Why do we use $\lambda$ in the equation $P_1 + \lambda P_2 = 0$?

The parameter $\lambda$ acts as a selector. Since there are infinitely many planes passing through the line of intersection, each unique value of $\lambda$ corresponds to one specific plane from that infinite family. By using a point constraint, we 'lock' $\lambda$ to the specific value that defines our required plane.

Can we use $P_2 + \lambda P_1 = 0$ instead?

Yes, absolutely! You can use $P_2 + \lambda P_1 = 0$ as your starting point. The value of $\lambda$ you calculate will be different (it will be the reciprocal of the $\lambda$ found using the other method), but the final equation of the plane will be identical. The math is consistent regardless of the order.

JEE Main 2019 (8 April Shift 1)

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Animated Solution for Mathematics - Three Dimensional Geometry: The equation of a plane containing the line of intersection of the planes $2 x - y - 4 = 0$ and $y + 2 z - 4 = 0$ and passing through the point $(1, 1, 0)$ is :

Q: What if the plane we are looking for is $P_2$ itself?

The family equation $P_1 + \lambda P_2 = 0$ represents all planes passing through the intersection, except for $P_2$ itself. Mathematically, $P_2$ corresponds to the case where $\lambda$ approaches infinity. In JEE problems, you rarely encounter this edge case, but it is a great conceptual detail to keep in mind!

Select Answer:

Visualized Solution

Visualize the Given Planes

Given Planes:
$P_{1} : 2 x - y - 4 = 0$
$P_{2} : y + 2 z - 4 = 0$

The Family of Planes Concept

Concept: Family of Planes passing through the intersection of $P_{1}$ and $P_{2}$ is given by:
$P_{1} + λ P_{2} = 0$
where $λ$ is a real parameter.

Setting up the Equation

Substituting $P_{1}$ and $P_{2}$ :
$(2 x - y - 4) + λ (y + 2 z - 4) = 0$

Identifying the Passing Point

The required plane passes through the point $(1, 1, 0)$ .
This point must satisfy the family equation.

Substituting the Point

Substitute $x = 1, y = 1, z = 0$ :
$(2 (1) - 1 - 4) + λ (1 + 2 (0) - 4) = 0$

Simplifying to find $λ$

Simplify the terms:
$(2 - 1 - 4) + λ (1 + 0 - 4) = 0$
$- 3 + λ (- 3) = 0$

Solving for $λ$

Solve for $λ$ :
$- 3 λ = 3$
$λ = - 1$

Final Equation Substitution

Substitute $λ = - 1$ back into the family equation:
$(2 x - y - 4) - 1 (y + 2 z - 4) = 0$

Expanding the Equation

Expand the brackets:
$2 x - y - 4 - y - 2 z + 4 = 0$

The Final Answer

Combine like terms:
$2 x - 2 y - 2 z = 0$
Divide by $2$ :
$x - y - z = 0$

00:00 / 00:00

The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

We are examining the intersection of two planes,

P_{1} : 2 x - y - 4 = 0

and

P_{2} : y + 2 z - 4 = 0

. When these two planes meet, they form an infinite line of intersection.

Imagine standing in a room where two walls meet at a corner; that corner represents our line of intersection.

The Book Analogy

The Family of Planes

Think of this line of intersection as the spine of a book. Every page of that book is a plane, and all those pages share the same spine. This collection is known as a Family of Planes.

To represent all possible planes passing through this line with one elegant algebraic expression, we use the concept of a linear combination:

P_{1} + λ P_{2} = 0

Here,

λ

is our magic parameter. By changing

λ

, we rotate through every possible plane in the family, effectively flipping through the pages of our book.

Locking the Target

The Power of a Point

We are searching for the specific plane that passes through the point

(1, 1, 0)

. Because this point lies on the required plane, its coordinates must satisfy the equation of the plane.

We substitute our planes into the family equation:

(2 x - y - 4) + λ (y + 2 z - 4) = 0

Now, we substitute

x = 1

y = 1

, and

z = 0

into this equation:

(2 (1) - 1 - 4) + λ (1 + 2 (0) - 4) = 0

The Final Calculation

Elegant Simplification

Simplifying the terms inside the brackets, we get:

(2 - 1 - 4) + λ (1 - 4) = 0

- 3 - 3 λ = 0

Solving for

λ

, we find:

- 3 λ = 3 \Rightarrow λ = - 1

Now, we substitute

λ = - 1

back into our family equation:

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

Expanding this expression yields:

2 x - y - 4 - y - 2 z + 4 = 0

The constant terms

- 4

and

+ 4

cancel out, leaving us with:

2 x - 2 y - 2 z = 0

Dividing the entire equation by

2

, we arrive at the final result:

$x - y - z = 0$

Common Traps & FAQs

Why do we use

λ

in the equation

P_{1} + λ P_{2} = 0

The parameter

λ

acts as a selector. Since there are infinitely many planes passing through the line of intersection, each unique value of

λ

corresponds to one specific plane from that infinite family. By using a point constraint, we 'lock'

λ

to the specific value that defines our required plane.

