Why can't $L_1$ or $L_2$ be the intersection line?

If $L_1$ were the intersection line, it would lie on both $P_1$ and $P_2$. But the problem states $L_1 \cap L_2 = \{O\}$. If $L_1$ were the intersection line, it would intersect $L_2$ at more than just the origin (unless $L_2$ is also the intersection line, which contradicts the setup).

How do we know $C \notin P_1$?

Because $C$ is on $L_2$, and $L_2$ only intersects $P_1$ at the origin. Since $C \neq O$, $C$ cannot be on $P_1$.

Is the permutation $(C, B, A)$ the only solution?

Not necessarily, but it is the most elegant one! The problem asks to show that *there exists* a permutation, so finding one valid configuration is sufficient.

JEE Advanced 2004

LEVELJEE Advanced

Animated Solution for Mathematics - Three Dimensional Geometry: $P_{1}$ and $P_{2}$ are planes passing through origin. $L_{1}$ and $L_{2}$ are two line on $P_{1}$ and $P_{2}$ respectively such that their intersection is origin. Show that there exists points $A, B, C$ whose permutation $A^{'}, B^{'}, C^{'}$ can be chosen such that (i) $A$ is on $L_{1}, B$ on $P_{1}$ but not on $L_{1}$ and $C$ not on $P_{1}$ (ii) $A^{'}$ is on $L_{2}, B^{'}$ on $P_{2}$ but not on $L_{2}$ and $C^{'}$ not on $P_{2}$ .

Visualized Solution

Visualizing the Intersecting Planes $P_{1}$ and $P_{2}$

Let $P_{1}$ and $P_{2}$ be two distinct planes passing through the origin $O (0, 0, 0)$ .
In three-dimensional space, any two non-parallel planes intersecting through the origin must intersect along a unique straight line.
Let this line of intersection be $L_{in t} = P_{1} \cap P_{2}$ .

Defining the Lines $L_{1} \subset P_{1}$ and $L_{2} \subset P_{2}$

We are given a line $L_{1}$ lying entirely on plane $P_{1}$ ( $L_{1} \subset P_{1}$ ).
Similarly, a line $L_{2}$ lies entirely on plane $P_{2}$ ( $L_{2} \subset P_{2}$ ).
The problem states that their intersection is exactly the origin: $L_{1} \cap L_{2} = {O}$ .
This implies that neither $L_{1}$ nor $L_{2}$ can be identical to the intersection line $L_{in t}$ (except at the origin).

Selecting Point $B$ on the Intersection Line

To satisfy the conditions, let's choose point $B$ on the intersection line $L_{in t}$ such that $B \neq = O$ .
Since $B \in L_{in t}$ and $L_{in t} \subset P_{1}$ , it follows that $B \in P_{1}$ .
Since $L_{1} \cap L_{in t} = {O}$ and $B \neq = O$ , it must be that $B \in / L_{1}$ .
Thus, $B$ is on $P_{1}$ but not on $L_{1}$ , satisfying condition (i) for $B$ .

Choosing Point $A$ on Line $L_{1}$

Let's choose point $A$ to lie on the line $L_{1}$ such that $A \neq = O$ .
This directly satisfies the first part of condition (i): $A$ is on $L_{1}$ .

Choosing Point $C$ on Line $L_{2}$

Let's choose point $C$ to lie on the line $L_{2}$ such that $C \neq = O$ .
Since $C \in L_{2}$ and $L_{2} \subset P_{2}$ , we have $C \in P_{2}$ .
Since $P_{1} \cap P_{2} = L_{in t}$ and $L_{2} \cap L_{in t} = {O}$ , any point on $L_{2}$ other than $O$ cannot lie on $P_{1}$ .
Therefore, $C \in / P_{1}$ , satisfying the third part of condition (i).

Constructing the Permutation $(A^{'}, B^{'}, C^{'})$

We have chosen three distinct points: $A \in L_{1}$ , $B \in P_{1} ∖ L_{1}$ , and $C \in / P_{1}$ .
Now, let's define a permutation $(A^{'}, B^{'}, C^{'})$ of these points.
Let $A^{'} = C$ , $B^{'} = B$ , and $C^{'} = A$ .
We must now verify if this permutation satisfies condition (ii).

Verifying $A^{'}$ on $L_{2}$ and $B^{'}$ on $P_{2} ∖ L_{2}$

First, $A^{'} = C$ . Since we chose $C \in L_{2}$ , $A^{'}$ is indeed on $L_{2}$ .
Second, $B^{'} = B$ . Since $B \in L_{in t}$ and $L_{in t} \subset P_{2}$ , we have $B^{'} \in P_{2}$ .
Since $L_{2} \cap L_{in t} = {O}$ and $B \neq = O$ , we have $B^{'} \in / L_{2}$ .
Thus, $B^{'}$ is on $P_{2}$ but not on $L_{2}$ .

Verifying $C^{'} \in / P_{2}$ and Final Proof

Third, $C^{'} = A$ . Since $A \in L_{1}$ and $A \neq = O$ .
Since $L_{1} \subset P_{1}$ and $L_{1} \cap P_{2} = {O}$ , any point on $L_{1}$ other than $O$ cannot lie on $P_{2}$ .
Therefore, $C^{'} \in / P_{2}$ .
All conditions are satisfied, completing the proof.

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The Sigma Insight: Equation of a Plane

The Geometry of Intersecting Worlds

Imagine you are standing in a vast, empty three-dimensional space. You have two infinite sheets,

P_{1}

and

P_{2}

, both slicing through the origin

O (0, 0, 0)

They are not parallel, so they must meet. They meet along a single, unique line, which we call the intersection line,

L_{in t} = P_{1} \cap P_{2}

. This line is our anchor, our bridge between these two worlds.

