Parametric equations for trajectory

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The jet's trajectory starts at the point (1, 1, 0) and moves with a constant speed of v = (−5, 0, 1). The position of the jet in three dimensions is given by the equation (1 - 5t, 1, t). To find the trajectory as seen in the yz-plane from the observer at (1, 0, 0), the line from this point to the jet's position is expressed in parameter s. The intersection with the yz-plane occurs when x = 0, leading to the projection point (0, 1 + 1/5t, 1/5). This provides the formulas for the trajectory on the screen as y(t) = 1 + 1/5t and z(t) = 1/5t.
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Homework Statement


A jet takes off from (1, 1, 0) at time t = 0 and moves with constant speed v = (−5, 0, 1).
In a flight simulator, the trajectory of the jet is displayed in the yz-plane as it would appear to an observer at the point (1, 0, 0). Find the formula (in the form y = y(t), z = z(t)) for the trajectory on the screen.

Homework Equations



The Attempt at a Solution


y(t) = 1
z(t) = t
Is that right?
 
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That would be the "parallel projection" onto the yz-plane, not "it would appear to an observer at the point (1, 0, 0)". The position of the jet, in 3 dimensions, is (1- 5t, 1, t). A straight line from (1, 0, 0) through that, in parameter s, would be (1- 5st, 1+ s, ts). That passes through the "yz-plane" (x= 0) when 1- 5st= 0 or s= 1/5t: (0, 1+1/5t, 1/5).
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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