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I just want to know if what I know about vector equations is true, and if there is more.

Let's say I have the following vector equations.

$\left[\begin{array}{c} x \\ y \\ z \end{array}\right]$ $= \left[\begin{array}{c} a \\ b \\ c \end{array}\right]$ $+ t\left[\begin{array}{c} d \\ e \\ f \end{array}\right]$ and $\left[\begin{array}{c} x_2 \\ y_2 \\ z_2 \end{array}\right]$ $= \left[\begin{array}{c} a_2 \\ b_2 \\ c_2 \end{array}\right]$ $+ t\left[\begin{array}{c} d_2 \\ e_2 \\ f_2 \end{array}\right]$

If I let $t = 1$, $\left[\begin{array}{c} d \\ e \\ f \end{array}\right]$ is a direction vector

$\left[\begin{array}{c} a \\ b \\ c \end{array}\right]$ is a point on the line, but since it starts from the origin, is also a vector

Changing different values of $t$ give me different points on the line by solving for $x, y, z$

The two vectors form a plane, so if I cross product, $\left[\begin{array}{c} a \\ b \\ c \end{array}\right]$ $\times$ $\left[\begin{array}{c} a_2 \\ b_2 \\ c_2 \end{array}\right]$, I get the normal to the plane.

Once I have the normal to the plane, I use it, along with one known point and one unknown point to find a scalar equation.

$\quad\quad(bc_2-cb_2)(x-a) + (ca_2-ac_2)(x-b) + (ab_2-ba_2)(x-c) = 0$

If I drew out a vector equation, it appears to form a plane on its own. Does that mean a vector equation of a line is also a vector equation of a plane? If not, can I cross $\left[\begin{array}{c} a \\ b \\ c \end{array}\right]$ with $t\left[\begin{array}{c} d \\ e \\ f \end{array}\right]$ to get the normal, then do the above again?

fossdeep
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    $\pmatrix{d \ e \ f}$ is the direction vector regardless of the value of $t$ you evaluate the function at. The equation of a line is the equation of a line. The general equation of a plane is $\vec r(s,t) = \vec r_0 + s\vec u + t\vec v$. –  Nov 19 '15 at 03:43
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    A linear equation in $x$, $y$ and $z$ defines a plane. To get a line you need two equations. You are essentially defining a line by an intersection of two planes. You may now argue that an equation such as $r(t)=t\hat i+2t\hat j + \hat k$ is one equation but it is not. It defines three coordinate functions; eliminate the $t$ and you will be left with two equations. – John Douma Nov 19 '15 at 03:45
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    @fossdeep Perhaps this answer will help you. –  Nov 19 '15 at 03:51
  • Thanks guys, for the clarifications. And for the link to the other answer. – fossdeep Nov 19 '15 at 04:16

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