# An equation for a pencil of planes

A family of planes intersecting in a straight line is called a pencil of planes. Any two nonparallel planes are part of a pencil of planes. If the two nonparallel planes have equations $$A_1x + B_1y + C_1z = D_1 \quad \text{and} \quad A_2x + B_2y + C_2z = D_2,$$ then, for any value of $$\lambda \in \mathbb{R}$$, the equation $$A_1x + B_1y + C_1z - D_1 + \lambda(A_2x + B_2y + C_2z - D_2) = 0$$ represents a plane in the pencil. To see this, observe that the equation is linear, and so represents a plane, and that any point $$(x, y, z)$$ satisfying the equations of both planes (i.e. any point on the line of intersection) also satisfies this equation.

How can it be proved that any plane in the pencil, except $$A_2x + B_2y + C_2z = D_2$$, can be obtained by suitably choosing the value of $$\lambda$$?

To prove that it is possible to obtain any plane in the pencil, except $$A_2x + B_2y + C_2z = D_2$$, by suitably choosing the value of $$\lambda$$ is equivalent to proving that a plane through any point $$(a, b, c)$$, that does not satisfy $$A_2x + B_2y + C_2z = D_2$$, is obtainable from $$A_1x + B_1y + C_1z - D_1 + \lambda(A_2x + B_2y + C_2z - D_2) = 0 \quad(*)$$ by appropriately choosing $$\lambda$$, as all planes can be defined by a line and a point.
Let $$A_1a + B_1b + C_1c - D_1 + \lambda(A_2a + B_2b + C_2c - D_2) = 0.$$ Then, the plane in the pencil that satisfies $$(a, b, c)$$ is given by choosing $$\lambda = \frac{-(A_1a + B_1b + C_1c - D_1)}{A_2a + B_2b + C_2c - D_2}.$$
Note that if $$(a, b, c)$$ would satisfy $$A_2a + B_2b + C_2c - D_2$$ we could not obtain $$\lambda$$ as it would require division by zero.
This proves that $$(*)$$ represents all planes in the pencil, except $$A_2a + B_2b + C_2c - D_2$$.