Equation of Plane

Equation of Plane describes its position and orientation in three-dimensional space, typically represented in the form (ax + by + cz + d = 0), where (a), (b), and (c) are coefficients representing the plane’s normal vector, and (d) is the distance from the origin along the normal vector.

In this article, we will learn about the what is the equation of a plane, its definition and general form the equation, the equation of a plane in 3D Space, a Cartesian form of an equation of a plane, the equation of a plane in intercept and parametric form, etc. At the end of this article, you will see some examples of solved problems that will provide a better understanding of the topic.

What is Plane?

A plane is a two-dimensional flat surface that extends infinitely in all directions. It is characterized by its width and length but has no thickness. In geometry, a plane is defined by an infinite number of points and can be described using equations involving variables such as (x), (y), and (z) in three-dimensional space.

What is the Equation of Plane?

The equation of a plane describes its position and orientation in three-dimensional space. It’s like a rule that tells us where all the points on the plane are located.

Suppose, you imagine a flat surface in space, its equation helps define it. This equation is like ( ax + by + cz + d = 0 ). Each letter (a, b, c, and d) represents a number, and ( x ), ( y ), and ( z ) stand for the coordinates of any point on the plane. When you put in specific numbers for ( a ), ( b ), ( c ), and ( d ), you get the equation that corresponds to a particular plane.

Some common forms of equations of plane are:

• Standard Form
• Vector Form
• Point-Normal Form
• Parametric Form

Definition of Equation of Plane

An equation of a plane is a mathematical expression that describes the relationship between the coordinates of points lying on the plane.

General Form of Equation of a Plane

The general form of the equation of a plane is ( ax + by + cz + d = 0 ), where ( a ), ( b ), and ( c ) are constants representing the plane’s normal vector, and ( d ) is a constant representing the plane’s distance from the origin.

The equation of a plane in three-dimensional space can be expressed in various forms, each serving different purposes. Here are the main forms of the equation of a plane:

• Cartesian Form: Ax + By + Cz = D Where A, B, and C are the coefficients of x, y, and z respectively, and D is a constant.
• Parametric Form:[Tex]\vec{r}[/Tex] = [Tex]\vec{a}[/Tex] + s[Tex]\vec{u}[/Tex] + t[Tex]\vec{v}[/Tex]. Where r is a position vector, a is a point on the plane, and u and v are direction vectors parallel to the plane. s and t are scalar parameters.
• Normal Form: Ax + By + Cz + D = 0, Where A, B, and C are the coefficients of x, y, and z respectively, and (A,B,C) is the normal vector to the plane. D is the distance from the origin to the plane along the direction of the normal vector.

Examples of Planes in General Form

Some examples of planes represented in general form Ax + By + Cz + D = 0:

Horizontal Plane: z=0 (or equivalently, 0x + 0y + z + 0 = 0

• This equation represents the xy-plane, where all points have z-coordinate equal to zero.

Vertical Plane: x=2 (or equivalently, x + 0y + 0z − 2 = 0

• This equation represents a vertical plane parallel to the yz-plane, passing through the point (2, 0, 0).

Diagonal Plane: 2x − 3y + z − 5 = 0

• This equation represents a plane with coefficients A = 2, B = −3, C = 1, and D = −5, passing through the origin and inclined with respect to the coordinate axes.

Perpendicular Plane: 3x + 4y − 2z + 6 = 0

• This equation represents a plane perpendicular to the vector n=(3, 4, −2), passing through the point (−2, 3, 0).

Equation of a Plane in Three Dimensional Space

The equation of a plane in three-dimensional space can be expressed in various forms, each serving different purposes. Here are the main forms of the equation of a plane

General Form: The general form of the equation of a plane is represented as:

Ax + By + Cz + D = 0

Where ( A ), ( B ), ( C ), and ( D ) are constants, and ( x ), ( y ), and ( z ) are the variables representing coordinates in three-dimensional space.

Point-Normal Form: The point-normal form of the equation of a plane is given by:

[Tex]\vec{n} \cdot (\vec{r} – \vec{r_0}) = 0[/Tex]

Where ( [Tex]\vec{n}[/Tex] ) is the normal vector to the plane, ( [Tex]\vec{r_0}[/Tex] ) is a known point on the plane, and ( [Tex]\vec{r}[/Tex] ) represents any point on the plane.

