Question

Show that the perpendicular distance of the point $\overrightarrow{c}$ from the line joining $\overrightarrow{a}$ and $\overrightarrow{b}$ is $\frac{|\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}+\overrightarrow{a}\times \overrightarrow{b}|}{|\overrightarrow{b}-\overrightarrow{a}|}$.

Answer 1

$\begin{array}{c}Given,Areaof\Delta ABC=\frac{1}{2}\left|\overrightarrow{AB}\right|\times P=\frac{1}{2}\left|\overrightarrow{AB}\times \overrightarrow{AC}\right|\\ P=\frac{\left|\overrightarrow{AB}\times \overrightarrow{AC}\right|}{\overrightarrow{AB}}\\ P=\frac{\left|(\overline{b}-\overline{a})\times (\overline{c}-\overline{a})\right|}{\left|\overline{b}-\overline{a}\right|}\\ p=\frac{\left|(\overline{b}-\overline{a})\times \overline{c}-(\overline{b}-\overline{a})\times \overline{a}\right|}{\left|\overline{b}-\overline{a}\right|}\\ p=\frac{\left|\overline{b}\times \overline{c}-\overline{a}\times \overline{c}-\overline{b}\times \overline{a}+\overline{a}\times \overline{a}\right|}{\left|\overline{b}-\overline{a}\right|}\\ p=\frac{\left|\overline{b}\times \overline{c}+\overline{c}\times \overline{a}+\overline{a}\times \overline{b}\right|}{\left|\overline{b}-\overline{a}\right|}\end{array}$

$\therefore {\perp }^{lar}$ distance from c to line joining $\overline{a}\&\overline{b}$.