JEE Main 2021 (27 Aug Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Three Dimensional Geometry: 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 :

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the two planes as two pages of an open book meeting at the spine. The 'spine' is the line of intersection. Any plane passing through this spine can be represented by the equation

P_{1} + λ P_{2} = 0

. This is the 'Family of Planes' concept. Think of

λ

as a parameter that rotates the plane around this fixed spine.

Hint 2

Hint 2: You have the general equation of the plane in terms of

λ

. Now, recall the distance formula from a point

(x_{0}, y_{0}, z_{0})

to a plane

A x + B y + C z + D = 0

, which is

d = \frac{∣ A x _{0} + B y _{0} + C z _{0} + D ∣}{A ^{2} + B ^{2} + C ^{2}}

. Since our point is the origin

(0, 0, 0)

, the formula simplifies significantly.

Hint 3

Hint 3: Substitute the coefficients

A, B, C,

and

D

(which are functions of

λ

) into the distance formula. Set this equal to

\frac{2}{21}

. Square both sides to eliminate the square root and the absolute value, leading to a quadratic equation in

λ

. Solve for

λ

and substitute it back into the family equation.

Visualized Solution

Visualizing the Intersecting Planes

Given Plane 1: $P_{1} : x - y - z - 1 = 0$
Given Plane 2: $P_{2} : 2 x + y - 3 z + 4 = 0$
These two planes intersect along a straight line.

Family of Planes

Any plane passing through the intersection of $P_{1}$ and $P_{2}$ belongs to a "family of planes".
Equation: $P_{1} + λ P_{2} = 0$
Here, $λ$ is a real parameter that defines the specific tilt of the plane.

Substituting the Plane Equations

Substitute $P_{1}$ and $P_{2}$ into the family equation:
$(x - y - z - 1) + λ (2 x + y - 3 z + 4) = 0$

Grouping the Variables

Group terms by $x, y,$ and $z$ :
$x (1 + 2 λ) + y (- 1 + λ) + z (- 1 - 3 λ) + (- 1 + 4 λ) = 0$
Standard form: $A x + B y + C z + D = 0$

The Distance Constraint

The problem states the plane is at a specific distance from the origin $(0, 0, 0)$ .
Given distance: $d = \frac{2}{21}$

Distance Formula from Origin

The perpendicular distance $d$ from the origin $(0, 0, 0)$ to a plane $A x + B y + C z + D = 0$ is:
$d = \frac{∣ D ∣}{A ^{2} + B ^{2} + C ^{2}}$

Applying the Distance Formula

Substitute $A = (1 + 2 λ)$ , $B = (λ - 1)$ , $C = - (1 + 3 λ)$ , $D = (4 λ - 1)$ :
$\frac{∣4 λ - 1∣}{( 1 + 2 λ ) ^{2} + ( λ - 1 ) ^{2} + ( - ( 1 + 3 λ ) ) ^{2}} = \frac{2}{21}$

Squaring Both Sides

To eliminate the square roots and absolute value, square both sides:
$\frac{( 4 λ - 1 ) ^{2}}{( 1 + 2 λ ) ^{2} + ( λ - 1 ) ^{2} + ( 1 + 3 λ ) ^{2}} = \frac{2}{21}$

Expanding the Denominator

Expand the terms in the denominator:
$(1 + 4 λ^{2} + 4 λ) + (λ^{2} - 2 λ + 1) + (1 + 9 λ^{2} + 6 λ)$
Combine like terms:
$14 λ^{2} + 8 λ + 3$

Expanding the Numerator and Cross-Multiplying

Expand the numerator: $(4 λ - 1)^{2} = 16 λ^{2} - 8 λ + 1$
The equation becomes: $\frac{16 λ ^{2} - 8 λ + 1}{14 λ ^{2} + 8 λ + 3} = \frac{2}{21}$
Cross-multiply:
$21 (16 λ^{2} - 8 λ + 1) = 2 (14 λ^{2} + 8 λ + 3)$

Forming the Quadratic Equation

Distribute the constants:
$336 λ^{2} - 168 λ + 21 = 28 λ^{2} + 16 λ + 6$
Bring all terms to one side:
$(336 - 28) λ^{2} + (- 168 - 16) λ + (21 - 6) = 0$
$308 λ^{2} - 184 λ + 15 = 0$

Solving for $λ$

Factorize the quadratic equation $308 λ^{2} - 184 λ + 15 = 0$ :
Split the middle term: $- 184 λ = - 154 λ - 30 λ$
$308 λ^{2} - 154 λ - 30 λ + 15 = 0$
$154 λ (2 λ - 1) - 15 (2 λ - 1) = 0$
$(2 λ - 1) (154 λ - 15) = 0$
Roots: $λ = \frac{1}{2}$ or $λ = \frac{15}{154}$

Substituting $λ$ Back

Let's check the options. They have simple integer coefficients, suggesting $λ = \frac{1}{2}$ .
Substitute $λ = \frac{1}{2}$ into the grouped plane equation:
$x (1 + 2 (\frac{1}{2})) + y (- 1 + \frac{1}{2}) + z (- 1 - 3 (\frac{1}{2})) + (- 1 + 4 (\frac{1}{2})) = 0$

Simplifying the Final Equation

Simplify the terms inside the brackets:
$x (1 + 1) + y (- \frac{1}{2}) + z (- 1 - \frac{3}{2}) + (- 1 + 2) = 0$
$2 x - \frac{1}{2} y - \frac{5}{2} z + 1 = 0$

Final Plane Equation

Multiply the entire equation by $2$ to remove fractions:
$4 x - y - 5 z + 2 = 0$
This matches Option 4.

