Another equation of another line

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Discussion Overview

The discussion revolves around finding the parametric and symmetric equations for a line that passes through the point (1,-1,1) and is parallel to a given equation represented in a symmetric form. Participants express confusion and seek clarification on the steps involved in deriving these equations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Some participants inquire about the process of finding the parametric and symmetric equations for a line through a specific point and parallel to another line.
  • There is a suggestion that parallel lines share the same direction vectors, prompting a request for the direction vector of the given line equation.
  • One participant initially proposes a direction vector of v=(-2,2,3) but later questions its correctness, indicating uncertainty in their reasoning.
  • A later reply encourages thinking in reverse from vector form to symmetric form, suggesting a method to derive the equations by separating x, y, and z components.
  • One participant indicates they figured out the solution, but the details of their resolution are not shared.

Areas of Agreement / Disagreement

Participants express confusion and uncertainty about the direction vector and the process of deriving the equations. There is no consensus on the correct direction vector, and the discussion remains unresolved regarding the specific steps to take.

Contextual Notes

Participants have not fully resolved the assumptions regarding the direction vector or the method for deriving the equations, leading to potential gaps in understanding the problem-solving strategy.

ineedhelpnow
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Please help, how do I do this?
Find the parametric and symmetric equation for a line through (1,-1,1) and parallel to $x+2= \frac{1}{2}y=z-3$

So confused
 
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ineedhelpnow said:
Please help, how do I do this?
Find the parametric and symmetric equation for a line through (1,-1,1) and parallel to $x+2= \frac{1}{2}y=z-3$

So confused

Parallel lines have the same direction vectors.

Can you get a direction vector for $\displaystyle \begin{align*} x + 2 = \frac{1}{2}y = z - 3 \end{align*}$?
 
That's what I'm stuck on.at first I though it would be v=(-2,2,3) but then I came to the conclusion that it was wrong so I am stuck again.
 
ineedhelpnow said:
That's what I'm stuck on.at first I though it would be v=(-2,2,3) but then I came to the conclusion that it was wrong so I am stuck again.

You need to get better at the problem solving strategy of thinking in reverse. How would you go from a vector form of a line to the symmetric form?

Start by writing out the x, y, z components separately to get the parametric equations.

Solve for t in each equation.

Realise since they are all equal to t, they are equal to each other.

Now how do you think you can go in reverse to get back to the vector equation?
 
i figured it out :)
 

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