Let P be the plane, which contains the line of intersection of the planes, x + y + z - 6 = 0 and 2x

Q: Wait, why do we use the family of planes equation $P_1 + \lambda P_2 = 0$ instead of finding the line of intersection first?

You absolutely could find the line of intersection first, but that involves solving a system of equations to get the line in symmetric form, which is much more time-consuming. The family of planes method is a powerful shortcut in JEE problems because it lets you jump straight to the equation of the required plane using only one parameter, $\lambda$.

Q: Why does the $z$-coefficient become zero when the plane is perpendicular to the $xy$-plane?

Great observation! If a plane is perpendicular to the $xy$-plane, it must be a 'vertical' plane. A vertical plane is parallel to the $z$-axis. Since the normal vector of a plane is perpendicular to the plane itself, if the plane is parallel to the $z$-axis, its normal vector must be perpendicular to the $z$-axis. This means the $z$-component of the normal vector must be zero.

Q: Does the point $(0, 0, 256)$ being on the $z$-axis make the distance calculation easier?

Yes, it does! When you plug $(0, 0, 256)$ into the distance formula $D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$, the $x$ and $y$ terms vanish because the coordinates are zero. It simplifies the numerator significantly, leaving you with just the constant term and the $z$ term (if $z$ exists in the plane equation). It's a nice little gift from the problem setter!

Question

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

Options:

(A) $63\sqrt{5}$

(B) $205\sqrt{5}$

(C) $17/\sqrt{5}$

(D) $11/\sqrt{5}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a room with two large sheets of paper representing the planes $P_1: x + y + z - 6 = 0$ and $P_2: 2x + 3y + z + 5 = 0$. Where these two sheets cross, they form a sharp, straight line of intersection.

We are tasked with finding a new plane, $P$, that contains this exact line and is also perpendicular to the $xy$-plane. This is a classic JEE Advanced problem that tests your ability to bridge the gap between abstract algebra and geometric intuition.

### The Power of the Family of Planes

Instead of laboriously finding the line of intersection, we use the elegant concept of the 'family of planes.' Any plane passing through the intersection of $P_1$ and $P_2$ can be represented by the equation $P_1 + \lambda P_2 = 0$, where $\lambda$ is a scalar parameter.

By substituting our planes, we obtain:

$$(x + y + z - 6) + \lambda(2x + 3y + z + 5) = 0$$

This equation represents every possible plane that could pass through the line of intersection. Our goal is to find the specific value of $\lambda$ that satisfies our perpendicularity condition.

### The Perpendicularity Constraint

To find $\lambda$, we first group the terms in our equation:

$$(1 + 2\lambda)x + (1 + 3\lambda)y + (1 + \lambda)z + (5\lambda - 6) = 0$$

The normal vector of this plane is $\vec{n}_P = (1+2\lambda, 1+3\lambda, 1+\lambda)$. The problem states that our plane $P$ is perpendicular to the $xy$-plane.

The $xy$-plane has a normal vector of $\vec{n}_{xy} = (0, 0, 1)$. For two planes to be perpendicular, their normal vectors must be perpendicular, meaning their dot product must be zero: $\vec{n}_P \cdot \vec{n}_{xy} = 0$.

Calculating this, we get:

$$(1+2\lambda)(0) + (1+3\lambda)(0) + (1+\lambda)(1) = 0$$

This simplifies beautifully to $1 + \lambda = 0$, which gives us $\lambda = -1$.

### Final Calculation

With $\lambda = -1$, we substitute it back into our grouped equation:

$$(1 - 2)x + (1 - 3)y + (1 - 1)z + (-5 - 6) = 0$$

This simplifies to $-x - 2y - 11 = 0$, or equivalently, $x + 2y + 11 = 0$.

Now, we calculate the distance of the point $(0, 0, 256)$ from this plane. Using the perpendicular distance formula:

$$D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$$

Substituting our values:

$$D = \frac{|1(0) + 2(0) + 0(256) + 11|}{\sqrt{1^2 + 2^2 + 0^2}}$$

The numerator becomes $|11|$ and the denominator is $\sqrt{5}$. Thus, the final distance is:

**$$D = \frac{11}{\sqrt{5}}$$**

Animated Solution for Mathematics - Three Dimensional Geometry: 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 :-

Select Answer:

Visualized Solution

Visualizing the Intersection

The Family of Planes Equation $P_{1} + λ P_{2} = 0$

Substituting the Planes

Grouping the Variables

The Perpendicularity Condition $n_{P} \cdot n_{x y} = 0$

Applying the Dot Product

Solving for $λ$

Finding the Equation of Plane $P$

The Target Point $(0, 0, 256)$

The Distance Formula $D$

Substituting the Values

Final Calculation

Conclusion and Key Takeaway

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Power of the Family of Planes

The Perpendicularity Constraint

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

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 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 $∣ d ∣$ .

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 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

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:

Let the equation of the plane passing through the line $x - 2 y - z - 5 = 0 = x + y + 3 z - 5$ and parallel to the line $x + y + 2 z - 7 = 0 = 2 x + 3 y + z - 2$ be $a x + b y + cz = 65$ . Then the distance of the point $(a, b, c)$ from the plane $2 x + 2 y - z + 16 = 0$ 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 :

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 :

Select Answer:

Visualized Solution

Visualizing the Intersection

The Family of Planes Equation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P1​+λP2​=0

Substituting the Planes

Grouping the Variables

The Perpendicularity Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } nP​⋅nxy​=0

Applying the Dot Product

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

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

The Target Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (0,0,256)

The Distance Formula .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Substituting the Values

Final Calculation

Conclusion and Key Takeaway

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Power of the Family of Planes

The Perpendicularity Constraint

Final Calculation

Similar Questions

.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; } 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; } 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; } 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 :

The Family of Planes Equation $P_{1} + λ P_{2} = 0$

The Perpendicularity Condition $n_{P} \cdot n_{x y} = 0$

Solving for $λ$

Finding the Equation of Plane $P$

The Target Point $(0, 0, 256)$

The Distance Formula $D$

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 $∣ d ∣$ .

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

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:

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 :