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Math Help - scalar equation of a line -> parametric

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    scalar equation of a line -> parametric

    The question is to find the direction cosines & angles of each line, but one of them is in scalar form, like this:

    3x + 5y + 2 = 0

    How can I find the direction of the vector of the line from scalar form? Or, how can I get this into vector form?
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    Quote Originally Posted by theowne View Post
    The question is to find the direction cosines & angles of each line, but one of them is in scalar form, like this:

    3x + 5y + 2 = 0

    How can I find the direction of the vector of the line from scalar form? Or, how can I get this into vector form?
    Draw the line. A vector in the direction of the line should now be obvious ......
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  3. #3
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    Quote Originally Posted by theowne View Post
    The question is to find the direction cosines & angles of each line, but one of them is in scalar form, like this:

    3x + 5y + 2 = 0

    How can I find the direction of the vector of the line from scalar form? Or, how can I get this into vector form?
    Rewrite the equation:

    3x+5y +2 = 0 ~\implies~\langle 3,5 \rangle \cdot \langle x,y \rangle + 2 = 0

    \vec n = \langle 3,5 \rangle is the normal vector of the line.

    You get the direction vector of the line if you change the order of the components and change the sign at one component:

    The direction vector is now: \vec d = \langle 5,-3 \rangle or \vec d = \langle -5,3 \rangle

    Remark: This procedure works only in 2D.
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  4. #4
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    Quote Originally Posted by theowne View Post
    ...

    3x + 5y + 2 = 0

    ... how can I get this into vector form?
    Change this equation into parametric form:

    \left| \begin{array}{l}x =0+ t \\y = -2 - \frac35 t\end{array}\right.

    Now you can rewrite this form into an equation in vector form:

    \langle x,y \rangle = \langle 0,-2 \rangle + t \cdot \langle 1~,~-\frac35 \rangle
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