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Zenmo
 one year ago
(A) Find a set of parametric equations and (B) symmetric equations that passes through the given two points of a line.
(1/2, 2, 1/2), (1, 1/2, 0)
Zenmo
 one year ago
(A) Find a set of parametric equations and (B) symmetric equations that passes through the given two points of a line. (1/2, 2, 1/2), (1, 1/2, 0)

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1how are defined symmetric equations?

dan815
 one year ago
Best ResponseYou've already chosen the best response.2probably just meants change them into z=f(x,y)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so we have to eliminate the parameter t, right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please keep in mind that a line, in the euclidean space, is given as an intersection between two planes

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so, all what we can do, is to solve the third equation for t, and then substitute that value of t into the first and second equation

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1namely: \[\Large t = \frac{1}{2}  z\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I got: \[\begin{gathered} x = 1  3z \hfill \\ 2y =  1 + 10z \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1So, your line is given by these equations: \[\Large \left\{ \begin{gathered} x + 3z = 1 \hfill \\ 2y  10z =  1 \hfill \\ \end{gathered} \right.\] As you can see those equations are the equations of two planes

dan815
 one year ago
Best ResponseYou've already chosen the best response.2u get which this is a line right, in euclian space

dan815
 one year ago
Best ResponseYou've already chosen the best response.2u were initially given the equation of a line in 3D <f(t),g(t),h(t)>

dan815
 one year ago
Best ResponseYou've already chosen the best response.2<x(t),y(t),z(t)> in that form

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.0x=(1/2)+3t, y=25t, z=(1/2)t. Those are the solutions to part (A).

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.0I want to eliminate the parameter, so X = Y = Z.

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.0The solution to symmetric is: \[\frac{ 2x+1 }{ 6 }=\frac{ y2 }{ 5 }=\frac{ 2z1 }{ 2 }\]. I don't know how to fix the x and z variable

dan815
 one year ago
Best ResponseYou've already chosen the best response.2well what looks like this is, it was isolated for t and set to each other

dan815
 one year ago
Best ResponseYou've already chosen the best response.2ah u removed the parametric form soo starting over then

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.0Ok, I actually found it out. I just had to multiply 2 to the numerator and denominator of x and z to get rid of the fraction at the numerator

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i didnt know u were being picking about the simplification

dan815
 one year ago
Best ResponseYou've already chosen the best response.2i thought u just wanted to know how to get that form

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.0yea i wanted to know how to convert x=(1/2)+3t, y=25t, z=(1/2)t into symmetric equations of X = Y = Z.

dan815
 one year ago
Best ResponseYou've already chosen the best response.2and set them to each other

Zenmo
 one year ago
Best ResponseYou've already chosen the best response.0Yea, I just figured it out. Fractions are evil.
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