text: # finding parametric & symmetric equations of line
  
  • Jun 8th 2008, 01:26 AM
  Craka
  finding parametric & symmetric equations of line
  Have a question which is as above for line through points (0, 1/2, 1) and (2,1, -3)
  
  They way I understand it this is what you do. Get a directional vector between the two points ie let (0, 1/2, 1) be point A and (2, 1, -3) be point B. Thus vector AtoB is < (2-0), (1-1/2), (-3-1)> giving < 2, 1/2, -4>
  
  Now parametrics would be x= (0+2t) , y=( 1/2+1/2t) and z = (1+ (-4t))
  
  giving symetric as being x/2 = (2y-1) = (z-1)/-4
  
  • Jun 8th 2008, 07:08 AM
  Soroban
  Hello, Craka!
  
  Quote:
  
  Find the equations for line through points $(0,\frac{1}{2}, 1)\text{ and }(2,1, -3)$
  
  The way I understand it this is what you do.
  Get a directional vector between the two points $A(0,\frac{1}{2}, 1)\text{ and }B(2, 1, -3)$
  Thus: $\vec{v} \:=\:\overrightarrow{AB} \:=\:\langle 2-0,\:1-\frac{1}{2},\:-3-1\rangle \:=\:\langle 2,\:\frac{1}{2},\:-4\rangle$
  
  The parametrics would be: . $\begin{Bmatrix} x&=& 0+2t \\ y& =& \frac{1}{2}+\frac{1}{2}t \\ z &=& 1-4t \end{Bmatrix}$
  
  giving symmetric as: . $\frac{x}{2} \;=\; \frac{2y-1}{1} \;= \;\frac{z-1}{-4}$
  
  But it looks like I'm wrong . . . . WHO said you're wrong?
  
  There is an unavoidable problem with the equations of a line.
  . . I wish textbook authors and math teachers would be own up to it.
  
  There are a zillion possible forms of the equations of a line.
  . . There is no "one right answer."
  
  You used point $A(0,\frac{1}{2},1)$ and $\vec{v} = \langle 2,\frac{1}{2},1\rangle$ . . . and your answers are correct.
  
  Suppose you used $B(2,1,-3)\text{ and }\vec{v} = \langle 2,\frac{1}{2},1\rangle$
  You would have: . $\begin{Bmatrix} x &=& 2 + 2t \\ y &=& 1 + \frac{1}{2}t \\ z &=& \text{-}3 -4t \end{Bmatrix}\;\;\text{ and }\;\;\frac{x-2}{2} \:=\:\frac{2y-2}{1} \:=\:\frac{z+3}{-4}$ . . also correct.
  
  $\text{Instead of }\,\vec{v} = \langle 2,\frac{1}{2},-4\rangle$, we can use a parallel vector: . $2\vec{v} = \langle 4,1,-8\rangle$
  
  With point $B(2,1,-3)$, we would have: . $\begin{Bmatrix}x &=& 2 +4t \\ y &=& 1 + t \\ z&=& \text{-}3 -8t \end{Bmatrix}\;\;\text{ and }\;\;\frac{x-2}{4} \:=\:\frac{y-1}{1} \:=\:\frac{z+3}{-8}$
  
  You see, the variations are endless . . .
  
  • Jun 8th 2008, 07:22 AM
  Craka
  Thankyou. That was causing me alot of anguish, as all the different sources I looked at seemed to have the method I was using, however the answer in the rear of my book was different to what I was getting.
explanation: The document is a forum post where a user named Craka asks for help with finding the parametric and symmetric equations of a line through two given points. The user provides their solution and expresses confusion because they believe their answer is incorrect. Another user, Soroban, responds by confirming that Craka's solution is correct and explains that there can be multiple correct forms of the equations for a line, depending on the point and direction vector chosen.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # finding parametric & symmetric equations of line
  
  • Jun 8th 2008, 01:26 AM
  Craka
  finding parametric & symmetric equations of line
  Have a question which is as above for line through points (0, 1/2, 1) and (2,1, -3)
  
  They way I understand it this is what you do. Get a directional vector between the two points ie let (0, 1/2, 1) be point A and (2, 1, -3) be point B. Thus vector AtoB is < (2-0), (1-1/2), (-3-1)> giving < 2, 1/2, -4>
  
  Now parametrics would be x= (0+2t) , y=( 1/2+1/2t) and z = (1+ (-4t))
  
  giving symetric as being x/2 = (2y-1) = (z-1)/-4
  
  • Jun 8th 2008, 07:08 AM
  Soroban
  Hello, Craka!
  
  Quote:
  
  Find the equations for line through points $(0,\frac{1}{2}, 1)\text{ and }(2,1, -3)$
  
  The way I understand it this is what you do.
  Get a directional vector between the two points $A(0,\frac{1}{2}, 1)\text{ and }B(2, 1, -3)$
  Thus: $\vec{v} \:=\:\overrightarrow{AB} \:=\:\langle 2-0,\:1-\frac{1}{2},\:-3-1\rangle \:=\:\langle 2,\:\frac{1}{2},\:-4\rangle$
  
  The parametrics would be: . $\begin{Bmatrix} x&=& 0+2t \\ y& =& \frac{1}{2}+\frac{1}{2}t \\ z &=& 1-4t \end{Bmatrix}$
  
  giving symmetric as: . $\frac{x}{2} \;=\; \frac{2y-1}{1} \;= \;\frac{z-1}{-4}$
  
  But it looks like I'm wrong . . . . WHO said you're wrong?
  
  There is an unavoidable problem with the equations of a line.
  . . I wish textbook authors and math teachers would be own up to it.
  
  There are a zillion possible forms of the equations of a line.
  . . There is no "one right answer."
  
  You used point $A(0,\frac{1}{2},1)$ and $\vec{v} = \langle 2,\frac{1}{2},1\rangle$ . . . and your answers are correct.
  
  Suppose you used $B(2,1,-3)\text{ and }\vec{v} = \langle 2,\frac{1}{2},1\rangle$
  You would have: . $\begin{Bmatrix} x &=& 2 + 2t \\ y &=& 1 + \frac{1}{2}t \\ z &=& \text{-}3 -4t \end{Bmatrix}\;\;\text{ and }\;\;\frac{x-2}{2} \:=\:\frac{2y-2}{1} \:=\:\frac{z+3}{-4}$ . . also correct.
  
  $\text{Instead of }\,\vec{v} = \langle 2,\frac{1}{2},-4\rangle$, we can use a parallel vector: . $2\vec{v} = \langle 4,1,-8\rangle$
  
  With point $B(2,1,-3)$, we would have: . $\begin{Bmatrix}x &=& 2 +4t \\ y &=& 1 + t \\ z&=& \text{-}3 -8t \end{Bmatrix}\;\;\text{ and }\;\;\frac{x-2}{4} \:=\:\frac{y-1}{1} \:=\:\frac{z+3}{-8}$
  
  You see, the variations are endless . . .
  
  • Jun 8th 2008, 07:22 AM
  Craka
  Thankyou. That was causing me alot of anguish, as all the different sources I looked at seemed to have the method I was using, however the answer in the rear of my book was different to what I was getting.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a forum post where a user named Craka asks for help with finding the parametric and symmetric equations of a line through two given points. The user provides their solution and expresses confusion because they believe their answer is incorrect. Another user, Soroban, responds by confirming that Craka's solution is correct and explains that there can be multiple correct forms of the equations for a line, depending on the point and direction vector chosen.
  
  Label: question-and-answer