Now, place two lines on these planes:

L_{1}

P_{1}

and

L_{2}

P_{2}

. The problem states that

L_{1} \cap L_{2} = {O}

This is a profound constraint! It means neither

L_{1}

nor

L_{2}

can be the intersection line

L_{in t}

. If they were, they would share more than just the origin. They are, in a sense, 'skew' to the intersection line.

The Strategic Selection

To solve this, we need to be architects of space. We need to pick three points,

A, B,

and

C

, that satisfy the conditions.

Let us start with

B

. We place

B

on the intersection line

L_{in t}

, but away from the origin ($B eq O$).

Because

B

is on

L_{in t}

, it belongs to both

P_{1}

and

P_{2}

. But because

L_{1}

and

L_{2}

only touch

L_{in t}

at the origin,

B

cannot be on

L_{1}

L_{2}

This is perfect!

B

is on

P_{1}

but not on

L_{1}

, and

B

is on

P_{2}

but not on

L_{2}

. We have our anchor point.

Next, we pick

A

L_{1}

(with $A eq O$) and

C

L_{2}

(with $C eq O$). Now, let us check the conditions for

(A, B, C)

A

is on

L_{1}

, so that is satisfied.

B

is on

P_{1}

but not on

L_{1}

, which we just established.

What about

C

C

is on

L_{2}

. Since

L_{2}

only meets

P_{1}

at the origin, and $C eq O$,

C

cannot be on

P_{1}

. Thus, $C otin P_1$. We have satisfied condition (i)!

The Permutation

Now, the magic happens. We define a permutation

(A^{'}, B^{'}, C^{'})

where

A^{'} = C

B^{'} = B

, and

C^{'} = A

Let us verify condition (ii). Is

A^{'}

L_{2}

? Yes, because

A^{'} = C

and

C \in L_{2}

B^{'}

P_{2}

but not on

L_{2}

? Yes, because

B^{'} = B

, which is on

L_{in t} \subset P_{2}

, and $B otin L_2$.

Finally, is

C^{'}

not on

P_{2}

? Yes, because

C^{'} = A

, and

A \in L_{1}

. Since

L_{1}

only meets

P_{2}

at the origin, and $A eq O$,

A

cannot be on

P_{2}

We have done it! We have navigated the geometry of these planes and lines, finding the perfect points to satisfy the conditions.

This problem is not just about points and lines; it is about understanding how objects in 3D space relate to one another. Keep this visualization in your mind, and you will never fear 3D geometry again.

Visualized Solution

Visualizing the Intersecting Planes $P_{1}$ and $P_{2}$

Defining the Lines $L_{1} \subset P_{1}$ and $L_{2} \subset P_{2}$

Selecting Point $B$ on the Intersection Line

Choosing Point $A$ on Line $L_{1}$

Choosing Point $C$ on Line $L_{2}$

Constructing the Permutation $(A^{'}, B^{'}, C^{'})$

Verifying $A^{'}$ on $L_{2}$ and $B^{'}$ on $P_{2} ∖ L_{2}$

Verifying $C^{'} \in / P_{2}$ and Final Proof

The Sigma Insight: Equation of a Plane

The Geometry of Intersecting Worlds

The Strategic Selection

The Permutation

Similar Questions

Let the lines $\frac{x - 1}{λ} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x + 26}{- 2} = \frac{y + 18}{3} = \frac{z + 28}{λ}$ be coplanar and $P$ be the plane containing these two lines. Then which of the following points does NOT lies on $P$ ?

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Let $P$ be the plane passing through the point $(1, 2, 3)$ and the line of intersection of the planes $r \cdot (\hat{i} + \hat{j} + 4 \hat{k}) = 16$ and $r \cdot (- \hat{i} + \hat{j} + \hat{k}) = 6$ . Then which of the following points does NOT lie on $P$ ?

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Let a plane $P$ contain two lines $r = \hat{i} + λ (\hat{i} + \hat{j}), λ \in R$ and $r = - \hat{j} + μ (\hat{j} - \hat{k}), μ \in R$ . If $Q (α, β, γ)$ is the foot of the perpendicular drawn from the point $M (1, 0, 1)$ to $P$ , then $3 (α + β + γ)$ equals

A plane $P$ contains the line $x + 2 y + 3 z + 1 = 0 = x - y - z - 6$ , and is perpendicular to the plane $- 2 x + y + z + 8 = 0$ . Then which of the following points lies on $P$ ?

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 :

Visualized Solution

Selecting Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B on the Intersection Line

Choosing Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A on Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Choosing Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C on Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Constructing the Permutation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (A′,B′,C′)

Verifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C′∈/P2​ and Final Proof

The Sigma Insight: Equation of a Plane

The Geometry of Intersecting Worlds

The Strategic Selection

The Permutation

Similar Questions

List-I

List-II

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane P contains the line x+2y+3z+1=0=x−y−z−6, and is perpendicular to the plane −2x+y+z+8=0. Then which of the following points lies on P?

Selecting Point $B$ on the Intersection Line

Choosing Point $A$ on Line $L_{1}$

Choosing Point $C$ on Line $L_{2}$

Constructing the Permutation $(A^{'}, B^{'}, C^{'})$

Verifying $C^{'} \in / P_{2}$ and Final Proof

A plane $P$ contains the line $x + 2 y + 3 z + 1 = 0 = x - y - z - 6$ , and is perpendicular to the plane $- 2 x + y + z + 8 = 0$ . Then which of the following points lies on $P$ ?