If L1 is the line of intersection of the planes 2x - 2y + 3z - 2 = 0, x - y + z + 1 = 0 and L2 is th

Q: Why can we just set $y=0$ to find a point on the line? Isn't that arbitrary?

It is arbitrary, but that is the beauty of 3D geometry! A line is a collection of infinite points. By setting $y=0$, you are simply choosing the specific point where the line pierces the $xz$-plane. Since the line lies entirely on the plane we are looking for, any point on that line is a valid point for our plane equation. Choosing $y=0$ is just a strategic move to simplify your algebra.

Q: Wait, why do we use the cross product of the direction vectors to find the normal of the plane?

Think of the plane as a flat sheet of paper. If two lines lie on that paper, their direction vectors lie on that paper too. A vector perpendicular to the paper (the normal) must, by definition, be perpendicular to every vector on that paper—including our two lines. The cross product is the perfect tool to find a vector orthogonal to two others. It is the geometric key that unlocks the orientation of the plane.

Q: Why can't we just use the distance formula between the two lines directly?

You could, but that is a much more complex path! The distance between two skew lines is a standard formula, but the question asks for the distance from the origin to the *plane* containing them. By finding the plane first, you reduce a complex 3D spatial problem into a simple point-to-plane distance calculation, which is much less prone to calculation errors.

Question

If $L_1$ is the line of intersection of the planes $2x - 2y + 3z - 2 = 0, x - y + z + 1 = 0$ and $L_2$ is the line of intersection of the planes $x + 2y - z - 3 = 0, 3x - y + 2z - 1 = 0$, then the distance of the origin from the plane, containing the lines $L_1$ and $L_2$, is :

Options:

(A) 1/\sqrt{2}

(B) 1/(4\sqrt{2})

(C) 1/(3\sqrt{2})

(D) 1/(2\sqrt{2})

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are not just solving a problem; we are constructing a world. We are given two lines, $L_1$ and $L_2$, floating in 3D space, each defined by the intersection of two planes.

Our goal is to find the distance from the origin to the unique plane that contains both of these lines. This is a classic JEE Advanced problem that tests your ability to visualize spatial relationships and your command over vector algebra.

### The Anatomy of the Lines

A line defined by the intersection of two planes is like the crease where two pages of a book meet. The direction of this crease is perpendicular to the normal vectors of both pages.

For $L_1$, defined by $2x - 2y + 3z - 2 = 0$ and $x - y + z + 1 = 0$, we find its direction vector $\vec{d}_1$ by taking the cross product of the normals $\vec{n}_{11} = \langle 2, -2, 3 angle$ and $\vec{n}_{12} = \langle 1, -1, 1 angle$:

$$
\vec{d}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & 3 \ 1 & -1 & 1 \end{vmatrix} = \hat{i} + \hat{j}
$$

We repeat this for $L_2$, defined by $x + 2y - z - 3 = 0$ and $3x - y + 2z - 1 = 0$. Using the same cross-product logic:

$$
\vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -1 \ 3 & -1 & 2 \end{vmatrix} = 3\hat{i} - 5\hat{j} - 7\hat{k}
$$

### Finding the Anchor

A plane is defined by a point and a normal vector. We have the orientation, but we need an anchor—a point that lies on our plane. Since our plane contains $L_1$, any point on $L_1$ is a point on our plane.

To find one, we set $y=0$ in the equations for $L_1$. This gives us the system:

$$
2x + 3z = 2 \quad 	ext{and} \quad x + z = -1
$$

Solving this system, we find $x = -5$ and $z = 4$. Thus, our anchor point $P$ is $(-5, 0, 4)$.

### The Master Normal

Now, we need the normal vector $\vec{N}$ of our target plane. This plane contains both $L_1$ and $L_2$, so the normal vector must be perpendicular to both $\vec{d}_1$ and $\vec{d}_2$:

$$
\vec{N} = \vec{d}_1 	imes \vec{d}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 0 \ 3 & -5 & -7 \end{vmatrix} = -7\hat{i} + 7\hat{j} - 8\hat{k}
$$

This vector $\vec{N}$ is the 'compass' that defines the tilt of our plane.

### Final Calculation

With our point $P(-5, 0, 4)$ and our normal $\vec{N} = \langle -7, 7, -8 angle$, we write the equation of the plane using the point-normal form:

$$
-7(x + 5) + 7(y - 0) - 8(z - 4) = 0
$$

Simplifying this, we arrive at $-7x - 35 + 7y - 8z + 32 = 0$, which rearranges to $7x - 7y + 8z + 3 = 0$.

Finally, we calculate the perpendicular distance from the origin $(0,0,0)$ to this plane using the formula $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$:

$$
d = \frac{|3|}{\sqrt{7^2 + (-7)^2 + 8^2}} = \frac{3}{\sqrt{49 + 49 + 64}} = \frac{3}{\sqrt{162}}
$$

Simplifying $\sqrt{162}$ to $9\sqrt{2}$, we find the final result:

$$
d = \frac{3}{9\sqrt{2}} = \mathbf{\frac{1}{3\sqrt{2}}}
$$

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Visualized Solution

Problem Overview

Direction Vector of $L_{1}$

Calculating $d_{1}$

Direction Vector of $L_{2}$

Calculating $d_{2}$

Finding a Point on the Plane

Calculating Point $P$

Normal Vector to the Plane

Calculating Normal $N$

Equation of the Plane

Deriving the Plane Equation

Distance from Origin

Final Calculation

Final Result

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Anatomy of the Lines

Finding the Anchor

The Master Normal

Final Calculation

Similar Questions

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

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:

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 $∣ d ∣$ .

List-I

List-II

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 :

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:

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

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

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Visualized Solution

Problem Overview

Direction Vector of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } d1​

Direction Vector of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

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

Finding a Point on the Plane

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

Normal Vector to the Plane

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

Equation of the Plane

Deriving the Plane Equation

Distance from Origin

Final Calculation

Final Result

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Anatomy of the Lines

Finding the Anchor

The Master Normal

Final Calculation

Similar Questions

.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:

.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∣.

List-I

List-II

Direction Vector of $L_{1}$

Calculating $d_{1}$

Direction Vector of $L_{2}$

Calculating $d_{2}$

Calculating Point $P$

Calculating Normal $N$

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:

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 $∣ d ∣$ .