In $\mathbb{R}^3$, consider the planes $P_1 : y = 0$ and $P_2 : x + z = 1$. Let $P_3$ be the plane, different from $P_1$ and $P_2$, which passes through the intersection of $P_1$ and $P_2$. If the distance of the point $(0, 1, 0)$ from $P_3$ is 1 and the distance of a point $(\alpha, \beta, \gamma)$ from $P_3$ is 2, then which of the following relations is (are) true? Options: (A) $2\alpha + \beta + 2\gamma + 2 = 0$ (B) $2\alpha - \beta + 2\gamma + 4 = 0$ (C) $2\alpha + \beta - 2\gamma - 10 = 0$ (D) $2\alpha - \beta + 2\gamma - 8 = 0$

### Analyzing the Setup Imagine you are standing in a three-dimensional space with two planes, $P_1$ and $P_2$. $P_1$ is the $xz$-plane, defined by $y = 0$, and $P_2$ is the plane $x + z = 1$. Where they meet, they form a line of intersection. A third plane, $P_3$, is anchored to this line and can rotate around it. We represent this family of planes using the equation $P_2 + \lambda P_1 = 0$. Substituting our given planes, we obtain: $$(x + z - 1) + \lambda(y) = 0$$ This simplifies to the general form: $$x + \lambda y + z - 1 = 0$$ Here, $\lambda$ is the parameter that dictates the specific tilt of the plane $P_3$. ### The Anchor: Using the Distance Constraint We are given a specific point $A(0, 1, 0)$ and told that the perpendicular distance from this point to $P_3$ is exactly $1$. We use the perpendicular distance formula: $$d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$$ Substituting point $A$ and the coefficients from $P_3$ into the formula, we get: $$1 = \frac{|1(0) + \lambda(1) + 1(0) - 1|}{\sqrt{1^2 + \lambda^2 + 1^2}}$$ The numerator simplifies to $|\lambda - 1|$ and the denominator becomes $\sqrt{\lambda^2 + 2}$. Setting this ratio to $1$ and squaring both sides yields: $$(\lambda - 1)^2 = \lambda^2 + 2$$ Expanding the left side, the $\lambda^2$ terms cancel out, leaving the linear equation: $$-2\lambda + 1 = 2 \Rightarrow \lambda = -\frac{1}{2}$$ ### The Final Equation and the Absolute Value Trap Substituting $\lambda = -\frac{1}{2}$ back into the family equation, we get $x - \frac{1}{2}y + z - 1 = 0$. Multiplying by $2$ to clear the fraction, we arrive at the specific equation for $P_3$: $$2x - y + 2z - 2 = 0$$ Now, consider a general point $B(\alpha, \beta, \gamma)$ whose distance to $P_3$ is $2$. Applying the distance formula again: $$\frac{|2\alpha - \beta + 2\gamma - 2|}{\sqrt{2^2 + (-1)^2 + 2^2}} = 2$$ The denominator simplifies to $\sqrt{9} = 3$. Thus, we have: $$\frac{|2\alpha - \beta + 2\gamma - 2|}{3} = 2 \Rightarrow |2\alpha - \beta + 2\gamma - 2| = 6$$ This absolute value splits into two distinct cases: $$2\alpha - \beta + 2\gamma - 2 = 6 \quad \text{and} \quad 2\alpha - \beta + 2\gamma - 2 = -6$$ These simplify to the final relations: **$2\alpha - \beta + 2\gamma - 8 = 0$** and **$2\alpha - \beta + 2\gamma + 4 = 0$**

JEE Advanced 2015

LEVELJEE Advanced

Animated Solution for Mathematics - Three Dimensional Geometry: In $R^{3}$ , consider the planes $P_{1} : y = 0$ and $P_{2} : x + z = 1$ . Let $P_{3}$ be the plane, different from $P_{1}$ and $P_{2}$ , which passes through the intersection of $P_{1}$ and $P_{2}$ . If the distance of the point $(0, 1, 0)$ from $P_{3}$ is 1 and the distance of a point $(α, β, γ)$ from $P_{3}$ is 2, then which of the following relations is (are) true?

Select Answer:

* Multiple Correct

Progressive Hints

Hint 1

Hint 1: Visualize the intersection of the two planes

P_{1}

and

P_{2}

. Any plane

P_{3}

passing through this intersection line can be expressed using the family of planes equation:

P_{2} + λ P_{1} = 0

. This parameter

λ

is the key to unlocking the orientation of

P_{3}

Hint 2

Hint 2: Use the standard formula for the perpendicular distance from a point

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

to a plane

a x + b y + cz + d = 0

, which is

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

. Apply this to the point

(0, 1, 0)

and the plane

P_{3}

to find the value of

λ

Hint 3

Hint 3: Once you have the specific equation for

P_{3}

, use the distance formula again for the point

(α, β, γ)

with the given distance of

2

. This will result in an absolute value equation:

∣2 α - β + 2 γ - 2∣ = 6

, which leads to two possible linear relations.