Intercept Form: The intercept form of the equation of a plane is expressed as:

[Tex]\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1[/Tex]

Where ( a ), ( b ), and ( c ) are the intercepts of the plane on the ( x ), ( y ), and ( z ) axes respectively.

Vector Form: The vector form of the equation of a plane is represented as:

[Tex]\vec{r} = \vec{r_0} + s\vec{v} + t\vec{w}[/Tex]

Where ( [Tex]\vec{r_0}[/Tex] ) is a known point on the plane, and ( [Tex]\vec{v}[/Tex] ) and ( [Tex]\vec{w}[/Tex] ) are two non-parallel vectors lying on the plane, and s and t are scalar parameters.

Equation for Intersection of Planes

The equation for the intersection of two planes can be found by solving their respective equations simultaneously.

Consider two planes represented by the equations (ax + by + cz + d₁ = 0) and (ex + fy + gz + d₂ = 0). To find the intersection line or point, solve these equations together to determine values for (x), (y), and (z) that satisfy both planes.

For example, if we have planes with equations (2x + 3y – z + 5 = 0) and (4x – y + 2z – 8 = 0), solving them simultaneously gives the values (x = 2), (y = 1), and (z = 3). These values represent a point of intersection for the two planes.

Methods to Find Equation of a Plane

There are several methods to find the equation of a plane, each depending on the available information about the plane.

• The equation of a plane at a distance (d) from the origin with a unit normal vector ( [Tex]\hat{n} [/Tex]) is given by ( [Tex]\vec{r} \cdot \hat{n} = d [/Tex]).
• If a plane is perpendicular to a given vector ( [Tex]\vec{N}[/Tex] ) and passes through a point ( [Tex]\vec{a}[/Tex] ), its equation is ( [Tex](\vec{r} – \vec{a}) \cdot \vec{N} [/Tex]= 0 ).
• When a plane passes through three non-collinear points ( [Tex]\vec{a} [/Tex]), ( [Tex]\vec{b} [/Tex]), and ( [Tex]\vec{c}[/Tex]), its equation is given by ([Tex] (\vec{r} – \vec{a}) \cdot [(\vec{b} – \vec{a}) \times (\vec{c} – \vec{a})] [/Tex]= 0).
• If a plane passes through the intersection of two planes with normal vectors ( [Tex]\vec{n}_1[/Tex]) and ( [Tex]\vec{n}_2 [/Tex]) at distances ( d₁ ) and ( d₂ ) from the origin respectively, its equation is ( [Tex]\vec{r} \cdot (\vec{n}_1 + \lambda \vec{n}_2) = d_1 + \lambda d_2 [/Tex]).

Equation of a Plane in Normal Form

Consider a plane in three-dimensional space, defined by a point (P(x₁, y₁, z₁)) and a normal vector ([Tex]\vec{n} = \langle a, b, c \rangle[/Tex]).

The equation of a plane in normal form is given by ([Tex]\vec{r} \cdot \vec{n}[/Tex] = d), where ([Tex]\vec{r} = \langle x, y, z \rangle[/Tex]) represents any point on the plane, and (d) is the perpendicular distance from the origin to the plane.

Now, let ([Tex]\vec{r_0} = \langle x_1, y_1, z_1 \rangle[/Tex]) be a point on the plane.

The vector connecting a general point ([Tex]\vec{r}[/Tex]) on the plane to the given point ([Tex]\vec{r_0}[/Tex]) is ([Tex]\vec{r} – \vec{r_0}[/Tex]).

For a point ([Tex]\vec{r}[/Tex]) to lie on the plane, the normal vector ([Tex]\vec{n}[/Tex]) must be perpendicular to the vector ([Tex]\vec{r} – \vec{r_0})[/Tex]. Therefore, the dot product of these vectors should be zero.

Mathematically, ([Tex]\vec{r} – \vec{r_0}) \cdot \vec{n}[/Tex] = 0).

Expanding this, we get ([Tex]\langle x, y, z \rangle – \langle x_1, y_1, z_1 \rangle) \cdot \langle a, b, c \rangle[/Tex] = 0).

Simplifying further, we have (a(x – x₁) + b(y – y₁) + c(z – z₁) = 0).

