I was trying to give a geometrical description of the the motion of Q, and the motion of Q can be described by the equation: $Q=cos(t+\frac{1}{4}\pi)[\frac{3}{2}\vec{i}+\frac{3\sqrt{3}}{2} \vec{k}] +3sin(t+\frac{1}{4}\pi)\vec{j}$.
By calculating |Q|, we know that the distance from the origin is constant; hence, the particle would travel at a circular path.
After proving that, I took a look at the mark scheme for this question, and the mark scheme indicated that:”it is evident that $\sqrt{3}x - z =0$, and so this defines the plane in which the motion of Q takes place”. I can’t understand what $\sqrt{3}x -z =0$ tells us, and why does it relate to the plane of motion, from my point of view, we can see that the equation of motion has variables in x-y-z, so it is obviously moves in x-y-z.
Thank you very much for you guy’s reply.
You can write the equation of motion for $Q$ as: $$ \begin{cases} x=\frac{3}{2}\cos(t+\frac{\pi}{4})\\ y=3\sin(t+\frac{\pi}{4})\\ z=\frac{3\sqrt{3}}{2}\cos(t+\frac{\pi}{4})\\ \end{cases} $$
so, multiplying the first equation by $\sqrt{3}$ ad subtracting the third we find the equation: $$ \sqrt{3}x-z=0 $$ This is the equation of a plane in $\mathbb{R}^3$ and the fact that the coordinates of $Q$ satisfies this equation means that the $Q$ is a point on the plane that has this equation.