Family of Planes Through Intersection of Two Planes

Every plane through the line of intersection of planes P₁=0 and P₂=0, for varying λ. One additional condition (e.g., passes through a given point, is perpendicular to a given plane) determines λ.

Let $P_1 \equiv a_1x+b_1y+c_1z+d_1=0$ and $P_2 \equiv a_2x+b_2y+c_2z+d_2=0$.

Consider $P_1 + \lambda P_2 = 0$ for $\lambda \in \mathbb{R}$:

$$(a_1+\lambda a_2)x+(b_1+\lambda b_2)y+(c_1+\lambda c_2)z+(d_1+\lambda d_2)=0$$

This is a first-degree equation in $x,y,z$ — a plane.

It passes through the line of intersection: Any point $(x_0,y_0,z_0)$ on the line of intersection satisfies both $P_1(x_0,y_0,z_0)=0$ and $P_2(x_0,y_0,z_0)=0$. Therefore $P_1+\lambda P_2$ is also zero at that point, for any $\lambda$.

This generates all planes through the intersection line (except $P_2=0$ itself, which requires $\lambda\to\infty$; handle by using $\mu P_1+P_2=0$ with $\mu=0$).

How to find $\lambda$: Impose one condition — the plane passes through a given point $(p,q,r)$:

$$P_1(p,q,r)+\lambda P_2(p,q,r) = 0 \implies \lambda = -\frac{P_1(p,q,r)}{P_2(p,q,r)}$$

Or: the plane is perpendicular to a given plane $P_3=0$:

$$(a_1+\lambda a_2)a_3+(b_1+\lambda b_2)b_3+(c_1+\lambda c_2)c_3 = 0$$

Solve for $\lambda$.

Analogy with 2D: This is the 3D version of the family of lines $L_1+\lambda L_2=0$ through the intersection of two lines — the structure is identical.