JEE Main 2023 (24 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: Let the plane containing the line of intersection of the planes $P 1 : x + (λ + 4) y + z = 1$ and $P 2 : 2 x + y + z = 2$ pass through the points $(0, 1, 0)$ and $(1, 0, 1)$ . Then the distance of the point $(2 λ, λ, - λ)$ from the plane $P 2$ is

Q: Wait, why do we get two different expressions for $k$?

Actually, we don't! We get a relationship between $k$ and $\lambda$ from the first point, and a fixed value for $k$ from the second point. By combining these, we lock down the value of $\lambda$. Think of it as tightening the screws on a hinge until the plane is perfectly positioned.

Select Answer:

Visualized Solution

The Intersecting Planes

Given planes: $P_{1} : x + (λ + 4) y + z = 1$
And $P_{2} : 2 x + y + z = 2$
These planes intersect along a straight line.

Family of Planes

Equation of any plane passing through this intersection is:
$P_{1} + k P_{2} = 0$
Let's call this new plane $P_{3}$ .

Setting up the Equation

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

Applying the First Point

Plane $P_{3}$ passes through $(0, 1, 0)$ .
Substitute $x = 0, y = 1, z = 0$ :
$(0 + (λ + 4) (1) + 0 - 1) + k (2 (0) + 1 + 0 - 2) = 0$

First Equation for $k$

Simplify the expression:
$(λ + 4 - 1) + k (1 - 2) = 0$
$λ + 3 - k = 0 ⟹ k = λ + 3$

Applying the Second Point

Plane $P_{3}$ also passes through $(1, 0, 1)$ .
Substitute $x = 1, y = 0, z = 1$ :
$(1 + (λ + 4) (0) + 1 - 1) + k (2 (1) + 0 + 1 - 2) = 0$

Second Equation for $k$

Simplify the expression:
$(1 + 0 + 1 - 1) + k (2 + 0 + 1 - 2) = 0$
$1 + k (1) = 0 ⟹ k = - 1$

Finding $λ$

Equate the two expressions for $k$ :
$λ + 3 = - 1$
$λ = - 4$

The Target Point

The target point is given as $(2 λ, λ, - λ)$ .
Substitute $λ = - 4$ :
Point $Q = (2 (- 4), - 4, - (- 4))$
$Q = (- 8, - 4, 4)$

Distance Formula

Distance $d$ of point $(x_{1}, y_{1}, z_{1})$ from plane $a x + b y + cz + d = 0$ is:
$d = \frac{∣ a x _{1} + b y _{1} + c z _{1} + d ∣}{a ^{2} + b ^{2} + c ^{2}}$

Substituting Values

Point $Q (- 8, - 4, 4)$ and Plane $P_{2} : 2 x + y + z - 2 = 0$
$d = \frac{∣2 ( - 8 ) + 1 ( - 4 ) + 1 ( 4 ) - 2∣}{2 ^{2} + 1 ^{2} + 1 ^{2}}$

Calculating the Distance

Numerator: $∣ - 16 - 4 + 4 - 2∣ = ∣ - 18∣ = 18$
Denominator: $4 + 1 + 1 = 6$
$d = \frac{18}{6}$

Final Answer

Rationalize the denominator:
$d = \frac{18}{6} \times \frac{6}{6} = \frac{18 6}{6}$
$d = 36$
Correct Option: (3)

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The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

The problem involves two intersecting planes,

P_{1} : x + (λ + 4) y + z - 1 = 0

and

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

. Any plane passing through the line of intersection of these two planes can be represented by the Family of Planes equation:

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

This equation represents all possible planes passing through the intersection line, where

k

is a parameter that determines the specific orientation of the plane.

The Constraint Hunt

To identify the specific plane, we utilize the two given points

(0, 1, 0)

and

(1, 0, 1)

that lie on the required plane. Substituting the point

(0, 1, 0)

into the family equation:

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

This simplifies to:

λ + 3 - k = 0 \Rightarrow k = λ + 3

Next, we substitute the point

(1, 0, 1)

into the family equation:

(1 + 0 + 1 - 1) + k (2 + 0 + 1 - 2) = 0

This simplifies to:

1 + k = 0 \Rightarrow k = - 1

The Parameter Reveal

By equating the two expressions found for

k

, we solve for the unknown parameter

λ

λ + 3 = - 1 \Rightarrow λ = - 4

With

λ

determined, we can identify the coordinates of the target point

Q (2 λ, λ, - λ)

. Substituting

λ = - 4

Q = (2 (- 4), - 4, - (- 4)) = (- 8, - 4, 4)

The Final Distance

We now calculate the perpendicular distance from point

Q (- 8, - 4, 4)

to the plane

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

using the standard distance formula:

d = \frac{∣ a x _{0} + b y _{0} + c z _{0} + d ∣}{a ^{2} + b ^{2} + c ^{2}}

Plugging in the values:

d = \frac{∣2 ( - 8 ) + 1 ( - 4 ) + 1 ( 4 ) - 2∣}{2 ^{2} + 1 ^{2} + 1 ^{2}}

The numerator simplifies as follows:

∣ - 16 - 4 + 4 - 2∣ = ∣ - 18∣ = 18

The denominator is:

4 + 1 + 1 = 6

Thus, the distance is:

d = \frac{18}{6} = \frac{18 6}{6} = 36

The final distance is $36$ .

