Vectors — Maths STD 12 Science — Question

Let $\overline{\mathrm{AC}}=\overline{\mathrm{CB}}=\bar{a}$ and $\overline{\mathrm{CP}}=\bar{r}$.

Then $|\bar{a}|=|\bar{r}| \ldots(1)$

$\overline{\mathrm{AP}}=\overline{\mathrm{AC}}+\overline{\mathrm{CP}}=\bar{a}+\bar{r}=\bar{r}+\bar{a}$

$\overline{\mathrm{BP}}=\overline{\mathrm{BC}}+\overline{\mathrm{CP}}=-\overline{\mathrm{CB}}+\overline{\mathrm{CP}}=-\bar{a}+\bar{r}$

$\begin{aligned} \therefore \overline{\mathrm{AP}} \cdot \overline{\mathrm{BP}}= & (\bar{r}+\bar{a}) \cdot(\bar{r}-\bar{a}) \\ & \bar{r} \cdot \bar{r}-\bar{r} \cdot \bar{a}+\bar{a} \cdot \bar{r}-\bar{a} \cdot \bar{a}\end{aligned}$

$=|\vec{r}|^2-|\bar{a}|^2=0 \quad \ldots(\because \bar{r} \cdot \bar{a}=\bar{a} \cdot \bar{r})$

$\therefore \overline{\mathrm{AP}} \perp \overline{\mathrm{BP}} \quad \therefore \angle \mathrm{APB}$ is a right angle.

Hence, the angle subtended on a semicircle is the right angle.