Intersection of 2 planes (HL)

See also intersection of 3 planes (HL)

Two planes can be identical (coincident, the same), parallel, or intersecting.

When the two normal vectors are scalar multiples of each other, the planes are either identical or parallel.

For example, $x + 2y - 3z = 4$ is identical to ${-3x - 6y + 9z = -12}$, but parallel to ${x + 2y - 3z = 5}$, where the constant is changed to any other value other than $4$.

Otherwise, the two planes intersect at a line.

Example: Let’s find the intersection between

Note that $(3, 4, -1)$ is not a scalar multiple of $(2, 3, -4)$, so there must be an intersection line.

Contents

• Method using cross product
  • step 1: direction vector
  • step 2: a point
• Method using row reduction

Method using cross product

The line of intersection is perpendicular to both normals $\bm {n_1}$ and $\bm {n_2}$.

step 1: direction vector

The direction vector of the line is $(l, m, n) = \bm {n_1} \times \bm {n_2}$.

$(3, 4, -1)\times(2, 3, -4) = (-13, 10, 1)$

step 2: a point

For the component of $(l, m, n)$ that is closest to $0$, but is not $0$, guess the corresponding component of a point.

In our example, since $n = 1$ is the closest component to $0$, let’s guess $z_0 = 0$. (You can also guess $z_0 = 1$, for instance, but $0$ is often easier)

This means

which has solution $x = 59$, $y = -41$

The final answer is $\bm r = (-13, 10, 1)\lambda + (59, -41, 0) \qed$

Method using row reduction

In this context, scalar multiplication means multiplying the row by a non-zero constant.

The rules are

1. Each elementary row operation uses some or all rows from the previous step.
2. Two rows may be swapped.
3. Each changed row must involve the corresponding row from previous step, unless two rows are swapped. Eg the new row 2 must involve a non-zero multiple of the old row 2, unless swapping row 2 with another.
4. Each row can become a scalar multiple of itself, and/or added a scalar multiple of any other row(s). Eg new row 1 can be $-2$ times old row 1 plus half times old row 3.

The goal is to get the it to look like

where at most a single column on the left side does not contain a one and a zero.

Our starting point is

Generally speaking we focus on getting one column into the desired state, in a left-to-right order.

We can first subtract row 2 from row 1

Subtract twice of row 1 from row 2

Subtract row 2 from row 1

This corresponds to the system

$z$ can be used to parametrize both $x$ and $y$, as it appears in both equations.

Let $z = t$

This means

which are the same equations as before, but in a different form. In both cases we assumed $z_0 = 0$ for the point. Had we used different assumptions, we would end up with different points.