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.