What is the equation of plane parallel to $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}=d_1$ ?

The equation of plane parallel to $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}=d_1$ is of the form $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}=d_2$, where $d_2$ can be determined by using the given conditions

What is the equation of a plane passing through the intersection of planes $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}_1=d_1$ and $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}_2=d_2$ ?

The equation of a plane passing through the intersection of planes $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}_1=d_1$ and $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}_2=d_2$ is $\overrightarrow{\mathbf{r}} \cdot\left(\overrightarrow{\mathbf{n}}_1+\lambda \overrightarrow{\mathbf{n}}_2\right)=d_1+\lambda d_2$

Family of Plane: Definition and Examples

Q: In cartesian form, if $a x+b y+c z+d=0$ is the given plane, then what is the plane parallel to this plane?

In cartesian form, if $a x+b y+c z+d=0$ is the given plane, then the plane parallel to this plane is $a x+b y+c z+k=0$,

Q: What is the vector equation of the plane?

The vector equation of the plane is given by $\overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}}_1=d_1$

Family of Planes

Edited By Komal Miglani | Updated on Feb 15, 2025 12:05 AM IST

For any three points that do not all lie on the same line, there is a unique plane that passes through these points. Given two distinct, intersecting lines, there is exactly one plane containing both lines. A plane is also determined by a line and any point that does not lie on the line. In real life, we use planes to measure the circumference, area, and volume.

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This Story also Contains

What is a plane parallel to a given plane?
Equation of plane parallel to a given plane in Vector Form
Derivation of Equation of plane parallel to a given plane
Equation of plane parallel to a given plane in Cartesian Form
Solved Examples Based on Family of Plane

Family of Planes

In this article, we will cover the concept of the Family of Plane. This topic falls under the broader category of Three Dimensional Geometry, which is a crucial chapter in Class 12 Mathematics. This is very important not only for board exams but also for competitive exams, which even include the Joint Entrance Examination Main and other entrance exams: SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE. A total of twenty-five questions have been asked on this topic in JEE Main from 2013 to 2023 including four in 2019, nine in 2021, eight in 2022, and two in 2023.

What is a plane parallel to a given plane?

Parallel planes have the same normal vectors, so the equation of plane parallel to $\vec{r} \cdot \vec{n} = d_{1}$ is of the form $\vec{r} \cdot \vec{n} = d_{2}$ , where $d_{2}$ can be determined by using the given conditions.

In cartesian form, if $a x + b y + c z + d = 0$ is the given plane, then the plane parallel to this plane is $a x + b y + c z + k = 0$ , where $k$ is any scalar.

The plane passes through the intersection of two given planes

Equation of plane parallel to a given plane in Vector Form

The equation of a plane passing through the intersection of planes $\vec{r} \cdot {\vec{n}}_{1} = d_{1}$ and $\vec{r} \cdot {\vec{n}}_{2} = d_{2}$ is

$\vec{r} \cdot ({\vec{n}}_{1} + λ {\vec{n}}_{2}) = d_{1} + λ d_{2}$

Derivation of Equation of plane parallel to a given plane

Let $π_{1}$ and $π_{2}$ be two planes with equations $\vec{r} \cdot {\vec{n}}_{1} = d_{1}$ and $\vec{r} \cdot {\vec{n}}_{2} = d_{2}$ respectively. The position vector of any point on the line of intersection must satisfy both equations.

If $\vec{P}$ is the position vector of a point $P$ on the line, then $\vec{p} \cdot {\hat{n}}_{1} = d_{1}$ and $\vec{p} \cdot {\hat{n}}_{2} = d_{2}$

Therefore, for all real values of $λ$ , we have

$\vec{p} \cdot ({\hat{n}}_{1} + λ {\hat{n}}_{2}) = d_{1} + λ d_{2}$
Since $p$ is arbitrary, it satisfies any point on the line.
Hence, the equation $\vec{r} \cdot ({\vec{n}}_{1} + λ {\vec{n}}_{2}) = d_{1} + λ d_{2}$ represents a plane $π_{3}$ which is such that if any vector $\vec{r}$ satisfies both the equations $π_{1}$ and $π_{2}$ , it also satisfies the equation $π_{3}$ i.e. any plane passing through the intersection of the planes.

Equation of plane parallel to a given plane in Cartesian Form

In the Cartesian system, let

$\begin{aligned} {\vec{n}}_{1} & = a_{1} \hat{i} + b_{2} \hat{j} + c_{1} \hat{k} \\ {\vec{n}}_{2} & = a_{2} \hat{i} + b_{2} \hat{j} + c_{2} \hat{k} \\ \vec{r} & = x \hat{i} + y \hat{j} + z k \end{aligned}$

then the vector equation, $\vec{r} \cdot ({\vec{n}}_{1} + λ {\vec{n}}_{2}) = d_{1} + λ d_{2}$ become

$\begin{matrix} x (a_{1} + λ a_{2}) + y (b_{1} + λ b_{2}) + z (c_{1} + λ c_{2}) = d_{1} + λ d_{2} \\ or (a_{1} x + b_{1} y + c_{1} z - d_{1}) + λ (a_{2} x + b_{2} y + c_{2} z - d_{2}) = 0 \end{matrix}$

which is the required Cartesian form of the equation of the plane passing through the intersection of the given planes for each value of $λ$ .

