Equation of a line from Parametric equations

AI Thread Summary
To find the equation of a line from the given parametric equations x(t) = 1 + 2t, y(t) = -1 + 3t, and z(t) = 4 + t, the correct approach involves using the vector form of the line, which is expressed as <x_0, y_0, z_0> + t<a, b, c>. The confusion arises in determining the direction vector <a, b, c>, which corresponds to the coefficients of t in the parametric equations, specifically <2, 3, 1>. For the equation of a plane containing this line and the point (1, -1, 5), the normal vector N can be derived from the direction vector and another vector formed from the point on the line to the given point. Clarification on these steps is essential for accurately deriving both the line and the plane equations.
Spectre32
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How does one find the equation of a line from parametric equations?

In spefiic I'm looking at this: x(t) = 1+2t , y(t) = -1 + 3t , z(t) = 4+t... I think i got to use something liek x-1/a = y-1/b=z-1/c or something like that. If what i just said is true, then I'm lost on what to do next.


Thanks
 
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t = (x-1)/2 = (y+1)/3 = z-4
 
or wait... do i uses this:

<x_0,y_0,z_0>+ t<a,b,c> ...


now I'm really confused.
 
Yeah tide, i did that, but then isn't there something else i got to do? I don't know I'm kinda confused about this stuff. Also allow me to add the rest of the question. IT's find the equation of a plane containing that line, and the point (1,-1,5). Sooo i need the N =<a,b,c> vector and a point ot get my equation of a plane.. So could i use use the second half of that equation i posted aboce ( t<a,b,c>) to get what i need for my N vector?
 
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