In the triangle PQR, PQ
= 2a and QR
= 2 b . The mid-point of PR is M. Find following
vectors in terms of a and b .
(i) PR
(ii) PM
(iii) QM
Answers
Answer:
In △PQR,PQ=2a and QR=2b
\textbf{To find:}To find:
\text{$\vec{PR}$ and $\vec{PM}$}PR and PM
\textbf{Solution:}Solution:
\text{Let O be the origin}Let O be the origin
\vec{PQ}=2\vec{a}PQ=2a
\vec{OQ}-\vec{OP}=2\vec{a}OQ−OP=2a ....(1)
\vec{QR}=2\vec{b}QR=2b
\vec{OR}-\vec{OQ}=2\vec{b}OR−OQ=2b ....(2)
\text{Adding (1) and (2), we get}Adding (1) and (2), we get
\vec{OQ}-\vec{OP}+\vec{OR}-\vec{OQ}=2\vec{a}+2\vec{b}OQ−OP+OR−OQ=2a+2b
\vec{OR}-\vec{OP}=2(\vec{a}+\vec{b})OR−OP=2(a+b)
\implies\boxed{\bf\vec{PR}=2(\vec{a}+\vec{b})}⟹PR=2(a+b)
\text{Since M is the midpoint of PR, we have}Since M is the midpoint of PR, we have
\vec{OM}=\dfrac{\vec{OP}+\vec{OR}}{2}OM=2OP+OR
\text{Now,}Now,
\vec{PM}=\vec{OM}-\vec{OP}PM=OM−OP
\vec{PM}=(\dfrac{\vec{OP}+\vec{OR}}{2})-\vec{OP}PM=(2OP+OR)−OP
\vec{PM}=\dfrac{\vec{OP}+\vec{OR}-2\vec{OP}}{2}PM=2OP+OR−2OP
\vec{PM}=\dfrac{\vec{OR}-\vec{OP}}{2}PM=2OR−OP
\vec{PM}=\dfrac{\vec{PR}}{2}PM=2PR
\vec{PM}=\dfrac{2(\vec{a}+\vec{b})}{2}P
Step-by-step explanation:
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Answer:
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