Question

Prove that angle in a semi circle is a right angle by vector method.

Answer 1

To prove: $\angle APB={90}^0$

$\overrightarrow{AB}$ is the diameter and $\overrightarrow{O}$ is the centre

By the law of triangle of vectors:

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

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

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

$\overrightarrow{PA}.\overrightarrow{PB}=(\overrightarrow{PO}+\overrightarrow{OA}).(\overrightarrow{PO}-\overrightarrow{OA})$

$={\left|\overrightarrow{PO}\right|}^2-{\left|\overrightarrow{OA}\right|}^2$

$(\because \left|\overrightarrow{PO}\right|=\left|\overrightarrow{OA}\right|=radius)$

$\therefore \overrightarrow{PA}.\overrightarrow{PB}=0$

Dot product of $\overrightarrow{PA}and\overrightarrow{PB}=0$

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

Hence, $\angle APB={90}^0$