Question

Show that the four points A, B, C, D with position vectors $\overrightarrow{a,}\overrightarrow{b,}\overrightarrow{c,}\overrightarrow{d}$respectively such that $3\overrightarrow{a}-2\overrightarrow{b}+5\overrightarrow{c}-6\overrightarrow{d}=0,$ are coplanar. Also, find the position vector of the point of intersection of the line segments AC and BD.

Answer 1

Let $AC$ and $BD$ intersects at a point $P$.
We have,
$$\begin{gathered} 3\overrightarrow{a}-2\stackrel{\rightharpoonup }{b}+5\stackrel{\rightharpoonup }{c}-6\stackrel{\rightharpoonup }{d}=\overrightarrow{0}. \\ \Rightarrow 3\overrightarrow{a}+5\overrightarrow{c}=2\overrightarrow{b}+6\overrightarrow{d} \\ \mathrm{Since}\mathrm{sum}\mathrm{of}\mathrm{coefficients}\mathrm{on}\mathrm{both}\mathrm{sides}\mathrm{of}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{is}8. \\ \mathrm{so}\mathrm{we}\mathrm{divide}\mathrm{the}\mathrm{equation}\mathrm{on}\mathrm{both}\mathrm{sides}\mathrm{by}8. \\ \Rightarrow \frac{3\overrightarrow{a}+5\overrightarrow{c}}{8}=\frac{2\overrightarrow{b}+6\overrightarrow{d}}{8} \\ \Rightarrow \frac{3\overrightarrow{a}+5\overrightarrow{c}}{3+5}=\frac{2\overrightarrow{b}+6\overrightarrow{d}}{2+6} \end{gathered}$$

Therefore, $P$ divides $AC$ in the ratio of $3:5$ and $P$ divides $BD$ in the ratio of 2:6.
Therefore, position vector of the point of intersection of AC and BD will be
$\frac{3\overrightarrow{a}+5\overrightarrow{c}}{8}=\frac{2\overrightarrow{b}+6\overrightarrow{d}}{8}$