Q69 / 160 · AP EAMCET 2024 (19 May Shift 2) · Mathematics
If $\mathrm{L}, \mathrm{M}, \mathrm{N}$ are the mid points of the sides $\mathrm{PQ}, \mathrm{QR}$ and RP of $\triangle \mathrm{PQR}$ respectively. then
$\overrightarrow{Q M}+\overrightarrow{L N}+\overrightarrow{M L}+\overrightarrow{R N}-\overrightarrow{M N}-\overrightarrow{Q L}=$
- A$\overrightarrow{P Q}+\overrightarrow{Q R}+\overrightarrow{L M}+\overrightarrow{M N}$
- B$\overrightarrow{L P}+\overrightarrow{P M}+\overrightarrow{M Q}$
- Correct answerC$\overrightarrow{P Q}+\overrightarrow{Q R}-\overrightarrow{P R}$
- D$\overrightarrow{L M}+\overrightarrow{M N}+\overrightarrow{N R}$
Solution
In $\triangle P Q R, M$ and $N$ are mid-points of $\overrightarrow{Q R}$ and $\overrightarrow{R P}$
So, $\overrightarrow{M N}=\frac{1}{2} \overrightarrow{P Q} \Rightarrow \overrightarrow{M N}=\overrightarrow{L P}=\overrightarrow{L Q}$ So, $\overrightarrow{N M}=\overrightarrow{P L}=\overrightarrow{L Q}$
Similarly, we get $\overrightarrow{M L}=\overrightarrow{R N}=\overrightarrow{N P}$ and $\overrightarrow{L N}=\overrightarrow{Q M}=\overrightarrow{M R}$
Now, $\overrightarrow{Q M}+\overrightarrow{L N}+\overrightarrow{M L}+\overrightarrow{R N}-\overrightarrow{M N}-\overrightarrow{Q L}$
$=\overrightarrow{Q R}+\overrightarrow{R P}-(-\overrightarrow{P Q})=\overrightarrow{P Q}+\overrightarrow{Q R}-\overrightarrow{P R} .$
Hence, the correct answer is option: 'C'.