Check Your Solution: Parametric Equations for a Line through Two Points P and Q

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To check the solution for parametric equations of a line through points P=(-2,0,3) and Q=(3,5,2), one can substitute values for the parameter t into the equation r(t)=<5t-2, 5t, 3-5t>. By solving for t when r(t) equals each of the given points, it confirms that the points lie on the same line. The relationship between the points can also be expressed using the formula (1-t)P + tQ, which describes the line segment between P and Q. This method allows for determining the position of points along the line based on the value of t. Understanding these concepts is essential for verifying solutions in parametric equations.
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Would somebody be so kind as to tell how I can check my solution to this problem(s) below:

Find parametric equations for the line through the points P=(-2,0,3) and Q=(3,5,2).

More specifically, I feel that I should be able to substitute some value for t or (x,y,z) to confirm my solutions. Thanks!
 
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I moved your thread to our homework section.

What did you get as solution?
randy_1981 said:
More specifically, I feel that I should be able to substitute some value for t or (x,y,z) to confirm my solutions. Thanks!
You should be able to plug in some values of t into your solution to get your two points.
 
If you have parametric equations, take any two values for the parameter and check if the points you get are on the same lie as the two given points. Depending upon how you find the parametric equations, it should be easy to choose the parameters so you get the given points. Of course, a line is determined by two points so you only need to check two.
 
Ok, I'm about to reveal my ignorance but how do I determine what the parameter values are?
 
My solution is r(t)=< 5t-2, 5t , 3-5t >
 
randy_1981 said:
Ok, I'm about to reveal my ignorance but how do I determine what the parameter values are?

randy_1981 said:
My solution is r(t)=< 5t-2, 5t , 3-5t >
Is there some value of the parameter t so that r(t) = (-2, 0, 3)? Is there another value of t so that r(t) = (3, 5, 2)? If the answers to these questions aren't obvious by inspection, set up an equation with <5t - 2, 5t, 3 - 5t> on one side, and either of the given points on the other side, and solve for t.
 
Gottca, thanks for the help! I'm undergoing a self study program and some things that should be obvious aren't at times.
 
randy_1981 said:
Gottca, thanks for the help! I'm undergoing a self study program and some things that should be obvious aren't at times.

Basically, that is what "through" means in this type of problem When you say the line passes through the points P=(-2,0,3) and Q=(3,5,2), this means that for some value of t you will have <5t - 2, 5t, 3 - 5t> = <-2,0,3>, while for some other value of t you will have <5t - 2, 5t, 3 - 5t> = <3,5,2>. Componentwise, <a,b,c> = <d,e,f> means a = d, b = e and c = f.

BTW: another (easier?) way is to note that the line through ##P = \langle -2,0,3 \rangle## and ##Q = \langle 3,5,2 \rangle## has the form
(1-t) P + t Q = (1-t) \langle -2,0,3 \rangle + t \langle 3,5,2 \rangle
For ## 0 \leq t \leq 1## the points are on the line-segment joining P and Q (with point = P when t = 0 and point = Q when t = 1). For t < 0 the points are on the line "before" P (that is, on the side away from Q), while for t > 1 they are on the line "after" Q (that is, on the side away from P).
 
If you are referring to something like "3x- 7\le 2 so 3x\le 9, x\le 3, you can't do that for functions that are not "one to one".
 
  • #10
You can always simplify ##3x- 7\le 2## to ##x \le 3## (assuming real numbers).
And all the functions that might come up here are injective.
 

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