Question

If $S$ is the circumcentre, $O$ is the orthocentre of $△ABC$, then $\overline{SA}+\overline{SB}+\overline{SB}$ equals

• A. $\overline{SO}$
• B. $2\overline{SO}$
• C. $\overline{OS}$
• D. $2\overline{OS}$

Answer 1

The correct option is A $\overline{SO}$

Given, 'S' is the circumcentre

'O' is the orthocentre

$\overrightarrow{SA}=\overrightarrow{a}$

$\overrightarrow{SB}=\overrightarrow{b}$

$\overrightarrow{SC}=\overrightarrow{c}$

$\overrightarrow{SO}=\overrightarrow{s}$

$\overrightarrow{SP}||\overrightarrow{AO}$

$\overrightarrow{SP}=\frac{\overrightarrow{SB}+\overrightarrow{SC}}{2}=\frac{\overrightarrow{b}+\overrightarrow{c}}{2}$

$\overrightarrow{SP}=\lambda \overrightarrow{AO}$

$\Rightarrow \frac{\overrightarrow{b}+\overrightarrow{c}}{2}=\lambda (\overrightarrow{s}-\overrightarrow{a})$

$\Rightarrow \overrightarrow{s}=\frac{\overrightarrow{b}+\overrightarrow{c}}{2\lambda }+\overrightarrow{a}$ …….$\left(1\right)$

$\overrightarrow{SQ}=\mu \overrightarrow{BO}$

$\frac{\overrightarrow{a}+\overrightarrow{c}}{2}=\mu (\overrightarrow{s}-\overrightarrow{b})$

$\Rightarrow \overrightarrow{s}=\frac{\overrightarrow{a}+\overrightarrow{c}}{2\mu }+\overrightarrow{b}$ ……….$\left(2\right)$

$\overrightarrow{a}+\frac{\overrightarrow{b}+\overrightarrow{c}}{2\lambda }=\overrightarrow{b}+\frac{\overrightarrow{a}+\overrightarrow{c}}{2\mu }$

$\Rightarrow \overrightarrow{a}+\frac{\overrightarrow{b^2}}{2\lambda }+\frac{\overrightarrow{c}}{2\lambda }=\overrightarrow{b}+\frac{\overrightarrow{a}}{2\mu }+\frac{\overrightarrow{c}}{2\mu }$

$\therefore 1=\frac{1}{2\mu }$

$\Rightarrow \mu =\frac{1}{2}$

$\therefore \overrightarrow{s}=\overrightarrow{a}+\overrightarrow{c}+\overrightarrow{b}$

$\therefore \overrightarrow{SO}=\overrightarrow{SA}+\overrightarrow{SB}+\overrightarrow{SC}$.