Question

In the triangle pqr PQ=2a and QR=2b. The mid point of PR is M. Find following vector in terms of a and b

Answer 1

$\textbf{Given:}$

$\text{In \triangle PQR,\;\vec{PQ}=2\vec{a} and \vec{QR}=2\vec{b} }$

$\textbf{To find:}$

$\text{ \vec{PR} and \vec{PM} }$

$\textbf{Solution:}$

$\text{Let O be the origin}$

$\vec{PQ}=2\vec{a}$

$\vec{OQ}-\vec{OP}=2\vec{a}$ ....(1)

$\vec{QR}=2\vec{b}$

$\vec{OR}-\vec{OQ}=2\vec{b}$ ....(2)

$\text{Adding (1) and (2), we get}$

$\vec{OQ}-\vec{OP}+\vec{OR}-\vec{OQ}=2\vec{a}+2\vec{b}$

$\vec{OR}-\vec{OP}=2(\vec{a}+\vec{b})$

$\implies\boxed{\bf\vec{PR}=2(\vec{a}+\vec{b})}$

$\text{Since M is the midpoint of PR, we have}$

$\vec{OM}=\dfrac{\vec{OP}+\vec{OR}}{2}$

$\text{Now,}$

$\vec{PM}=\vec{OM}-\vec{OP}$

$\vec{PM}=(\dfrac{\vec{OP}+\vec{OR}}{2})-\vec{OP}$

$\vec{PM}=\dfrac{\vec{OP}+\vec{OR}-2\vec{OP}}{2}$

$\vec{PM}=\dfrac{\vec{OR}-\vec{OP}}{2}$

$\vec{PM}=\dfrac{\vec{PR}}{2}$

$\vec{PM}=\dfrac{2(\vec{a}+\vec{b})}{2}$

$\implies\boxed{\bf\vec{PM}=\vec{a}+\vec{b}}$