Solved Examples Based on Family of Plane

Example 1: If the equation of the plane passing through the line of intersection of the planes $2 x - y + z = 3, 4 x - 3 y + 5 z + 9 = 0$ and parallel to the line $\frac{x + 1}{- 2} = \frac{y + 3}{4} = \frac{z - 2}{5}$ is $a x + b y + c z + 6 = 0$ , ,then $a + b + c$ is equal to :
[JEE MAINS 2023]

Solution:
Using a family of plane

$\begin{aligned} P : P_{1} + λ P_{2} = 0 \Rightarrow P (2 + 4 λ) x - (1 + 3 λ) y + (1 + 5 λ) z = (3 - 9 λ) \\ P is ‖ to \frac{x + 1}{- 2} = \frac{y + 3}{4} = \frac{z - 2}{5} \end{aligned}$

Then for $λ : {\vec{n}}_{p} \cdot {\vec{v}}_{L} = 0$

$\begin{aligned} - 2 (2 + 4 λ) - 4 (1 + 3 λ) + 5 (1 + 5 λ) = 0 \\ - 3 + 5 λ = 0 \Rightarrow λ = \frac{3}{5} \end{aligned}$

Hence : $P : 22 x - 14 y + 20 z = - 12$

$\begin{aligned} P : 11 x - 7 y + 10 z + 6 = 0 \\ \Rightarrow a = 11 \\ b = - 7 \\ c = 10 \\ \Rightarrow a + b + c = 14 \end{aligned}$

Hence, the answer is 14

Example 2: Let the equation of the plane passing through the line of intersection of the planes $x + 2 y + a z = 2$ and $x - y + z = 3$ be $5 x - 11 y + bz = 6 a - 1$ . For $c ϵ Z$ , if the distance of this plane from the point $(a, - c, c)$ is $\frac{2}{\sqrt{a}}$ , then $\frac{a + b}{c}$ is equal
[JEE MAINS 2023]

Solution:
$(x + 2 y + a z - 2) + λ (x - y + z - 3) = 0$

$\begin{aligned} \frac{1 + λ}{5} = \frac{2 - λ}{- 11} = \frac{a + λ}{b} = \frac{2 + 3 λ}{6 a - 1} \\ λ = - \frac{7}{2}, a = 3, b = 1 \\ \frac{2}{\sqrt{a}} = | \frac{5 a + 11 c + b c - 6 a + 1}{\sqrt{25 + 121 + 1}} | \\ c = - 1 \\ ∴ \frac{a + b}{c} = \frac{3 + 1}{- 1} = - 4 \end{aligned}$

Hence, the answer is -4

Example 3: Let $P_{1} : \vec{r} \cdot (2 \hat{i} + \hat{j} - 3 \hat{k}) = 4$ be a plane. Let $P_{2}$ be another plane that passes through the points $(2, - 3, 2), (2, - 2, - 3)$ and $(1, - 4, 2)$ . If the direction ratios of the line of intersection of $P_{1}$ and $P_{2}$ be $16, α, β$ , then the value of $α + β$ is equal to
[JEE MAINS 2022]

Solution:
Let $A (2, - 3, 2), B (2, - 2, - 3), C (1, - 4, 2)$

$\begin{aligned} \vec{n_{1}} & = 2 i + j - 3 k \\ {\vec{n}}_{2} & = vector perpendicular to 2^{nd} plane \\ = \vec{A B} \times \vec{B C} \\ = | \begin{array}{ccc} i & j & k \\ 0 & 1 & - 5 \\ - 1 & - 2 & 5 \end{array} | \\ = (5 - 10) i - (- 5) j + (1) k \\ = & - 5 i + 5 j + k \end{aligned}$
To find the Direction ratios of the line of intersection, we first find ${\vec{n}}_{1} \times {\vec{n}}_{2}$

$\begin{aligned} {\vec{n}}_{1} \times {\vec{n}}_{2} & = | \begin{array}{ccc} i & j & k \\ 2 & 1 & - 3 \\ - 5 & 5 & 1 \end{array} | \\ = (16) i - (- 13) j + (15) k \\ = 16 i + 13 j + 15 k \end{aligned}$

$\begin{aligned} ∴ α = 13, β = 15 \\ \Rightarrow α + β = 28 \end{aligned}$

Hence, the answer is 28

Example 4: The acute angle between the planes $P_{1}$ and $P_{2}$ , when $P_{1}$ and $P_{2}$ are the planes pass through the intersection of the planes $5 x + 8 y + 13 z - 29 = 0$ and $8 x - 7 y + z - 20 = 0$ and the points $(2, 1, 3)$ and $(0, 1, 2)$ , respectively, is
[JEE MAINS 2022]

