MCQ
Given $ABC$ is an equilateral triangle of side length $1$ unit and $P$ be any arbitrary point on the circumcircle of triangle $ABC,$ then $|\vec PA|^2+|\vec PB|^2+|\vec PC|^2$ is equal to
- A$3$
- B$1$
- ✓$2$
- D$2\sqrt 3$
✓
Answer
Correct option: C.
$2$
c
$|\vec a| = |\vec b| = |\vec c| = \left| {\vec r} \right| = \frac{1}{{2\sin {{60}^\circ }}} = \frac{1}{{\sqrt 3 }}$
$|\vec a| = |\vec b| = |\vec c| = \left| {\vec r} \right| = \frac{1}{{2\sin {{60}^\circ }}} = \frac{1}{{\sqrt 3 }}$
$\therefore|\overrightarrow{\mathrm{PA}}|^{2}+|\overrightarrow{\mathrm{PB}}|^{2}+|\overrightarrow{\mathrm{PC}}|^{2} $
$=|\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{a}}|^{2}+|\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{b}}|^{2}+|\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}}|^{2} $
$=3|\overrightarrow{\mathrm{r}}|^{2}+3|\overrightarrow{\mathrm{a}}|^{2}-2 \overrightarrow{\mathrm{r}} \cdot(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) $
$=1+1-2 \cdot 0=2(\because \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0})$
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