Let $\vec a,\vec b,\vec c$ be the position vectors of three non-collinear points A, B and C respectively in a plane.

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

Statement-2: Length of the perpendicular from vertex A on BC is $\frac{|\vec a×\vec b+ \vec b ×\vec c +\vec c × \vec a|}{|\vec b-\vec c|}$

Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1. 

Clearly,

Δ = Area of ΔABC = $\frac{1}{2}|\vec{AB}×\vec{AC}|$

⇒ Δ = Area of ΔABC = $\frac{1}{2}|(\vec b-\vec a)×(\vec c-\vec a)|$

$⇒ Δ =\frac{1}{2}|\vec b×\vec c-\vec b×\vec a-\vec a×\vec c+\vec a×\vec a|$

$⇒ Δ =\frac{1}{2}|\vec a×\vec b+\vec b×\vec c+\vec c×\vec a|$  ...(i)

So, statement-1 is true.

Let p be the length of perpendicular from vertex A on BC.

Then,

$Δ=\frac{1}{2}(BC×p)$    ...(ii)

From (i) and (ii), we get

$\frac{1}{2}|\vec a×\vec b+\vec b×\vec c+\vec c×\vec a|=\frac{1}{2}|\vec c-\vec b|×p$

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

So, statement-2 is true.