JEE Main 2023 (01 February Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Three Dimensional Geometry: 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 :

Q: Why does the parameter $\lambda$ disappear after we use the dot product condition?

The parameter $\lambda$ is the 'degree of freedom' of your plane. By imposing the condition that the plane must be perpendicular to a third plane, you are effectively 'locking' the plane into one specific orientation. That is why the math forces $\lambda$ to take a single, unique value.

Q: Is it possible for the denominator in the distance formula to be zero?

Mathematically, the denominator is $\sqrt{a^2 + b^2 + c^2}$. Since $a, b,$ and $c$ are the components of the normal vector, and a normal vector cannot be the zero vector (otherwise it wouldn't define a plane), the denominator will always be positive and non-zero. You are safe!

Q: I got $\lambda = -8$. Does the sign of $\lambda$ matter?

The sign is crucial! If you had calculated $\lambda$ incorrectly, your plane equation would represent a completely different plane in space. Always double-check your arithmetic when expanding the dot product—it is the most common place for a sign error to creep in.

Select Answer:

Visualized Solution

The Family of Planes

Let $P_{1} : 2 x + 3 y - z - 2 = 0$
Let $P_{2} : x + 2 y + 3 z - 6 = 0$
Equation of any plane passing through their intersection:
$P_{1} + λ P_{2} = 0$

Substituting the Plane Equations

$(2 x + 3 y - z - 2) + λ (x + 2 y + 3 z - 6) = 0$

Grouping the Variables

Grouping $x$ , $y$ , and $z$ terms:
$(2 + λ) x + (3 + 2 λ) y + (- 1 + 3 λ) z - (2 + 6 λ) = 0$

Identifying the Normal Vector

The normal vector of plane $P$ is:
$n_{P} = (2 + λ) \hat{i} + (3 + 2 λ) \hat{j} + (- 1 + 3 λ) \hat{k}$

The Perpendicularity Condition

Plane $P$ is perpendicular to plane $P_{3}$ :
$P_{3} : 2 x + y - z + 1 = 0$
Normal of $P_{3}$ : $n_{3} = 2 \hat{i} + 1 \hat{j} - 1 \hat{k}$

Dot Product of Normals

For perpendicular planes, the dot product of their normals is zero:
$n_{P} \cdot n_{3} = 0$

Setting up the Dot Product

$(2 + λ) (2) + (3 + 2 λ) (1) + (- 1 + 3 λ) (- 1) = 0$

Solving for $λ$

$4 + 2 λ + 3 + 2 λ + 1 - 3 λ = 0$
$(2 λ + 2 λ - 3 λ) + (4 + 3 + 1) = 0$
$λ + 8 = 0$
$λ = - 8$

The Exact Equation of Plane $P$

Substitute $λ = - 8$ back into the grouped equation:
$(2 - 8) x + (3 - 16) y + (- 1 - 24) z - (2 - 48) = 0$
$- 6 x - 13 y - 25 z + 46 = 0$
$6 x + 13 y + 25 z - 46 = 0$

Distance from a Point to the Plane

We need the distance $d$ from point $Q (- 7, 1, 1)$ to plane $P$ .
Distance formula:
$d = \frac{∣ a x _{1} + b y _{1} + c z _{1} + D ∣}{a ^{2} + b ^{2} + c ^{2}}$

Substituting into the Distance Formula

$d = \frac{∣6 ( - 7 ) + 13 ( 1 ) + 25 ( 1 ) - 46∣}{6 ^{2} + 1 3 ^{2} + 2 5 ^{2}}$

Calculating the Numerator

Numerator $= ∣ - 42 + 13 + 25 - 46∣$
Numerator $= ∣ - 42 + 38 - 46∣$
Numerator $= ∣ - 50∣ = 50$

Calculating the Denominator

Denominator $= 36 + 169 + 625$
Denominator $= 205 + 625$
Denominator $= 830$

Final Calculation for $d^{2}$

$d = \frac{50}{830}$
The question asks for $d^{2}$ :
$d^{2} = \frac{2500}{830} = \frac{250}{83}$

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The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

The intersection of two planes

P_{1} : 2 x + 3 y - z - 2 = 0

and

P_{2} : x + 2 y + 3 z - 6 = 0

defines a line. Any plane passing through this line can be represented by the family of planes equation:

(2 x + 3 y - z - 2) + λ (x + 2 y + 3 z - 6) = 0

By rearranging the terms, we group the coefficients of

x

y

, and

z

(2 + λ) x + (3 + 2 λ) y + (- 1 + 3 λ) z - (2 + 6 λ) = 0

Here,

λ

acts as a parameter that rotates the plane around the fixed line of intersection.

