The perpendicular distance of the point $\vec c$ from the line joining $\vec a$ and $\vec b$, is

$\frac{|\vec b×\vec c +\vec c×\vec a+\vec a×\vec b|}{|\vec b-\vec a|}$

Let ABC be a triangle and let $\vec a, \vec b, \vec c$ be the position vectors of its vertices A, B, C respectively. Let CM be the perpendicular from C on AB. Then,

Area of ΔABC =$\frac{1}{2}(AB).CM=\frac{1}{2}|\vec{AB}|CM$

Also,

Area of ΔABC =$\frac{1}{2}|\vec a×\vec b +\vec b×\vec a+\vec c×\vec a|$

$∴\frac{1}{2}|\vec{AB}|CM=\frac{1}{2}|\vec a×\vec b +\vec b×\vec a+\vec c×\vec a|$

$⇒CM=\frac{|\vec a×\vec b +\vec b×\vec a+\vec c×\vec a|}{|\vec{AB}|}$

$⇒CM=\frac{|\vec a×\vec b +\vec b×\vec a+\vec c×\vec a|}{|\vec b-\vec a|}$