Three particles start from the origin at the same time, one with a velocity ${\mathrm{v}}_1$ along the $x$-axis, second along the negative $y$-axis with a velocity ${\mathrm{v}}_2$ and third particle moves along the line $\mathrm{x}=\mathrm{y}$. The velocity of third particle, so that three may always lie on the same line is :

• A

  $\frac{{\text{v}}_1+{\text{v}}_2}{2}$
• B

  $\sqrt{{\text{v}}_1+{\text{v}}_2}$
• C

  $\frac{{\text{v}}_1{\text{v}}_2}{{\text{v}}_1+{\text{v}}_2}$
• Correct answer
  D

  $\mathrm{v}=\frac{\sqrt{2}{\displaystyle }{\displaystyle {{\displaystyle \mathrm{v}}}_1}{\displaystyle {{\displaystyle \mathrm{v}}}_2}}{{\mathrm{v}}_2-{\mathrm{v}}_1}$
  

Solution

Slope of $\mathrm{AC}$ line = Slope of $\mathrm{BA}$ line

$\frac{\left({\displaystyle \frac{\mathrm{vt}}{\sqrt{2}}-0}\right)}{\left({\displaystyle \frac{\mathrm{vt}}{\sqrt{2}}-{\mathrm{v}}_1\mathrm{t}}\right)}=\frac{0-\left(-{\mathrm{v}}_2\mathrm{t}\right)}{\left({\mathrm{v}}_1\mathrm{t}-0\right)}$

$\mathrm{v}=\frac{\sqrt{2}{\displaystyle }{\displaystyle {{\displaystyle \mathrm{v}}}_1}{\displaystyle {{\displaystyle \mathrm{v}}}_2}}{{\mathrm{v}}_2-{\mathrm{v}}_1}$


 

Hence, the correct answer is option: 'D'.