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Assume that $p_1: a_1x+b_1y+c_1z=d_1$ and $p_2: a_2x+b_2y+c_2z=d_2$ are two planes in $\mathbb R^3$ such that $p_1$ and $p_2$ intersect each other.

Since these are just two algebraic equations with constants $a_1$, $a_2$, $b_1$, $b_2$, $c_1$, $c_2$, $d_1$ and $d_2$ why can't we use simple algebraic methods, such as solving a system of linear equations, to find the unique line that is contained in both planes?

Why do we calculate the normal vectors of the planes first, then calculate the vector product of them to find the direction vector of the line?

Note: I've tried using calculus to show that using simple algebra is not enough, but I can't come up with a solution.

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  • $\begingroup$ You can also solve the System $$a_1x+b_1y+c_1z+d_1=0$$ and $$a_2x+b_2y+c_2z+d_2=0$$ $\endgroup$ – Dr. Sonnhard Graubner Jul 14 '18 at 20:06
  • $\begingroup$ wow, why is there a downvote :) $\endgroup$ – user2312512851 Aug 1 '18 at 15:13
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You are correct.

We can find the line of intersection without finding the normals and the cross products.

All we need is two points on the intersection of the two planes.

To find a point on the intersection we simply assign an arbitrary value to one of the coordinates and find the other two coordinates.

For example if our planes are $$ 2x+3y-5z=10$$ and $$ 3x+y-z=5$$ we let $x=0$, and find $y$ and $z$ from the system $$ 3y-5z=10,y-z=5 $$ to find the point $P(0,7.5,2.5)$ on the intersection line.

Similarly we find the point $Q(\frac {25}{13},0, \frac {10}{13})$

Now all we have to do is to write the equation of the line passing through $P$ and $Q$.

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Because in 3-D, there one equation is not enough to determine a 1-D curve. You need parametric equations (or something equivalent.)

If you eliminate one variable by adding or subtracting, you just end up with another plane which happens to be parallel to the axis of whatever variable you eliminated.

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You'd be better if cross product the two corresponding characteristic vectors of the planes $$v_1=(a_1,b_1,c_1)\\v_2=(a_2,b_2,c_2)\\v=v_1\times v_2=(a,b,c)$$this because the line is perpendicular on both $v_1$ and $v_2$ so is parallel to their cross product.

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The cartesian form of a line in $\mathbb R^3$ is expressed by the equations of two planes containing the line; that is, it is expressed in the system

$$\begin{cases} a_1x+b_1y+c_1z+d_1=0\\\\ a_2x+b_2y+c_2z+d_2=0 \end{cases}$$

What we can find is the parametric form, indeed finding two points P and Q in the intersection (that is two solutions of the Cartesian system), we have that the line parametric equation is given by

$$P+t(Q-P)$$

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  • $\begingroup$ @amWhy Thanks for the editing! $\endgroup$ – gimusi Jul 17 '18 at 21:25
  • $\begingroup$ No problem! Your post was still understandable, prior to my edit. Just have too much time on my hands this afternoon (late afternoon, for me)! :-) $\endgroup$ – Namaste Jul 17 '18 at 21:28
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There are several good answers already, but there’s an important way of looking at these equations that I don’t think that any of them have really touched on.

In an important sense, the pair of plane equations that you’re starting with already specify the line: it’s the set of points that satisfy both equations—the intersection of the two planes. What you’re really being asked to do is to convert this implicit representation of the line into a different representation, namely, a set of parametric equation. The general way to perform this conversion is to solve the system of equations, as you suggest. When there is an infinite number of solutions, as there is in this case, you will end up with a parametric description of the line. All of that stuff with cross products is really just a geometric method of solving this system of equations that takes advantage of some special properties of three-dimensional space: the cross product of two vectors only really works in three dimensions, planes embedded in higher-dimensional spaces don’t have a unique normal vector, and so on.

These two ways of representing a line in three-dimensional space are an instance of a more general phenomenon. There are two basic ways to specify an $m$-dimensional subspace of an $n$-dimensional vector space. The first is to describe it as a span, that is, as the set of all linear combinations of some fixed set of vectors. Essentially, you build it up from lower-dimensional objects—the subspace is the join of the lower-dimensional objects. This is what the parametric equation of a line is. I’ll call this the span representation.

The other way is to give a set of $(n-1)$-dimensional subspaces—hyperspaces—whose intersection is the $m$-dimensional subspace being described. Basically, the subspace is the set of common points of some collection of higher-dimensional objects—the meet of these objects. These subspaces are usually themselves described by homogeneous linear equations, so you have a system of homogeneous equations whose solution set is the subspace being described. This obviously corresponds to the implicit system of linear equations that describe the line above. (When you learn about dual vector spaces, you’ll find that this is equivalent to giving a set of dual vectors that all annihilate the subspace.) Since this subspace is the null space of the matrix that represents the system of equations, I’ll call this the null space representation.

As I mentioned earlier, you can convert from the null space representation to the span representation by solving the system of equations. It turns out that you can convert in the other direction the same way: if you plug each of the spanning vectors into a linear equation with unknown coefficients, you end up with a system of linear equations in the coefficients. The solution to this system then gives you a system of linear equations that describe the same space.

Now, a flat is basically a vector subspace offset by some fixed vector, i.e., the set $\{\mathbf v+\mathbf u \mid \mathbf u\in U\}$, where $\mathbf v$ is a fixed vector and $U$ is a subspace of the enclosing vector space. You’ll often see this set denoted by $\mathbf v+U$. Obviously, if we have a span representation for $U$, then we have a span representation for the flat $\mathbf v+U$: just add $\mathbf v$ to every linear combination. By the same token, we also have a null space representation, but the hyperplanes being intersected don’t necessarily pass through the origin any more, so they are represented by inhomogeneous linear equations. The solution to such a system of equations, if you’ll recall, can be represented as the sum of a particular solution and the null space of the related homogeneous system, which is exactly what we want. (Actually, if you move to homogeneous coordinates, then the equations of the hyperspaces are also homogeneous, and we’re back to the solution being purely the null space of the coefficient matrix.)

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