Now, rearrange the terms to obtain the normal form of the plane equation: (ax + by + cz = ax₁ + by₁ + cz₁).

Thus, the point-normal form of the equation of a plane is given by:

[Tex]\vec{n} \cdot (\vec{r} – \vec{r_0}) = 0[/Tex]

Where,

• \vec{r} = ❬ x, y, z ❭ represents a generic point in the plane,
• \vec{r_0} = ❬ x₀, y₀, z₀ ❭ is a specific point in the plane, and
• \vec{n} = ❬ a, b, c ❭ is the normal vector to the plane.

This equation represents the plane in normal form, where (a, b, c) is the normal vector to the plane, and (d = ax₁ + by₁ + cz₁) is the perpendicular distance from the origin to the plane.

Examples of Planes in Point-Normal Form 

Here are examples of planes in point-normal form:

Plane Parallel to the XY Plane

• Equation: z – 3 = 0
• Normal Vector: ❬ 0, 0, 1 ❭
• Point in the Plane: ❬ x, y, 3 ❭

Inclined Plane in First Quadrant

• Equation: 2x – y + z – 5 = 0
• Normal Vector: ❬ 2, -1, 1 ❭
• Point in the Plane: ❬ x, 2x + 3, 3x – 2❭

Vertical Plane Passing Through the Y-Axis

• Equation: x + z + 4 = 0
• Normal Vector: ❬ 1, 0, 1 ❭
• Point in the Plane: ❬ -4, y, -4 ❭

Equation of a Plane Perpendicular to a Given Vector And Through a Point

The equation of a plane perpendicular to a given vector ( [Tex]\vec{N} = [/Tex] ❬a,b,c❭) and passing through a point ( P(x₁, y₁, z₁) ) can be derived as shown below:

The equation of the plane can be expressed as ( [Tex]\vec{r} – \vec{a}) \cdot \vec{N}[/Tex] = 0 ), where ( [Tex]\vec{r} = [/Tex] ❬x,y,z❭) represents any point on the plane, and ( [Tex]\vec{a} = [/Tex] ❬x₁,y₁,z₁❭) is the given point.

Expanding the dot product, we get ((❬x,y,z❭ – ❬x₁,y₁,z₁❭) · ❬a,b,c❭ = 0)

This simplifies to ( a(x – x₁) + b(y – y₁) + c(z – z₁) = 0 ).

Rearranging the terms, we obtain the equation of the plane: ( ax + by + cz = ax₁ + by₁ + cz₁ ).

This equation represents the plane that is perpendicular to the given vector ( [Tex]\vec{N} [/Tex]) and passes through the point ( P(x₁, y₁, z₁) ).

Equation of a Plane Passing Through Three Points

To derive the equation of a plane passing through three non-collinear points ( [Tex]\vec{a} = [/Tex] ❬x₁,y₁,z₁❭), ( [Tex]\vec{b} =[/Tex] ❬x₂, y₂, z_2❭), and ( [Tex]\vec{c} = [/Tex]❬x₃,y₃,z₃❭), we will use the cross product.

First, define two vectors that lie in the plane, formed by the given points ( [Tex]\vec{a}[/Tex] ), ( [Tex]\vec{b} [/Tex]), and ( [Tex]\vec{c}[/Tex] ). We’ll call these vectors ( [Tex]\vec{v_1}[/Tex] ) and ( [Tex]\vec{v_2}[/Tex]):

[Tex]\vec{v_1} = \vec{b} – \vec{a} = [/Tex]❬ x₂ – x₁, y₂ – y₁, z₂ – z₁ ❭

[Tex]\vec{v_2} = \vec{c} – \vec{a} = [/Tex]❬x₃ – x₁, y₃ – y₁, z₃ – z₁❭

Next, we take the cross product of ( [Tex]\vec{v_1}[/Tex] ) and ( [Tex]\vec{v_2}[/Tex] ) to find a vector that is perpendicular to the plane:

[Tex]\vec{n} = \vec{v_1} \times \vec{v_2} [/Tex]

[Tex]\Rightarrow \vec{n} =\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x_2 – x_1 & y_2 – y_1 & z_2 – z_1 \\ x_3 – x_1 & y_3 – y_1 & z_3 – z_1 \end{vmatrix} [/Tex]

