planes intersection line

Question

There are 4 points: $A(0, -1, 5)$, $B(1, 3, 3)$, $C(5, 4, 0)$ and $D(3, 0, 4)$. The first plane($\pi_1$) contains the points $A$, $B$ and $C$. The second plane ($\pi_2$) contains the points $A$, $B$ and $D$.

Find the Cartesian equation of the line where $\pi_2$ intersects $\pi_1$.

I just need the concept of how to find that line. Should that be a comparison between $\pi_1$ and $\pi_2$ cartesian equations?

Answer 1

Given the two equations of the lines, as you said (in a comment) that you have, in the form $Ex+Fy+Gz=H$ (to borrow Arturo's notation), you know that the vector $\langle E,F,G\rangle$ is orthogonal to the plane. You have one such vector for each plane. Since the line of intersection is in both planes, it is orthogonal to both of these vectors. That means that a vector that is orthogonal to both of the orthogonal-to-the-plane vectors is along the line. The cross-product of the two orthogonal-to-the-plane vectors is orthogonal to both. From this and finding one point that is on the line, you can write a parametric/vector equation of the line of intersection.

Answer 2

Each plane will have an equation of the form $$Ex + Fy + Gz = H$$ for some constants $E$, $F$, $G$, and $H$; a point will lie in the intersection of both planes if and only if it satisfies both equations at the same time, so you'll want to find all solutions to a system of two equations in three unknowns. The set of solutions will give you the line.

To find the equation for a plane given three points that are not collinear, $P$, $Q$, and $R$, you first find the normal vector to the plane. This is a vector that is perpendicular to any vector in the plane. Simplest way to find it is to find two non-collinear vectors that lie in the plane, for example, the vectors $\mathbf{v}_1 = P-Q$ and $\mathbf{v}_2 = R-Q$. With those two vectors, you can compute the normal vector of the plane with $\mathbf{n}=\mathbf{v}_1\times \mathbf{v}_2$ (cross product). If $\mathbf{n}=(a,b,c)$, then the equation of the plane is of the form $ax+by+cz = d$ for some constant $d$. Use the points that are in the plane to find the value of $d$.

You do this for each of the planes; this gives you the two equations of the planes. Then solve the system you get from the two equations; this gives you the line of intersection.