Can we use

P_{2} + λ P_{1} = 0

instead?

Yes, absolutely! You can use

P_{2} + λ P_{1} = 0

as your starting point. The value of

λ

you calculate will be different (it will be the reciprocal of the

λ

found using the other method), but the final equation of the plane will be identical. The math is consistent regardless of the order.

What if the plane we are looking for is

P_{2}

itself?

The family equation

P_{1} + λ P_{2} = 0

represents all planes passing through the intersection, except for

P_{2}

itself. Mathematically,

P_{2}

corresponds to the case where

λ

approaches infinity. In JEE problems, you rarely encounter this edge case, but it is a great conceptual detail to keep in mind!

Animated Solution for Mathematics - Three Dimensional Geometry: The equation of a plane containing the line of intersection of the planes $2 x - y - 4 = 0$ and $y + 2 z - 4 = 0$ and passing through the point $(1, 1, 0)$ is :

Select Answer:

Visualized Solution

Visualize the Given Planes

The Family of Planes Concept

Setting up the Equation

Identifying the Passing Point

Substituting the Point

Simplifying to find $λ$

Solving for $λ$

Final Equation Substitution

Expanding the Equation

The Final Answer

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Book Analogy

Locking the Target

The Final Calculation

Similar Questions

Equation of a plane at a distance $\frac{2}{21}$ from the origin, which contains the line of intersection of the planes $x - y - z - 1 = 0$ and $2 x + y - 3 z + 4 = 0$ is :

The equation of the plane passing through the line of intersection of the planes $r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $r \cdot (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4 = 0$ and parallel to the $x$ -axis is:

The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$ is :

The plane through the intersection of the planes $x + y + z = 1$ and $2 x + 3 y - z + 4 = 0$ and parallel to y-axis also passes through the point :

The vector equation of the plane passing through the intersection of the planes $r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $r \cdot (\hat{i} - 2 \hat{j}) = - 2$ , and the point $(1, 0, 2)$ is :

Equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

The equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is:

Let the equation of the plane, that passes through the point $(1, 4, - 3)$ and contains the line of intersection of the planes $3 x - 2 y + 4 z - 7 = 0$ and $x + 5 y - 2 z + 9 = 0$ , be $α x + β y + γ z + 3 = 0$ , then $α + β + γ$ is equal to :

The equation of the plane passing through the point $(1, 2, - 3)$ and perpendicular to the planes $3 x + y - 2 z = 5$ and $2 x - 5 y - z = 7$ , is:

The equation of the plane containing the line $2 x - 5 y + z = 3$ ; $x + y + 4 z = 5$ , and parallel to the plane, $x + 3 y + 6 z = 1$ , is:

Select Answer:

Visualized Solution

Visualize the Given Planes

The Family of Planes Concept

Setting up the Equation

Identifying the Passing Point

Substituting the Point

Simplifying to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Final Equation Substitution

Expanding the Equation

The Final Answer

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Book Analogy

Locking the Target

The Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the plane passing through the line of intersection of the planes r⋅(i^+j^​+k^)=1 and r⋅(2i^+3j^​−k^)+4=0 and parallel to the x-axis is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The vector equation of the plane through the line of intersection of the planes x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x−y+z=0 is :

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The vector equation of the plane passing through the intersection of the planes r⋅(i^+j^​+k^)=1 and r⋅(i^−2j^​)=−2, and the point (1,0,2) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of the plane containing the straight line 2x​=3y​=4z​ and perpendicular to the plane containing the straight lines 3x​=4y​=2z​ and 4x​=2y​=3z​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the plane passing through the point (1,2,−3) and perpendicular to the planes 3x+y−2z=5 and 2x−5y−z=7, is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of the plane containing the line 2x−5y+z=3; x+y+4z=5, and parallel to the plane, x+3y+6z=1, is:

Simplifying to find $λ$

Solving for $λ$

Equation of a plane at a distance $\frac{2}{21}$ from the origin, which contains the line of intersection of the planes $x - y - z - 1 = 0$ and $2 x + y - 3 z + 4 = 0$ is :

The equation of the plane passing through the line of intersection of the planes $r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $r \cdot (2 \hat{i} + 3 \hat{j} - \hat{k}) + 4 = 0$ and parallel to the $x$ -axis is:

The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$ is :

The plane through the intersection of the planes $x + y + z = 1$ and $2 x + 3 y - z + 4 = 0$ and parallel to y-axis also passes through the point :

The vector equation of the plane passing through the intersection of the planes $r \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $r \cdot (\hat{i} - 2 \hat{j}) = - 2$ , and the point $(1, 0, 2)$ is :

Equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

The equation of the plane passing through the point $(1, 2, - 3)$ and perpendicular to the planes $3 x + y - 2 z = 5$ and $2 x - 5 y - z = 7$ , is:

The equation of the plane containing the line $2 x - 5 y + z = 3$ ; $x + y + 4 z = 5$ , and parallel to the plane, $x + 3 y + 6 z = 1$ , is: