 3\overrightarrow {\rm{b}} )}}{{3 + 1}}$
Since, the position vector of a point $R$ dividing the line segment joining the points $P$ and $Q$,
whose position vectors are $\overrightarrow {\rm{p}}$ and $\overrightarrow {\rm{q}}$
in the ratio $m:n$ internally, is given by $\frac{{m\vec q + n\vec p}}{{m + n}}$.
$\therefore$ $R = \frac{{5\vec a}}{4}$
No comments yet — start the discussion.