Common Traps & FAQs

Why do we use

P_{1} + k P_{2} = 0

instead of finding the line of intersection first?

Finding the line of intersection is like finding the spine of the book. While possible, it is a long, winding road involving cross products and parametric forms. The 'Family of Planes' method is a powerful shortcut—it lets us work with the entire plane directly without ever needing to explicitly define the line.

Wait, why do we get two different expressions for

k

Actually, we don't! We get a relationship between

k

and

λ

from the first point, and a fixed value for

k

from the second point. By combining these, we lock down the value of

λ

. Think of it as tightening the screws on a hinge until the plane is perfectly positioned.

Is the distance formula always positive?

Yes! Distance is a physical reality, and it cannot be negative. The absolute value bars in the formula

∣ a x_{0} + b y_{0} + c z_{0} + d ∣

are there to ensure that no matter which side of the plane the point lies on, the distance remains a positive magnitude.

Select Answer:

Visualized Solution

The Intersecting Planes

Family of Planes

Setting up the Equation

Applying the First Point

First Equation for $k$

Applying the Second Point

Second Equation for $k$

Finding $λ$

The Target Point

Distance Formula

Substituting Values

Calculating the Distance

Final Answer

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Constraint Hunt

The Parameter Reveal

The Final Distance

Similar Questions

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

Let the plane $P$ pass through the intersection of the planes $2 x + 3 y - z = 2$ and $x + 2 y + 3 z = 6$ , and be perpendicular to the plane $2 x + y - z + 1 = 0$ . If $d$ is the distance of $P$ from the point $(- 7, 1, 1)$ , then $d^{2}$ is equal to :

If $L_{1}$ is the line of intersection of the planes $2 x - 2 y + 3 z - 2 = 0, x - y + z + 1 = 0$ and $L_{2}$ is the line of intersection of the planes $x + 2 y - z - 3 = 0, 3 x - y + 2 z - 1 = 0$ , then the distance of the origin from the plane, containing the lines $L_{1}$ and $L_{2}$ , is :

If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

If the equation of a plane $P$ , passing through the intersection of the planes $x + 4 y - z + 7 = 0$ and $3 x + y + 5 z = 8$ is $a x + b y + 6 z = 15$ for some $a, b \in R$ , then the distance of the point $(3, 2, - 1)$ from the plane $P$ is

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 :

A plane $E$ is perpendicular to the two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , and passes through the point $P (1, - 1, 1)$ . If the distance of the plane $E$ from the point $Q (a, a, 2)$ is $32$ , then $(P Q)^{2}$ is equal to

The distance of the point $(1, 3, - 7)$ from the plane passing through the point $(1, - 1, - 1)$ , having normal perpendicular to both the lines $\frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}$ and $\frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}$ , is:

Let P be the plane, which contains the line of intersection of the planes, $x + y + z - 6 = 0$ and $2 x + 3 y + z + 5 = 0$ and it is perpendicular to the xy-plane. Then the distance of the point $(0, 0, 256)$ from P is equal to :-

The perpendicular distance from the origin to the plane containing the two lines, $\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}$ and $\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}$ , is:

Select Answer:

Visualized Solution

The Intersecting Planes

Family of Planes

Setting up the Equation

Applying the First Point

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

Applying the Second Point

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

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

The Target Point

Distance Formula

Substituting Values

Calculating the Distance

Final Answer

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Constraint Hunt

The Parameter Reveal

The Final Distance

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane which is perpendicular to two planes 2x−2y+z=0 and x−y+2z=4, passes through (1,−2,1). The distance of the plane from the point (1,2,2) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the distance between the plane Ax−2y+z=d and the plane containing the lines 2x−1​=3y−2​=4z−3​ and 3x−2​=4y−3​=5z−4​ is 6​, then find ∣d∣.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of a plane at a distance 212​​ from the origin, which contains the line of intersection of the planes x−y−z−1=0 and 2x+y−3z+4=0 is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The perpendicular distance from the origin to the plane containing the two lines, 3x+2​=5y−2​=7z+5​ and 1x−1​=4y−4​=7z+4​, is:

First Equation for $k$

Second Equation for $k$

Finding $λ$

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

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 perpendicular distance from the origin to the plane containing the two lines, $\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}$ and $\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}$ , is: