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 $\frac{|\vec a×\vec b+ \vec b ×\vec c +\vec c × \vec a|}{|\vec b-\vec c|}$
Answer & explanation
Correct answer: option 2
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.