Question

Show that the four points P, Q, R, S with position vectors , , , respectively such that 5−2+6−9=, are coplanar. Also, find the position vector of the point of intersection of the line segments PR and QS.

Answer 1

Let the point of intersection of the line segments $PR$ and $QS$ is A. Then
$$\begin{gathered} 5\overrightarrow{p}-2\overrightarrow{q}+6\overrightarrow{r}-9\overrightarrow{s}=\overrightarrow{0}. \\ \Rightarrow 5\overrightarrow{p}+6\overrightarrow{r}=2\overrightarrow{q}+9\overrightarrow{s} \\ \mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{coefficients}\mathrm{on}\mathrm{both}\mathrm{the}\mathrm{sides}\mathrm{of}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{is}11. \\ \mathrm{So},\mathrm{we}\mathrm{divide}\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{with}11. \\ \Rightarrow \frac{5\overrightarrow{p}+6\overrightarrow{r}}{11}=\frac{2\overrightarrow{q}+9\overrightarrow{s}}{11} \end{gathered}$$

$\frac{5\overrightarrow{p}+6\overrightarrow{r}}{5+6}=\frac{2\overrightarrow{q}+9\overrightarrow{s}}{2+9}$



Therefore, A divides $PR$ in the ratio of $5:6$ and $QS$ in the ratio of $2:9$.
The position vector of the point of intersection of the line segment is $\frac{5\overrightarrow{p}+6\overrightarrow{r}}{11},\frac{2\overrightarrow{q}+9\overrightarrow{s}}{11}$.