Intersection of two planes on a line

Question

The intersection of the planes defined by v $\cdot\begin{pmatrix} 8\\1\\-12 \end{pmatrix} = 35$ and v $\cdot \begin{pmatrix} 6\\7\\-9 \end{pmatrix} = 70$ is a line. Find the equation of this line.

Any hints or solutions will be appreciated.

Answer 1

The line will be perpendicular to both (8,1,-12) and (6,7,-9). (Do you understand why?) Therefore, taking the cross product of those two vectors is your first step.

Answer 2

The equation of the plane denoted by v $\cdot\begin{pmatrix} 8\\1\\-12 \end{pmatrix} = 35$ is

$$\mathbb{v}\cdot\begin{pmatrix} 8\\1\\-12 \end{pmatrix} = \begin{pmatrix} x\\y\\z \end{pmatrix}\cdot\begin{pmatrix} 8\\1\\-12 \end{pmatrix} = 8x+y-12z = 35$$

Similarly for the second plane, $$\mathbb{v}\cdot\begin{pmatrix} 6\\7\\-9 \end{pmatrix} = \begin{pmatrix} x\\y\\z \end{pmatrix}\cdot\begin{pmatrix} 6\\7\\-9 \end{pmatrix} = 6x+7y-9z = 70$$

Solve the set of the equations in terms of any variable to obtain the equation of the line.

OR

Observe that the equation of the planes are provided in the form $$ \mathbb{r}\cdot\mathbb{n} = \mathbb{r_0}\cdot\mathbb{n}$$ where $\mathbb{n}$ denotes the normal vector to the plane. Since the line will lie on the plane, its direction vector will be orthogonal to the normal vector of each planes.

Thus we can state that for line $l$:- $$l=a_0+at=a_0+(\mathbb{n_1}\times\mathbb{n_2})t$$ where $a_0$ is a point on the line, $t$ is a scalar and $\mathbb{n_1},\mathbb{n_2}$ denotes the normal vectors of the planes.

Answer 3

Let $\vec v=(x,y,z)$ then the equations for the two planes are

• $8x+y-12z = 35$
• $6x+7y-9z = 70$

which also represent the cartesian equation for the line.

What we can find is the parametric equation for the line $P(t)=P_0+t\vec w$ where $P_0$ is a point which belongs to the line and $\vec w$ is the direction vector.

To find the parametric equation we can follow (at least) two different methods:

1. Find one point $P_0$ as a solution of the system of equations of the two planes and direction vector $\vec w$ by cross product of the normal vectors of the two planes, that is $\vec w=\vec v_1\times \vec v_2$

2. Find two different points $P$ and $Q$ from the system of equations of the two planes and assume $P_0=P$ and the direction vector $\vec w=Q-P$

Answer 4

Taking a projective-geometric perspective, the two planes are represented by the homogeneous vectors $\mathbf\pi_1 = [8:1:-12:-35]$ and $\mathbf\pi_2=[6:7:-9:-70]$. The points on each plane are the solutions to the equations $\mathbf\pi_1\cdot\mathbf x=0$ and $\mathbf\pi_2\cdot\mathbf x=0$, respectively, so the intersection of the planes is the null space of the matrix $$\begin{bmatrix}\mathbf\pi_1\\\mathbf\pi_2\end{bmatrix} = \begin{bmatrix}8&1&-12&-35\\6&7&-9&-70\end{bmatrix}.$$ Using any of the usual methods, such as Gaussian elimination, you can find that the null space is spanned by $\mathbf p_1 =[7:14:0:2]$ and $\mathbf p_2=[3:0:2:0]$, i.e., every point $\mathbf p$ in the intersection is a linear combination $s\mathbf p_1+t\mathbf p_2$ of these two points. Fix $s=\frac12$ and convert back to inhomogeneous Cartesian coordinates to get a parametric equation for the line of intersection.

Note that this method is essentially equivalent to solving the system of linear of equations of the two planes.