Locking the Orientation

We require this plane to be perpendicular to the third plane

P_{3} : 2 x + y - z + 1 = 0

. The normal vector of our target plane is

n_{P} = (2 + λ, 3 + 2 λ, - 1 + 3 λ)

, and the normal vector of

P_{3}

n_{3} = (2, 1, - 1)

Since the planes are perpendicular, the dot product of their normal vectors must be zero:

n_{P} \cdot n_{3} = 0

Substituting the components, we obtain:

(2 + λ) (2) + (3 + 2 λ) (1) + (- 1 + 3 λ) (- 1) = 0

Expanding this expression yields:

4 + 2 λ + 3 + 2 λ + 1 - 3 λ = 0

Solving for

λ

, we find:

λ + 8 = 0 \Rightarrow λ = - 8

Determining the Plane Equation

Substituting

λ = - 8

back into the family of planes equation:

(2 - 8) x + (3 - 16) y + (- 1 - 24) z - (2 - 48) = 0

This simplifies to the final equation of the plane:

- 6 x - 13 y - 25 z + 46 = 0 or 6 x + 13 y + 25 z - 46 = 0

Final Calculation

We now calculate the perpendicular distance

d

from the point

Q (- 7, 1, 1)

to the plane

6 x + 13 y + 25 z - 46 = 0

using the formula:

d = \frac{∣ a x _{1} + b y _{1} + c z _{1} + D ∣}{a ^{2} + b ^{2} + c ^{2}}

Substituting the coordinates of

Q

and the coefficients of the plane:

d = \frac{∣6 ( - 7 ) + 13 ( 1 ) + 25 ( 1 ) - 46∣}{6 ^{2} + 1 3 ^{2} + 2 5 ^{2}}

d = \frac{∣ - 42 + 13 + 25 - 46∣}{36 + 169 + 625} = \frac{∣ - 50∣}{830} = \frac{50}{830}

The problem requires the value of

d^{2}

d^{2} = \frac{2500}{830} = \frac{250}{83}

The final result is $\frac{250}{83}$ .

Common Traps & FAQs

Wait, 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?

You absolutely could! You could find the direction vector of the line of intersection using the cross product of the normals of

P_{1}

and

P_{2}

, find a point on that line, and then construct the plane. However, the 'Family of Planes' method is the elegant shortcut—it encapsulates all possible planes passing through that line into one single algebraic expression, saving you from tedious vector geometry.

How do I know if the dot product of the normals is zero or if the cross product is involved?

Great question! Remember: if two planes are perpendicular, their normal vectors are also perpendicular. The condition for perpendicular vectors is that their dot product is zero. If the problem had asked for a plane *parallel* to another, you would look for proportional normal vectors.

Why does the parameter

λ

disappear after we use the dot product condition?

The parameter

λ

is the 'degree of freedom' of your plane. By imposing the condition that the plane must be perpendicular to a third plane, you are effectively 'locking' the plane into one specific orientation. That is why the math forces

λ

to take a single, unique value.

Is it possible for the denominator in the distance formula to be zero?

Mathematically, the denominator is

a^{2} + b^{2} + c^{2}

. Since

a, b,

and

c

are the components of the normal vector, and a normal vector cannot be the zero vector (otherwise it wouldn't define a plane), the denominator will always be positive and non-zero. You are safe!

I got

λ = - 8

. Does the sign of

λ

matter?

The sign is crucial! If you had calculated

λ

incorrectly, your plane equation would represent a completely different plane in space. Always double-check your arithmetic when expanding the dot product—it is the most common place for a sign error to creep in.

Select Answer:

Visualized Solution

The Family of Planes

Substituting the Plane Equations

Grouping the Variables

Identifying the Normal Vector

The Perpendicularity Condition

Dot Product of Normals

Setting up the Dot Product

Solving for $λ$

The Exact Equation of Plane $P$

Distance from a Point to the Plane

Substituting into the Distance Formula

Calculating the Numerator

Calculating the Denominator

Final Calculation for $d^{2}$

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Locking the Orientation

Determining the Plane Equation

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

The Family of Planes

Substituting the Plane Equations

Grouping the Variables

Identifying the Normal Vector

The Perpendicularity Condition

Dot Product of Normals

Setting up the Dot Product

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

The Exact Equation of Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Distance from a Point to the Plane

Substituting into the Distance Formula

Calculating the Numerator

Calculating the Denominator

Final Calculation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d2

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Locking the Orientation

Determining the Plane Equation

Final Calculation

Similar Questions

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The Exact Equation of Plane $P$

Final Calculation for $d^{2}$

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