A particle moving in the xy-plane experiences a velocity dependent force $$\vec{F} = k(v_y\hat{i} + v_x\hat{j})$$, where $$v_x$$ and $$v_y$$ are the x and y components of its velocity $$\vec{v}$$. If $$\vec{a}$$ is the acceleration of the particle, then which of the following statements is true for the particle?

We have a particle of mass $$m$$ moving only in the $$xy$$-plane, so its velocity may be written as $$\vec v = v_x \hat i + v_y \hat j$$ with $$v_x = \dfrac{dx}{dt}$$ and $$v_y = \dfrac{dy}{dt}$$. The force acting on it is given in the question as

$$\vec F = k\bigl(v_y\hat i + v_x\hat j\bigr).$$

By Newton’s second law $$\vec F = m\vec a$$, therefore the acceleration components satisfy

$$\begin{aligned} a_x &= \dfrac{F_x}{m} = \dfrac{k}{m}\,v_y,\\[4pt] a_y &= \dfrac{F_y}{m} = \dfrac{k}{m}\,v_x. \end{aligned}$$

So the two coupled differential equations for the velocity components are

$$\begin{aligned} \frac{dv_x}{dt} &= \frac{k}{m}\,v_y, \quad\quad (1)\\[6pt] \frac{dv_y}{dt} &= \frac{k}{m}\,v_x. \quad\quad (2) \end{aligned}$$

Now we examine each statement one by one, using these relations.

Testing the constancy of $$\vec v\times\vec a$$ (Option A).

Because motion is confined to the plane, both $$\vec v$$ and $$\vec a$$ have zero $$z$$-components, so their cross-product points along $$\hat k$$ and its magnitude equals the scalar expression

$$\vec v \times \vec a = (v_x\hat i + v_y\hat j)\times\bigl(a_x\hat i + a_y\hat j\bigr) = (v_x a_y - v_y a_x)\,\hat k.$$

Substituting $$a_x$$ and $$a_y$$ from above,

$$\begin{aligned} v_x a_y - v_y a_x &= v_x\left(\frac{k}{m}v_x\right) - v_y\left(\frac{k}{m}v_y\right)\\ &= \frac{k}{m}\left(v_x^2 - v_y^2\right). \end{aligned}$$

We differentiate this quantity with respect to time to see whether it changes:

$$\begin{aligned} \frac{d}{dt}\!\left(v_x a_y - v_y a_x\right) &=\frac{k}{m}\,\frac{d}{dt}\!\left(v_x^2 - v_y^2\right)\\[6pt] &=\frac{k}{m}\,\bigl(2v_x\dot v_x - 2v_y\dot v_y\bigr)\\[6pt] &=\frac{2k}{m}\,\Bigl(v_x\frac{dv_x}{dt} - v_y\frac{dv_y}{dt}\Bigr). \end{aligned}$$

Using Eqs. (1) and (2): $$\dot v_x = \dfrac{k}{m}v_y$$ and $$\dot v_y = \dfrac{k}{m}v_x$$, so

$$\begin{aligned} \frac{d}{dt}\!\left(v_x a_y - v_y a_x\right) &=\frac{2k}{m}\Bigl(v_x\left(\frac{k}{m}v_y\right) - v_y\left(\frac{k}{m}v_x\right)\Bigr)\\[6pt] &=\frac{2k}{m}\cdot\frac{k}{m}\,(v_x v_y - v_y v_x)\\[6pt] &=\frac{2k^2}{m^2}\,(0)=0. \end{aligned}$$

The time derivative is zero, so $$v_x a_y - v_y a_x$$, and hence $$\vec v\times\vec a$$, is a constant vector. Therefore statement A is correct.

Testing whether $$\vec F$$ could be a magnetic force (Option B).

A magnetic (Lorentz) force on a charge $$q$$ has the form $$\vec F = q\,\vec v\times\vec B$$, which is always perpendicular to $$\vec v$$, implying $$\vec v\cdot\vec F = 0$$. For the given force, however,

$$\vec v\cdot\vec F = v_x(kv_y) + v_y(kv_x) = 2k\,v_x v_y,$$

which is in general non-zero. Hence the given force cannot come purely from a magnetic field, so statement B is false.

Testing the constancy of kinetic energy (Option C).

Kinetic energy is $$K = \dfrac12 m v^2 = \dfrac12 m(v_x^2 + v_y^2).$$ Its time derivative is

$$\frac{dK}{dt} = m\,\vec v\cdot\vec a.$$

We have just computed $$\vec v\cdot\vec a = 2\frac{k}{m}v_x v_y$$, so

$$\frac{dK}{dt} = m\left(2\frac{k}{m}v_x v_y\right) = 2k\,v_x v_y,$$

which is not generally zero. Thus kinetic energy is not constant, and statement C is false.

Testing whether $$\vec v\cdot\vec a$$ itself is constant (Option D).

From above $$\vec v\cdot\vec a = 2\dfrac{k}{m}v_x v_y.$$ Differentiating using the product rule and Eqs. (1) and (2),

$$\begin{aligned} \frac{d}{dt}\!\bigl(\vec v\cdot\vec a\bigr) &=2\frac{k}{m}\Bigl(v_x\dot v_y + v_y\dot v_x\Bigr)\\[6pt] &=2\frac{k}{m}\Bigl(v_x\cdot\frac{k}{m}v_x + v_y\cdot\frac{k}{m}v_y\Bigr)\\[6pt] &=2\frac{k^2}{m^2}\bigl(v_x^2 + v_y^2\bigr), \end{aligned}$$

which is strictly positive (unless the particle is at rest). Therefore $$\vec v\cdot\vec a$$ is not constant, and statement D is false.

Only Option A stands up to scrutiny.

Hence, the correct answer is Option A.