I am a bit confused and probably mixing some terms, and I wish to ask for your assistance in putting things in the right place.

There are 4 ways to determine a plane:

  1. 3 points
  2. a point and a line
  3. 2 intersecting lines
  4. 2 parallel lines

I have a problem understanding the 4th way. If two lines are parallel, aren't they linearly dependent ? If so, how can they be a basis for this plane ? Can they still span it if they are parallel ?

This equation: x=(1,2,-3)+t(1,-2,4)+s(-2,4,-8) does not represent a plane. This is confusing, I mean, (1,-2-4) and (-2,4,-8) are parallel vectors, right ? They are linearly dependent, so according to the definition above they should determine a plane.

I know that a set of vectors in $R^{3}$ span the entire space if every vector can be written as a linear combination of the vectors. In addition if they are linearly independent they are also a basis. The elementary basis contains the vectors (1,0,0), (0,1,0) and (0,0,1). Now I am trying to translate this logic to planes in $R^{3}$, and to connect it to the parametric representation of planes and lines.

Any explanations will be most appreciated ! Thank you !

  • 1
    $\begingroup$ there is other way: a point and a vector, an orthogonal vector to the plane. $\endgroup$ – Masacroso Jul 18 '17 at 5:56
  • $\begingroup$ It seems that you’re thinking of parallel vectors, that is, of lines through the origin, in which case parallel lines are coincident. The lines in the methods that you list are not so restricted. E.g., the lines $x=0,z=0$ and $x=1,z=0$ uniquely determine the $x$-$y$ plane. $\endgroup$ – amd Jul 18 '17 at 6:34

In order to find the plane of 2 distinct parallel lines $l_1$ and $l_2$, take a point $P$ on one of the lines, for example $l_2$, and then determine the plane given by $P$ and $l_1$ (case 2). Note that $P\not\in l_1$. Show that this plane contains the whole line $l_2$.


The parallel lines need to be distinct as well as parallel. The equations \begin{align*} x(t) &= (1, 2, -3) + t(1, -2, 4) \\ y(s) &= (1, 2, -3) + s(2, -4, 8) \end{align*} describe the same line, and cannot be used to uniquely describe any of the infinitely many planes that contain the line. But, as soon as you make the lines distinct, by changing appropriately the initial vector, e.g. \begin{align*} x(t) &= (1, 2, -3) + t(1, -2, 4) \\ y(s) &= (0, 1, 0) + s(2, -4, 8), \end{align*} then the plane described is unique. Any direction vector from one point in the first line to another point in the second line, will not be parallel to the two parallel direction vectors. Thus, we will have two linearly independent directions, which in conjunction with any point from either line, will be enough to define a plane.

Taking the above example, we can form a new direction vector $(1, 2, -3) - (0, 1, 0) = (1, 1, -3)$. Then, our plane has a vector equation, $$p(s, t) = (0, 1, 0) + t(1, -2, 4) + s(1, 1, -3)$$ By constantly setting $s$ to $0$ or $1$, we recover the two lines that formed the plane.

Hope that helps.


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