Question

In $\Delta ABC,D,E,F$ are midpoints of the sides $BC,CA$ and $AB$ respectively. $O$' is the circumcentre, $G$' is the centroid, $H$' is the orthocentre and $P$ is any point.
Match the following

List I	List II
$1)\overrightarrow{PA}+\overrightarrow{PB}+\overrightarrow{PC}$	$a)$$0$
$2)\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}$	$b)\overrightarrow{OH}$
$3)\overrightarrow{AD}+\frac{2}{3}\overrightarrow{BE}+\frac{1}{3}\overrightarrow{CF}$	$c)\overrightarrow{PD}+\overrightarrow{PE}+\overrightarrow{PF}$
$4)\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}$	$d)\frac{1}{2}\overrightarrow{AC}$

• A. $1-a,2-b,3-c,4-d$
• B. $1-c,2-a,3-b,4-c$
• C. $1-c,2-a,3-d,4-b$
• D. $1-a,2-b,3-d,4-c$

Answer 1

Answer: (C)

$1-c,2-a,3-d,4-b$

1. Since there is an arbitrary point $P$ in the question, the answer needs to be dependent on $P$ and so, there's only one option which has that, implying the answer to be $c$.
2. $G=\frac{a+b+c}{3}$ where $a,b,c$ are the position vectors of $A,B,C$ respectively.
Thus, $\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=G-a+G-b+G-c=3G-a-b-c=0$
3. We have $D=\frac{b+c}{2},E=\frac{c+a}{2},F=\frac{a+b}{2}$
Thus, $\overrightarrow{AD}+\frac{2}{3}\overrightarrow{BE}+\frac{1}{3}\overrightarrow{CF}=D-a+\frac{2}{3}E-\frac{2}{3}b+\frac{1}{3}F-\frac{1}{3}c$
$=\frac{1}{6}[3b+3c-6a+2c+2a-4b+a+b-2c]=\frac{1}{2}[c-a]$, hence option $d$.
4. Option $b$ is the remaining option!