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Math Help - finding parametric & symmetric equations of line

  1. #1
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    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

    But it looks like I'm wrong can someone please advise what I am doing wrong here.
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  2. #2
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    Hello, Craka!


    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 . . .

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  3. #3
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    Thumbs up

    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.
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