Question

Prove by vector method that the angle in a semcircle is right amgle.

Answer 1

Let $AB$ be the diameter and $O$ be the centre of a circle.

Let $P$ be a point on the semi-circle.

Construction:- Join $PA,PB$ and $PO$.

To prove:- $\angle APB=90°$

By law of triangle of vectors,

$\overrightarrow{PA}=\overrightarrow{PO}+\overrightarrow{OA}$

$\overrightarrow{PB}=\overrightarrow{PO}+\overrightarrow{OB}$

$\overrightarrow{PB}=\overrightarrow{PO}-\overrightarrow{OA}(\because \overrightarrow{OB}=-\overrightarrow{OA})$

Consider,

$\overrightarrow{PA}\cdot \overrightarrow{PB}$ $=\overrightarrow{(PO+OA)}\cdot \overrightarrow{(PO-OA)}={\left|\overrightarrow{PO}\right|}^2-{\overrightarrow{\left|OA\right|}}^2$

$\because \left|\overrightarrow{PO}\right|=\left|\overrightarrow{OA}\right|=$ radius of the circle

Therefore,

$\overrightarrow{PA}\cdot \overrightarrow{PB}=0$

$\therefore \overrightarrow{PA}\perp \overrightarrow{PB}$

$\therefore \angle APB=90°$

Hence proved.