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(C) Given force $\overrightarrow{F} = m\overrightarrow{a} = k(v_y \hat{i} + v_x \hat{j})$.Thus,$a_x = \frac{dv_x}{dt} = \frac{k}{m} v_y$ and $a_y = \f

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The quantity $\overrightarrow{v} 	imes \overrightarrow{a}$ is constant in time.

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(C) Given force $\overrightarrow{F} = m\overrightarrow{a} = k(v_y \hat{i} + v_x \hat{j})$.Thus,$a_x = \frac{dv_x}{dt} = \frac{k}{m} v_y$ and $a_y = \frac{dv_y}{dt} = \frac{k}{m} v_x$.Dividing the two equations: $\frac{dv_y}{dv_x} = \frac{v_x}{v_y} \implies v_y dv_y = v_x dv_x$.Integrating both sides: $\int v_y dv_y = \int v_x dv_x \implies v_y^2 = v_x^2 + C \implies v_y^2 - v_x^2 = 	ext{constant}$.Now,calculate $\overrightarrow{v} 	imes \overrightarrow{a}$:$\overrightarrow{v} 	imes \overrightarrow{a} = (v_x \hat{i} + v_y \hat{j}) 	imes \frac{k}{m}(v_y \hat{i} + v_x \hat{j})$$= \frac{k}{m} [v_x^2 (\hat{i} 	imes \hat{j}) + v_y^2 (\hat{j} 	imes \hat{i})]$$= \frac{k}{m} [v_x^2 \hat{k} - v_y^2 \hat{k}] = \frac{k}{m} (v_x^2 - v_y^2) \hat{k}$.Since $v_y^2 - v_x^2$ is constant,$\overrightarrow{v} 	imes \overrightarrow{a}$ is constant in time.

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