 # Why does this show that the plane at which the motion of Q take place? 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+1/4 pi)[3/2 vec(i) +(3sqrt3)/2 vec(k) ]+3 sin(t+1/4 pi)vec(j) . Urijah Estes 2022-07-22 Answered
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: $Q=cos\left(t+\frac{1}{4}\pi \right)\left[\frac{3}{2}\stackrel{\to }{i}+\frac{3\sqrt{3}}{2}\stackrel{\to }{k}\right]+3sin\left(t+\frac{1}{4}\pi \right)\stackrel{\to }{j}$
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.
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You can write the equation of motion for Q as:
$\left\{\begin{array}{l}x=\frac{3}{2}\mathrm{cos}\left(t+\frac{\pi }{4}\right)\\ y=3\mathrm{sin}\left(t+\frac{\pi }{4}\right)\\ z=\frac{3\sqrt{3}}{2}\mathrm{cos}\left(t+\frac{\pi }{4}\right)\end{array}$
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.

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