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How does one convert the vector representation of a 3D-line to one represented as the intersection of two planes and vice versa?

$$\overline{r} = \overline{r}_0+t\overline{v}$$ $$a_1x+b_1y+c_1z=d_1\quad\text{and}\quad a_2x+b_2y+c_2z=d_2$$

They are two very different forms, and somehow I can't wrap my head around this.

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  • $\begingroup$ You need two scalar parameters , one for each basis vector, for a plane. $\endgroup$ – mathreadler Sep 20 '17 at 10:38
  • $\begingroup$ What you have written is the parametric equation for a line passing through $\vec r_0$. $\endgroup$ – mathreadler Sep 20 '17 at 10:39
  • $\begingroup$ @mathreadler: sorry, but what is the relevance of these remarks ? $\endgroup$ – Yves Daoust Sep 20 '17 at 10:40
  • $\begingroup$ @YvesDaoust : That it is impossible to answer given the equations in the question because the first equation does not give a plane but a line. $\endgroup$ – mathreadler Sep 20 '17 at 10:41
  • $\begingroup$ @mathreadler: I think you misunderstand the question. The solution is given below. $\endgroup$ – Yves Daoust Sep 20 '17 at 10:41
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From the vector representation of the line to planes:

Write $\bar r=(x,y,z)$. If $\bar r_0=(x_0,y_0,z_0)$ and $\bar v=(a_0,b_0,c_0)$, then $\bar r=\bar r_0+t\bar v$ is equivalent to three equations: \begin{align*} x&=x_0+t a_0, \\ y&=y_0+t b_0, \\ z&=z_0+t c_0. \\ \end{align*} Now, solve one of these equations for $t$ and put the result into the remaing two equations.

Example: Let the line be given by $\bar r_0=(2,-1,1)$ and $\bar v=(0,1,3)$. Then the three equations read \begin{align*} x&=2,\\ y&=-1+t, \\ z&=1+3t. \end{align*} To solve one of these equations for $t$, I choose the second one (you can also take the third, but not the first, because there is no $t$ anymore). This gives $t=y+1$, and putting this in the first and third gives \begin{align*} x=2,\qquad z=1+3(y+1)=3y+4. \end{align*} Rearrange these equations a little and obtain \begin{align*} x=2,\qquad 3y-z=-4. \end{align*} These are the equations for the planes.

From the planes to the vector representation of the line:

The two equations for the planes basically are a system of two linear equations. All you need to do is to solve this system for $x,y,z$. Since there are only two equations, two of the variables will eventually depend on the third. This third variable can now be relabeled into $t$. Now write everything as vectors, and that's it.

Example: Take the equations from the example above (we would like to check if we got correct planes) \begin{align*} x=2,\qquad 3y-z=-4. \end{align*} The first equation cannot be simplified anymore. The second can be solved for $y$ or $z$. I choose $z$, because then I don't have to divide by 3, which gives nicer numbers. But $y$ would also work. This gives $z=3y+4$. As above, we write this as vectors and obtain \begin{align*} \begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2\\y\\3y+4\end{pmatrix}=\begin{pmatrix}2\\0\\4\end{pmatrix}+y\begin{pmatrix}0\\1\\3\end{pmatrix}. \end{align*} Now you can relabel $y$ into $t$ if you like, but this obviously does not matter. Note that we got a different $\bar r_0$ this time, which does not matter, since the new $\bar r_0$ is also a point on the line given in the first example.

Remark: As has been pointed out in another answer, solutions to both processes are not unique (which is true as our second example shows).There are in fact infinitely many possibilities for the plane equations as well as for the line representation.

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1) expand the vector equation in three scalar equations; draw $t$ from the first and plug it in the remaining two.

2) choose two arbitrary values of $x$ (say $0$ and $1$) and solve the system of equations for $x$ and $y$ (twice). This gives you two points on the line, let $\bar r_0$ and $\bar r_1$. Then $\bar v=\bar r_1-\bar r_0$.

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  • $\begingroup$ What is v then? I'm also interested in the inverse of this method. So how does this result in the two equations for the plane? Or is it the result of the solved systems? $\endgroup$ – Felix Sep 20 '17 at 10:47
  • $\begingroup$ @Felix: are you seriously asking what $\bar v$ is ? Read your own post. My answer covers the two transformations: 1) and 2). $\endgroup$ – Yves Daoust Sep 20 '17 at 11:35
  • $\begingroup$ Ah, I misunderstood. What a joke. $\endgroup$ – Felix Sep 20 '17 at 11:48
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You can write top planes equation together on matrix-vector form $\bf Mv=d$. (One row each in $\bf M$). ${\bf v} = [x,y,z]^T$. What you can do is to $\bf v= M^\dagger d$ Where $^\dagger$ denotes some pseudoinverse. One representation is before the solution to the linear equation system and one is after.

$${\bf M} = \begin{bmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\end{bmatrix} , {\bf d} =\begin{bmatrix}d_1\\d_2\end{bmatrix}$$

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Say you have a line given as intersection of two planes $$a_1 x + b_1 y + c_1 z = d_1 \\ a_2 x + b_2 y + c_2 z = d_2 $$

The vector $(a_1, b_1, c_1)$ is perpendicular to the first plane, and $(a_2, b_2, c_2)$ is perpendicular to the second plane. Now the intersection line of these two planes has a directional vector $v$. $v$ has to be therefore perpendicular to both $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$. You can take $v$ to be their vector product. Now, if you just one point $v_0$ on the line, you got your parametric equation. For this, find a particular solution to the system above.

The other way: the line is given parametrically as $$x = x_0 + t u \\ y= y_0 + t v \\ z= z_0 + t w$$

So you can write this in an equivalent way (assume non-zero denominator, if zero, then the numerator must be $0$ also)

$$\frac{x-x_0}{u} = \frac{y-y_0}{v} = \frac{z-z_0}{w}$$ or, equivalently

$$\frac{x-x_0}{u} = \frac{y-y_0}{v}\\ \frac{x-x_0}{u} = \frac{z-z_0}{w}$$ (or other possible combinations of two linear equations).

Note that neither the pair of planes giving a line, nor a parametric equation, is unique, there is some freedom in choosing them.

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