A particle moving in the $x y$ plane experiences a velocity dependent force $\overrightarrow{\mathrm{F}}=\mathrm{k}\left(\mathrm{v}_{\mathrm{y}} \hat{\mathrm{i}}+\mathrm{v}_{x} \hat{\mathrm{j}}\right)$, where $\mathrm{v}_{\mathrm{x}}$ and $\mathrm{v}_{\mathrm{y}}$ are the $\mathrm{x}$ and $\mathrm{y}$ components of its velocity $\overrightarrow{\mathrm{v}}$. If $\overrightarrow{\mathrm{a}}$ Ls the acceleration of the particle, then which of the following statements is true for the particle ?
Correct Option: , 3
$\frac{\mathrm{dv}_{\mathrm{x}}}{\mathrm{dt}}=\frac{\mathrm{k}}{\mathrm{m}} \mathrm{v}_{\mathrm{y}}$
$\frac{\mathrm{dv}_{y}}{\mathrm{dt}}=\frac{\mathrm{k}}{\mathrm{m}} \mathrm{v}_{\mathrm{x}}$
$\frac{\mathrm{dv}_{y}}{\mathrm{dv}_{x}}=\frac{\mathrm{v}_{x}}{\mathrm{v}_{y}} \Rightarrow \int \mathrm{v}_{y} \mathrm{dv}_{y}=\int \mathrm{v}_{x} \mathrm{dv}_{x}$
$\mathrm{v}_{\mathrm{y}}^{2}=\mathrm{v}_{\mathrm{x}}^{2}+\mathrm{C}$
$\mathrm{v}_{\mathrm{y}}^{2}-\mathrm{v}_{\mathrm{x}}^{2}=\mathrm{cons} \tan \mathrm{t}$
Option (3)
$\vec{v} \times \vec{a}=\left(v_{x} \hat{i}+v_{y} \hat{j}\right) \times \frac{k}{m}\left(v_{y} \hat{i}+v_{x} \hat{j}\right)$
$=\left(\mathrm{v}_{\mathrm{x}}^{2} \hat{\mathrm{k}}-\mathrm{v}_{\mathrm{y}}^{2} \hat{\mathrm{k}}\right) \frac{\mathrm{k}}{\mathrm{m}}$
$=\left(\mathrm{v}_{\mathrm{x}}^{2}-\mathrm{v}_{\mathrm{y}}^{2}\right) \frac{\mathrm{k}}{\mathrm{m}} \hat{\mathrm{k}}$
$=$ Constant
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