00:00 / 00:00

The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

Imagine you are standing in a vast, empty 3D space. Before you, two flat, infinite sheets of paper—our planes

P_{1}

and

P_{2}

—intersect. They define a unique line, a 'spine' around which an infinite number of other planes could rotate, like pages in a book.

We are tasked with finding one specific page in that infinite book that sits at a precise distance of

\frac{2}{21}

from the origin. We utilize the 'Family of Planes' method, defined by the equation

P_{1} + λ P_{2} = 0

Think of

λ

as a tuning dial. As you turn it, the plane pivots around the intersection line. Our mission is to find the exact setting of this dial.

The Algebraic Bridge

We start by writing the family equation:

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

We group the terms by their variables to tame the expression:

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

This takes the form of a standard plane equation

A x + B y + C z + D = 0

, where:

A = 1 + 2 λ

B = λ - 1

C = - (1 + 3 λ)

, and

D = 4 λ - 1

The Distance Constraint

We are given that the plane is at a distance

d = \frac{2}{21}

from the origin

(0, 0, 0)

. The distance formula is our primary tool:

d = \frac{∣ D ∣}{A ^{2} + B ^{2} + C ^{2}}

Substituting our

λ

-dependent coefficients, we obtain:

\frac{∣4 λ - 1∣}{( 1 + 2 λ ) ^{2} + ( λ - 1 ) ^{2} + ( - ( 1 + 3 λ ) ) ^{2}} = \frac{2}{21}

By squaring both sides, we strip away the complexity of the modulus and the square root:

\frac{( 4 λ - 1 ) ^{2}}{( 1 + 2 λ ) ^{2} + ( λ - 1 ) ^{2} + ( 1 + 3 λ ) ^{2}} = \frac{2}{21}

The Final Grind

Expanding the denominator yields

14 λ^{2} + 8 λ + 3

, while the numerator is

16 λ^{2} - 8 λ + 1

. Cross-multiplying gives:

21 (16 λ^{2} - 8 λ + 1) = 2 (14 λ^{2} + 8 λ + 3)

After careful distribution and grouping, we arrive at the quadratic equation:

308 λ^{2} - 184 λ + 15 = 0

Factoring this quadratic, we find:

(2 λ - 1) (154 λ - 15) = 0

This yields two possible values for the parameter:

λ = \frac{1}{2}

λ = \frac{15}{154}

Final Calculation

Choosing

λ = \frac{1}{2}

and substituting it back into our grouped equation, we simplify the coefficients:

2 x - \frac{1}{2} y - \frac{5}{2} z + 1 = 0

Multiplying by

2

to clear the fractions, we obtain the final equation of the plane: $4 x - y - 5 z + 2 = 0$

We have arrived. The plane is found, the geometry is satisfied, and the math is complete. You have successfully navigated the intersection of two worlds—the visual and the algebraic.

Common Traps & FAQs

Why do we use the family of planes equation

P_{1} + λ P_{2} = 0

? Can't we just find the line of intersection first?

That is a great question! While you *could* find the line of intersection by setting one variable to zero or finding the direction vector, it is computationally tedious and prone to error. The 'Family of Planes' method is a powerful shortcut. It encapsulates all possible planes passing through that line into a single variable

λ

. It is the elegant way to solve this, saving you time and mental energy for the actual calculation.

Why do we get two values for

λ

? Does that mean there are two answers?

Exactly! Geometrically, there are two planes at a specific distance from the origin that pass through a given line of intersection—one on either side of the line. In this problem, we found two values for

λ

. Usually, one will match the options provided, or both might be valid. Always check the options to see which one is presented.

Why do we square both sides of the distance equation? Is there a risk of introducing extraneous roots?

Squaring is a standard algebraic maneuver to remove the modulus (absolute value) and the square root in the denominator. While squaring *can* introduce extraneous roots in some contexts, here we are solving for a parameter

λ

that defines a physical plane. If you get two valid real roots for

λ

, both are mathematically correct. If you are worried about extraneous roots, you can always plug your final

λ

values back into the distance formula to verify they yield the correct distance.

Animated Solution for Mathematics - Three Dimensional Geometry: 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 :

Select Answer:

Visualized Solution

Visualizing the Intersecting Planes

Family of Planes

Substituting the Plane Equations

Grouping the Variables

The Distance Constraint

Distance Formula from Origin

Applying the Distance Formula

Squaring Both Sides

Expanding the Denominator

Expanding the Numerator and Cross-Multiplying

Forming the Quadratic Equation

Solving for $λ$

Substituting $λ$ Back

Simplifying the Final Equation

Final Plane Equation

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Algebraic Bridge

The Distance Constraint

The Final Grind

Final Calculation

Similar Questions

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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

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Select Answer:

Visualized Solution

Visualizing the Intersecting Planes

Family of Planes

Substituting the Plane Equations

Grouping the Variables

The Distance Constraint

Distance Formula from Origin

Applying the Distance Formula

Squaring Both Sides

Expanding the Denominator

Expanding the Numerator and Cross-Multiplying

Forming the Quadratic Equation

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

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

Simplifying the Final Equation

Final Plane Equation

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Algebraic Bridge

The Distance Constraint

The Final Grind

Final Calculation

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Substituting $λ$ Back

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