Question

Let $\overline{a},\overline{b},\overline{c}$ and $\overline{d}$ be position vectors of four points $A,B,C$ and $D$ lying in a plane. If $(\overline{a}-\overline{d}).(\overline{b}-\overline{c})=0=(\overline{b}-\overline{d}).(\overline{c}-\overline{a}),$ then $\Delta ABC$ has $D$ as

• A. in-centre
• B. circum-centre
• C. ortho-centre
• D. centroid

Answer 1

The correct option is C ortho-centre

$(\overline{a}-\overline{d}).(\overline{b}-\overline{c})=0=(\overline{b}-\overline{d}).(\overline{c}-\overline{a})$
$\Rightarrow \overline{DA}$ is perpendicular to $\overline{CB}$ and $\overline{DB}$ is perpendicular to $\overline{AC}$
$\Rightarrow \overline{AD}$ and $\overline{BD}$ are along altitudes to $\overline{BC}$ and $\overline{AC}$ respectively.
$\therefore D$ is point of intersection of altitudes.
Hecne, option C.