Velocity of a particle with some condition

[itex]\left[[/itex]1. The problem statement, all variables and given/known data

Three particles start moving 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 three particles always lie on a straight line is,

1. ##\frac{v_1+v_2}{2}##
   
   
   
   
   


2. [itex]\sqrt{v_1v_2}s[/itex]
   
   
   
   
   


3. ##\frac{v_1v_2}{v_1+v_2}##
   
   
   
   
   


4. ##\sqrt{2}\frac{v_1v_2}{v_1+v_2}##
   
   
   

2. Relevant equations







3. The attempt at a solution



Let ##v_1=\frac{x_1}{t}##, ##v_2=\frac{x_2}{t}##, ##x_3=\frac{x_3}{t}##, where ##x_1, x_2, x_3## are displacement of three particles in ##t## seconds respectively.



##x_1=v_1t##



##x_2=v_2t##



##x_3=v_3t##



Displacement of third particle is the hypotenuse of the orange rectangle in the figure.



Displacement of particle A=2*adjacent side.



OB=Adjacent Side=##\frac{x_1}{2}##



Similarly, AB=Opposite side=##\frac{x_2}{2}##



##OA^2=x_3^2=\left[\frac{x_1}{2}\right]^2+\left[{\frac{x_2}{2}}\right]^2##



##x_3=\sqrt{\frac{x_1^2}{4}+\frac{x_2^2}{4}}##





##x_3=\frac{\sqrt{x_1^2+x_2^2}}{\sqrt{4}}##





##x_3=\frac{\sqrt{v_1^2t^2+v_2^2t^2}}{2}##





##x_3=\frac{\sqrt{(v_1^2+v_2^2)t^2}}{2}##





##x_3=\frac{t\sqrt{(v_1^2+v_2^2)}}{2}##





##v_3t=\frac{t\sqrt{(v_1^2+v_2^2)}}{2}##





##v_3=\frac{\sqrt{(v_1^2+v_2^2)}}{2}##





I checked my answer by substituting values for ##v_1## and ##v_2##. It's right.



I assume option D must be correct because it is the only answer which has the term 2.



I checked and D is the right answer but I should take



##v_3=\frac{\sqrt{(v_1^2+v_2^2)}}{2}##



to



##v3=\sqrt{2}\frac{v_1v_2}{v_1+v_2}##