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 $\lambda$?
Best Answer
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 $A_2a + B_2b + C_2c - D_2$.