Visualized Solution

Visualizing the Setup of Planes $P_{1}$ and $P_{2}$

We are given two planes in three-dimensional space:
Plane $P_{1} : y = 0$ (which is the $x z$ -plane).
Plane $P_{2} : x + z = 1$ (a plane parallel to the $y$ -axis).
These two planes intersect along a unique line, which we will call $L$ .
Any plane $P_{3}$ passing through this line of intersection belongs to the family of planes defined by $P_{1}$ and $P_{2}$ .

Formulating the Family of Planes $P_{3}$

The equation of any plane passing through the intersection of $P_{1} = 0$ and $P_{2} = 0$ is given by:
$P_{3} : P_{2} + λ P_{1} = 0$
Substituting the given equations:
$(x + z - 1) + λ (y) = 0$
Rearranging into standard form:
$x + λ y + z - 1 = 0$
Here, $λ$ is a real parameter that determines the orientation of $P_{3}$ as it rotates around the intersection line $L$ .

Setting up the Distance Constraint from $A (0, 1, 0)$

We are given that the perpendicular distance from the point $A (0, 1, 0)$ to $P_{3}$ is $1$ .
Recall the perpendicular distance formula from a point $(x_{1}, y_{1}, z_{1})$ to a plane $a x + b y + cz + d = 0$ :
$d = \frac{∣ a x _{1} + b y _{1} + c z _{1} + d ∣}{a ^{2} + b ^{2} + c ^{2}}$
Substituting $A (0, 1, 0)$ and the plane $P_{3} : 1 x + λ y + 1 z - 1 = 0$ :
$1 = \frac{∣1 ( 0 ) + λ ( 1 ) + 1 ( 0 ) - 1∣}{1 ^{2} + λ ^{2} + 1 ^{2}}$

Simplifying the Distance Equation

Let's simplify the numerator and denominator:
Numerator: $∣0 + λ + 0 - 1∣ = ∣ λ - 1∣$
Denominator: $1 + λ^{2} + 1 = λ^{2} + 2$
This gives us the equation:
$\frac{∣ λ - 1∣}{λ ^{2} + 2} = 1$
To eliminate the square root and absolute value, we square both sides:
$\frac{( λ - 1 ) ^{2}}{λ ^{2} + 2} = 1$

Solving for the Parameter $λ$

Cross-multiplying the terms:
$(λ - 1)^{2} = λ^{2} + 2$
Expanding the left-hand side:
$λ^{2} - 2 λ + 1 = λ^{2} + 2$
Subtracting $λ^{2}$ from both sides:
$- 2 λ + 1 = 2$
Solving for $λ$ :
$- 2 λ = 1 ⟹ λ = - \frac{1}{2}$

Determining the Equation of Plane $P_{3}$

Substitute $λ = - \frac{1}{2}$ back into the family of planes equation:
$x + (- \frac{1}{2}) y + z - 1 = 0$
To eliminate the fraction, multiply the entire equation by $2$ :
$2 x - y + 2 z - 2 = 0$
This is the unique equation of the plane $P_{3}$ .

Distance from Point $B (α, β, γ)$ to $P_{3}$

We are given that the distance from a general point $B (α, β, γ)$ to $P_{3}$ is $2$ .
Using the distance formula again with our plane $2 x - y + 2 z - 2 = 0$ :
$\frac{∣2 α - β + 2 γ - 2∣}{2 ^{2} + ( - 1 ) ^{2} + 2 ^{2}} = 2$
Simplifying the denominator:
$4 + 1 + 4 = 9 = 3$
Substituting back:
$\frac{∣2 α - β + 2 γ - 2∣}{3} = 2$

Deriving the Two Possible Relations

Multiply both sides by $3$ :
$∣2 α - β + 2 γ - 2∣ = 6$
This absolute value equation yields two distinct cases:
Case 1 (Positive sign):
$2 α - β + 2 γ - 2 = 6 ⟹ 2 α - β + 2 γ - 8 = 0$
Case 2 (Negative sign):
$2 α - β + 2 γ - 2 = - 6 ⟹ 2 α - β + 2 γ + 4 = 0$

Matching with Options and Conclusion

Let's compare our derived equations with the given options:
$2 α - β + 2 γ + 4 = 0$ matches Option B.
$2 α - β + 2 γ - 8 = 0$ matches Option D.
Therefore, the correct options are B and D.

00:00 / 00:00

The Sigma Insight: Equation of a Plane

Solution Diagram

Analyzing the Setup

Imagine you are standing in a three-dimensional space with two planes,

P_{1}

and

P_{2}

P_{1}

is the

x z

-plane, defined by

y = 0

, and

P_{2}

is the plane

x + z = 1

Where they meet, they form a line of intersection. A third plane,

P_{3}

, is anchored to this line and can rotate around it.

