JEE MainMathematicsToughVectors

In a triangle PQR, S and T are points on QR and PR respectively,
such that QS = 3SR and PT = 4TR. Let M be the point of intersection of
PS and QT. Find the ratio $\frac{QM}{MT}$
$\Delta PQR$ esa S rFkk T nks fcUnq Øe'k QR rFkk PR bl izdkj gS fdQS =
3SR, PT = 4TR
Ekkuk] M, PS rFkk QT dk izfrPNsnu

Show answer and solution

Let

\overset{\rightarrow}{QP} = \overset{\rightarrow}{a},\overset{\rightarrow}{QR} = \overset{\rightarrow}{b},QM:MT = \lambda:1,PM:MS = \mu:1

\Rightarrow \overset{\rightarrow}{QT} = \frac{4\overset{\rightarrow}{b} + \overset{\rightarrow}{a}}{5},\overset{\rightarrow}{QM} = \frac{\lambda}{\lambda + 1}\left( \frac{4\overset{\rightarrow}{b} + \overset{\rightarrow}{a}}{5} \right)

Also

\overset{\rightarrow}{QS} = \frac{3}{4}\overset{\rightarrow}{b},\overset{\rightarrow}{QM} = \frac{\frac{3\mu}{4}\overset{\rightarrow}{b} + \overset{\rightarrow}{a}}{\mu + 1}

\Rightarrow \frac{1}{\mu + 1} = \frac{\lambda}{5(\lambda + 1)}\ and\ \frac{4\lambda}{5(\lambda + 1)} = \frac{3\mu}{4(\mu + 1)} \Rightarrow \mu = \frac{16}{3},\lambda = \frac{15}{4}

Source: Grass-G / M-21 Mathematics Paper 19 Dec Hindi Eng.docx

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