Math, asked by bibekkumar3921, 8 months ago

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

Answers

Answered by MaheswariS
2

\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}}

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