Solution:
Let $P_{1} : (5 x + 2 y + 13 z - 29) + λ (8 x - 7 y + z - 20) = 0$
Passes through $(2, 1, 3)$

$\begin{aligned} \Rightarrow (10 + 8 + 39 - 29) + λ (16 - 7 + 3 - 20) = 0 \Rightarrow λ = \frac{7}{2} \\ \Rightarrow P_{1} : (10 x + 16 y + 26 z - 58) + (56 x - 49 y + 73 - 140) = 0 \\ \Rightarrow 66 x - 33 y + 33 z - 198 = 0 \Rightarrow 2 x - y + z - 6 = 0 - - - (1) \end{aligned}$
Let $P_{2} : (5 x + 8 y + 13 z - 29) + μ (8 x - 7 y + z - 20) = 0$
Passes through $(0, 1, 2)$

$\begin{aligned} \Rightarrow (0 + 8 + 26 - 29) + μ (0 - 7 + 2 - 20) = 0 \Rightarrow μ = 1 / 5 \\ \Rightarrow P_{2} : (25 x + 40 y + 65 z - 145) + 8 x - 7 y + z - 20 = 0 \Rightarrow 33 x + 33 y + 663 - 165 = 0 \\ \Rightarrow x + y + 2 z - 5 = 0 - - - (2) \\ \cos θ = | \frac{(2, - 1, 1) \cdot (1, 1, 2)}{\sqrt{2^{2} + 1^{2} + 1^{2}} \sqrt{1^{2} + 1^{2} + 2^{2}}} | = \frac{3}{6} = \frac{1}{2} \Rightarrow θ = π / 3 \end{aligned}$

Hence, the answer is $\frac{π}{3}$

Example 5: The vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1$ and $\vec{r} \cdot (\hat{i} - 2 \hat{j}) = - 2$ , and the point $(1, 0, 2)$ is:
[JEE MAINS 2021]

Solution

$\begin{aligned} \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 1 \\ \vec{r} \cdot (\hat{i} - 2 \hat{j}) = - 2 \\ point (1, 0, 2) \end{aligned}$

Equation of Plane

$\begin{aligned} \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) - 1 + λ {\vec{r} \cdot (\hat{i} - 2 \hat{j}) + 2} = 0 \\ \vec{r} \cdot {\hat{i} (1 + λ) + \hat{j} (1 - 2 λ) + \hat{k}} - 1 + 2 λ = 0 \\ Point: \hat{i} + 0 \hat{j} + 2 \hat{k} = \vec{r} \\ ∴ (\hat{i} + 2 \hat{k}) \cdot {\hat{i} (1 + λ) + \hat{j} (1 - 2 λ) + \hat{k}} - 1 + 2 λ = 0 \\ 1 + λ + 2 - 1 + 2 λ = 0 \\ λ = - \frac{2}{3} \\ ∴ \vec{r} \cdot [\hat{i} (\frac{1}{3}) + \hat{j} (\frac{7}{3}) + \hat{k}] = \frac{7}{3} \\ \vec{r} \cdot [\hat{i} + 7 \hat{j} + 3 \hat{k}] = 7 \\ Hence, the answer is \vec{r} \cdot (\hat{i} + 7 \hat{j} + 3 \hat{k}) = 7 \end{aligned}$

Frequently Asked Questions (FAQs)

1. What is the equation of plane parallel to

\vec{r} \cdot \vec{n} = d_{1}

The equation of plane parallel to $\vec{r} \cdot \vec{n} = d_{1}$ is of the form $\vec{r} \cdot \vec{n} = d_{2}$ , where $d_{2}$ can be determined by using the given conditions

2. What is the equation of a plane passing through the intersection of planes

\vec{r} \cdot {\vec{n}}_{1} = d_{1}

and

\vec{r} \cdot {\vec{n}}_{2} = d_{2}

The equation of a plane passing through the intersection of planes $\vec{r} \cdot {\vec{n}}_{1} = d_{1}$ and $\vec{r} \cdot {\vec{n}}_{2} = d_{2}$ is $\vec{r} \cdot ({\vec{n}}_{1} + λ {\vec{n}}_{2}) = d_{1} + λ d_{2}$

3. In cartesian form, if

a x + b y + c z + d = 0

is the given plane, then what is the plane parallel to this plane?

In cartesian form, if $a x + b y + c z + d = 0$ is the given plane, then the plane parallel to this plane is $a x + b y + c z + k = 0$ ,

4. What is the vector equation of the plane?

The vector equation of the plane is given by $\vec{r} \cdot {\vec{n}}_{1} = d_{1}$

5. What is the Cartesian form of the equation of the plane passing through the intersection of the given planes?

The Cartesian form of the equation of the plane passing through the intersection of the given planes is $(a_{1} x + b_{1} y + c_{1} z - d_{1}) + λ (a_{2} x + b_{2} y$ $+ c_{2} z - d_{1}) = 0$