Question

Let $ABC$ be a triangle whose centroid is $G$, orthocentre is $H$ and circumcentre is the origin '$O$'. If $D$ is any point in the plane of the triangle such that no three of $O,A,C$ and $D$ are collinear satisfying the relation $\overrightarrow{AD}+\overrightarrow{BD}+\overrightarrow{CH}+3\overrightarrow{HG}=\lambda \overrightarrow{HD}$, then what is the value of the scalar ${}^{\prime }{\lambda }^{\prime }$?

• A. $2$
• B. $3$
• C. $-1$
• D. $-2$

Answer 1

The correct option is A $2$

We know that in vector

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$ where O is the origin

SO from ques

$\overrightarrow{AD}+\overrightarrow{BD}+\overrightarrow{CH}+3\overrightarrow{HG}=\lambda \overrightarrow{HD}$

$\overrightarrow{OD}-\overrightarrow{OA}+\overrightarrow{OD}-\overrightarrow{OB}+\overrightarrow{OH}-\overrightarrow{OC}+3\overrightarrow{OG}-3\overrightarrow{OH}=\lambda \overrightarrow{HD}$

$2\overrightarrow{OD}-2\overrightarrow{OH}-\overrightarrow{OA}-\overrightarrow{OB}-\overrightarrow{OC}+3\overrightarrow{OG}=\lambda \overrightarrow{HD}$

$2(\overrightarrow{OD}-\overrightarrow{OH})-\overrightarrow{OA}-\overrightarrow{OB}-\overrightarrow{OC}+3\overrightarrow{OG}=\lambda \overrightarrow{HD}$

$2\overrightarrow{HD}-\overrightarrow{OA}-\overrightarrow{OB}-\overrightarrow{OC}+3\overrightarrow{OG}=\lambda \overrightarrow{HD}$

Comparing both sides

$\lambda =2$