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Math Help - point and direction of a line help

  1. #1
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    point and direction of a line help

    find the direction and one point on the following line

     x + 1 = -\frac{y-5}{3} = \frac{3-z}{2}

    am not sure how to do this ?

    I have simplified the equation and get

     6x+2y+3z-13 =0
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  2. #2
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    Re: point and direction of a line help

    Quote Originally Posted by Tweety View Post
    I have simplified the equation and get

     6x+2y+3z-13 =0
    The result of your simplification is a single equation, which defines a plane in 3D space (two variables are free and one depends on them). The original formula consists of two first-order equations. Each of them determines a plane, and together they determine the intersection of those planes, i.e., a line.

    Suppose the following equations are given.
    $$
    \frac{x-x_0}{A}=\frac{y-y_0}{B} =\frac{z-z_0}{C}
    $$
    If we denote the value of these three fractions by $t$ and express $x,y,z$ through $t$, we get a system of parametric equations
    $$
    \begin{aligned}
    x&=x_0+At\\
    y&=y_0+Bt\\
    z&=z_0+Ct
    \end{aligned}
    $$
    They determine a line passing through $(x_0,y_0,z_0)$ in the direction $(A,B,C)$.
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  3. #3
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    Re: point and direction of a line help

    Quote Originally Posted by Tweety View Post
    find the direction and one point on the following line

     x + 1 = -\frac{y-5}{3} = \frac{3-z}{2} [/tex]
    The way you have rewritten it is completely wrong. You have turned a line into a plane.

    This is the symmetric form $\dfrac{x-a}{n_1}=\dfrac{y-b}{n_2}=\dfrac{z-c}{n_3}$ of a line.
    The point $(a,b,c)$ is on that line, and the vector $\vec{D}=<n_1,n_2,n_3>$ is the direction vector.
    The vector form is $<a,b,c>+t\cdot \vec{D}$. The parametric form is $\begin{array}{*{20}{c}}
    {x = a + t{n_1}} \\
    {y = b + t{n_2}} \\
    {z = c + t{n_3}} \end{array}$


    Write this as $\dfrac{x + 1}{1} = \dfrac{y-5}{-3} = \dfrac{z-3}{-2}$ now we have a point $(-1,5,3)$ and direction vector $<1,-3,-2>$
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