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 = c o s ( t + 1 4 π ) [ 3 2 i + 3 3 2 k ] + 3 s i n ( t + 1 4 π ) 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 3 x z = 0, and so this defines the plane in which the motion of Q takes place”. I can’t understand what 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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Answers (1)

Raul Garrett
Answered 2022-07-23 Author has 14 answers
You can write the equation of motion for Q as:
{ x = 3 2 cos ( t + π 4 ) y = 3 sin ( t + π 4 ) z = 3 3 2 cos ( t + π 4 )
so, multiplying the first equation by 3 ad subtracting the third we find the equation:
3 x z = 0
This is the equation of a plane in 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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