Find the equation of plane through intersection of two planes

Question

Find the equation of the plane passing through the line of intersection of the planes $2x + 3y - z = 1$ and $x + y + 2z = 3$ , and also passing through the point $(1, 2, -1)$ .

(CBSE 2024, similar pattern)

Solution — Step by Step

Any plane through the intersection of planes $P_1: 2x + 3y - z - 1 = 0$ and $P_2: x + y + 2z - 3 = 0$ can be written as:

P_1 + \lambda P_2 = 0

(2x + 3y - z - 1) + \lambda(x + y + 2z - 3) = 0

This is a one-parameter family — we need one condition (the given point) to find $\lambda$ .

The plane passes through $(1, 2, -1)$ . Substitute $x = 1, y = 2, z = -1$ :

(2(1) + 3(2) - (-1) - 1) + \lambda(1 + 2 + 2(-1) - 3) = 0

(2 + 6 + 1 - 1) + \lambda(1 + 2 - 2 - 3) = 0

8 + \lambda(-2) = 0

\lambda = 4

Substitute $\lambda = 4$ back:

(2x + 3y - z - 1) + 4(x + y + 2z - 3) = 0

2x + 3y - z - 1 + 4x + 4y + 8z - 12 = 0

6x + 7y + 7z - 13 = 0

Answer: $\boxed{6x + 7y + 7z = 13}$

Check that $(1, 2, -1)$ satisfies the equation: $6(1) + 7(2) + 7(-1) = 6 + 14 - 7 = 13$ ✓

Check that the plane passes through the intersection line by verifying any point on the intersection satisfies our equation. If $(x_0, y_0, z_0)$ satisfies both original planes, then both $P_1 = 0$ and $P_2 = 0$ , so $P_1 + 4P_2 = 0$ ✓

Why This Works

The equation $P_1 + \lambda P_2 = 0$ represents the family of planes through the intersection of $P_1$ and $P_2$ . Why? Because any point on the line of intersection satisfies both $P_1 = 0$ and $P_2 = 0$ , so it automatically satisfies $P_1 + \lambda P_2 = 0$ for every value of $\lambda$ .

Different values of $\lambda$ give different planes, but all share the same intersection line. The additional condition (passing through a specific point, being parallel to a line, etc.) determines the unique value of $\lambda$ .

This technique works for any type of additional condition — not just a point. You could also be given that the plane is perpendicular to another plane, or that it makes a specific angle with a coordinate axis.

Alternative Method

You could find the direction of the intersection line first (cross product of normals), then find a point on the line, then use the point and direction with the given point to find the plane equation. But the $\lambda$ -method above is much faster and cleaner.

This $P_1 + \lambda P_2 = 0$ technique is one of the most powerful in 3D geometry for CBSE and JEE. It applies to planes, lines (in 2D: $L_1 + \lambda L_2 = 0$ gives circles/conics through intersection of two curves), and even second-degree curves. Master this one technique and you’ll handle a huge class of problems.

Common Mistake

When substituting the point to find $\lambda$ , students sometimes forget to move the constant term to the left side. The plane equation must be in the form $ax + by + cz + d = 0$ (with the constant on the left) before forming $P_1 + \lambda P_2$ . If you write $P_1: 2x + 3y - z = 1$ and substitute directly without moving the 1, your $\lambda$ value will be wrong. Always rewrite as $2x + 3y - z - 1 = 0$ first.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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