We represent this family of planes using the equation

P_{2} + λ P_{1} = 0

. Substituting our given planes, we obtain:

(x + z - 1) + λ (y) = 0

This simplifies to the general form:

x + λ y + z - 1 = 0

Here,

λ

is the parameter that dictates the specific tilt of the plane

P_{3}

The Anchor

Using the Distance Constraint

We are given a specific point

A (0, 1, 0)

and told that the perpendicular distance from this point to

P_{3}

is exactly

1

. We use the perpendicular distance formula:

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

Substituting point

A

and the coefficients from

P_{3}

into the formula, we get:

1 = \frac{∣1 ( 0 ) + λ ( 1 ) + 1 ( 0 ) - 1∣}{1 ^{2} + λ ^{2} + 1 ^{2}}

The numerator simplifies to

∣ λ - 1∣

and the denominator becomes

λ^{2} + 2

. Setting this ratio to

1

and squaring both sides yields:

(λ - 1)^{2} = λ^{2} + 2

Expanding the left side, the

λ^{2}

terms cancel out, leaving the linear equation:

- 2 λ + 1 = 2 \Rightarrow λ = - \frac{1}{2}

The Final Equation and the Absolute Value Trap

Substituting

λ = - \frac{1}{2}

back into the family equation, we get

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

. Multiplying by

2

to clear the fraction, we arrive at the specific equation for

P_{3}

2 x - y + 2 z - 2 = 0

Now, consider a general point

B (α, β, γ)

whose distance to

P_{3}

2

. Applying the distance formula again:

\frac{∣2 α - β + 2 γ - 2∣}{2 ^{2} + ( - 1 ) ^{2} + 2 ^{2}} = 2

The denominator simplifies to

9 = 3

. Thus, we have:

\frac{∣2 α - β + 2 γ - 2∣}{3} = 2 \Rightarrow ∣2 α - β + 2 γ - 2∣ = 6

This absolute value splits into two distinct cases:

2 α - β + 2 γ - 2 = 6 and 2 α - β + 2 γ - 2 = - 6

These simplify to the final relations:

$2 α - β + 2 γ - 8 = 0$ and $2 α - β + 2 γ + 4 = 0$

Common Traps & FAQs

Wait, why do we use the family of planes equation

P_{2} + λ P_{1} = 0

Think of the intersection line of two planes as a hinge. Just as a door rotates around its hinge, any plane passing through that line can be described by a linear combination of the two original planes. The parameter

λ

acts as a 'tuning knob' that rotates the plane

P_{3}

around that fixed line of intersection.

Why do we get two answers for the second part of the problem?

That is a brilliant observation! When we set the distance equal to

2

, we are essentially saying that the point

(α, β, γ)

lies on one of two parallel planes, each at a distance of

2

from

P_{3}

. One plane is on one side, and the other is on the opposite side. The absolute value in the distance formula naturally captures both possibilities.

Is there a way to avoid the square root in the distance formula?

While you can square both sides to eliminate the square root, you must be careful with the absolute value. Squaring is a great tool, but it can sometimes introduce extraneous solutions if you are not careful. In this case, it is perfectly valid, but always remember to check your final equations against the original constraints.

Select Answer:

Visualized Solution

Visualizing the Setup of Planes $P_{1}$ and $P_{2}$

Formulating the Family of Planes $P_{3}$

Setting up the Distance Constraint from $A (0, 1, 0)$

Simplifying the Distance Equation

Solving for the Parameter $λ$

Determining the Equation of Plane $P_{3}$

Distance from Point $B (α, β, γ)$ to $P_{3}$

Deriving the Two Possible Relations

Matching with Options and Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Anchor

The Final Equation and the Absolute Value Trap

Similar Questions

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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

List-I

List-II

Select Answer:

Visualized Solution

Visualizing the Setup of Planes .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P2​

Formulating the Family of Planes .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P3​

Setting up the Distance Constraint from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A(0,1,0)

Simplifying the Distance Equation

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

Determining the Equation of Plane .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P3​

Distance from Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B(α,β,γ) to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P3​

Deriving the Two Possible Relations

Matching with Options and Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Anchor

The Final Equation and the Absolute Value Trap

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the plane passing through the point (−1,0,−2) and perpendicular to each of the planes 2x+y−z=2 and x−y−z=3 be ax+by+cz+8=0. Then the value of a+b+c is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane passing through the points (0,−1,0) and (0,0,1) and making an angle π/4 with the plane y−z+5=0, also passes through the point

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List-I

List-II

Visualizing the Setup of Planes $P_{1}$ and $P_{2}$

Formulating the Family of Planes $P_{3}$

Setting up the Distance Constraint from $A (0, 1, 0)$

Solving for the Parameter $λ$

Determining the Equation of Plane $P_{3}$

Distance from Point $B (α, β, γ)$ to $P_{3}$

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