Parametric Equations for Line PQ: Find Solution

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The discussion focuses on finding the parametric equations for the line connecting points P = (1,2,-1) and Q = (5,7,5). The vector v is calculated as <4,5,6>, which is used in the parametric equations x = 1 + 4t, y = 2 + 5t, and z = -1 + 6t. The original poster's teacher marked the solution incorrect, stating that the vector represents the normal rather than the direction of the line. Participants express confusion over the teacher's feedback, as the calculations appear correct. The consensus is that the parametric equations derived from the points should be valid.
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1. Find parametric equations for the line joining the points P = (1,2,-1) and Q = (5,7,5).[/b]

2. x = x0+ta
y = y0+tb
z = z0+tc

3. v = <(5-1), (7-2), (5+1)>
so v = <4,5,6> and since v is a vector in the direction of the line and should be able to be placed in the above equations in place of <a,b,c> and either of the point (P or Q) should be placed in the values of x0, y0, and z0. This should result in the final parametric equations:

x = 1+4t
y = 2+5t
z = -1+6t

This is a problem on one of my tests but my teacher marked it wrong, noting that "line PQ = <4,5,6> is the normal" and marked off 6 points out of 10. I've got no clue what I did wrong.
 
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I don't see anything wrong with that.
 
Hm, I'm a student myself but that looks correct to me...

I would've done the same thing!
 
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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