Three particles start simultaneously from a point on a horizontal smooth plane. First particle moves with speed $v_{1}$ towards east, second particle moves towards north with speed $v_{2}$ and third-one moves towards north east. The velocity of the third particle, so that the three always lie on a line, is

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$\frac{v_{1} + v_{2}}{\sqrt{2}}$

$\sqrt{v_{1} v_{2}}$

$\frac{v_{1} v_{2}}{v_{1} + v_{2}}$

$\sqrt{2} (\frac{v_{1} v_{2}}{v_{1} + v_{2}})$

answer is D.

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Detailed Solution

Equation of line $R S$ is $y = - m x + C$

or $y = - (\frac{v_{2}}{v_{1}}) x + v_{2} t$
or $v_{1} y = - v_{2} x + v_{1} v_{2} t$ ….(i)
Equation of line $O P$ is
$y = x$ ….(ii)
Point $P$ is the point of intersection. So, we get
$x_{P} = y_{P} = \frac{v_{1} v_{2} t}{v_{1} + v_{2}}$
$∴ O P = \sqrt{x_{P}^{2} + y_{P}^{2}} = \frac{\sqrt{2} v_{1} v_{2} t}{v_{1} + v_{2}}$
or $v_{3} = \frac{O P}{t} = \frac{\sqrt{2} v_{1} v_{2}}{v_{1} + v_{2}}$