[Tex]\Rightarrow \vec{n} [/Tex] = (y₂ – y₁)(z₃ – z₁) – (z₂ – z₁)(y₃ – y₁) [Tex]\hat{i}[/Tex] – (x₂ – x₁)(z₃ – z₁) + (z₂ – z₁)(x₃ – x₁) [Tex]\hat{j}[/Tex] + (x₂ – x₁)(y₃ – y₁) – (y₂ – y₁)(x₃ – x₁) [Tex]\hat{k} [/Tex]

Now, the equation of the plane passing through the points ( [Tex]\vec{a}[/Tex] ), ( [Tex]\vec{b}[/Tex] ), and ( [Tex]\vec{c}[/Tex] ) is given by:

[Tex]\vec{n} \cdot (\vec{r} – \vec{a})[/Tex] = 0

Substituting the components of ( [Tex]\vec{n}[/Tex] ) and ( [Tex]\vec{a} [/Tex]) into the equation:

(y₂ – y₁)(z₃ – z₁) – (z₂ – z₁)(y₃ – y₁)(x – x₁) + (z₂ – z₁)(x₃ – x₁)(y – y₁) + (x₂ – x₁)(y₃ – y₁) – (y₂ – y₁)(x₃ – x₁)(z – z₁) = 0

This equation represents the plane passing through the three non-collinear points [Tex]\vec{a}[/Tex], [Tex]\vec{b}[/Tex], and [Tex]\vec{c} [/Tex].

Equation of a Plane Passing Through the Intersection of Two Given Planes

Consider two planes with normal vectors ( [Tex]\vec{n}_1 [/Tex]= ❬a₁, b₁, c₁❭) and ([Tex] \vec{n}_2 [/Tex]= ❬a₂, b₂, c₂❭ ), and distances from the origin ( d₁ ) and ( d₂ ) respectively.

The equation of the first plane is given by ( [Tex]\vec{r} \cdot \vec{n}_1 = d_1 [/Tex]), and the equation of the second plane is ( [Tex]\vec{r} \cdot \vec{n}_2 [/Tex]= d₂ ).

Now, find the equation of the plane passing through their intersection. Since the plane is passing through the intersection of both planes, any point ( [Tex]\vec{r} [/Tex]) lying on it must satisfy both equations simultaneously.

Therefore, we can take the sum of the equations of the two planes and introduce a parameter (ƛ) to account for all possible points of intersection:

[Tex]\vec{r} \cdot (\vec{n}_1 + \lambda \vec{n}_2) = d_1 + \lambda d_2 [/Tex]

On expanding this equation, we get:

(x,y,z) · (a₁ + ƛa₂, b₁ + ƛb₂, c₁ + ƛc₂ = d₁ + ƛd₂

⇒ (x(a₁ + ƛa₂) +y(b₁ + ƛb₂) +z(c₁ +ƛc₂)) = d₁ + ƛd₂

⇒ a₁x +a₂ ƛx + b₁y + b₂ ƛy + c₁z + c₂ ƛz = d₁ + ƛ d₂

⇒ (a₁x + b₁y + c₁z) + ƛ(a₂x + b₂y + c₂z) = d₁ + ƛ d₂

This equation represents the plane passing through the intersection of the two given planes. The coefficients (a₁, b₁, c₁) represent the normal vector of the first plane, (a₂, b₂, c₂) represent the normal vector of the second plane, and ( ƛ) is a parameter representing the different points of intersection along the line of intersection of the two planes.

Cartesian Form of Equation of a Plane

The point-normal form equation of a plane:

[Tex]\vec{n} \cdot (\vec{r} – \vec{r_0}) = 0[/Tex]

Where ( [Tex]\vec{n}[/Tex] = ❬a,b,c❭) is the normal vector to the plane, ( [Tex]\vec{r_0}[/Tex] = ❬ x₀, y₀, z₀ ❭ ) is a known point on the plane, and ( [Tex]\vec{r} [/Tex]= ❬x,y,z❭) represents any point on the plane.

Expanding the dot product, we get:

a(x – x₀) + b(y – y₀) + c(z – z₀) = 0

This equation can be rearranged to obtain the Cartesian form of the equation of the plane:

ax + by + cz = ax₀ + by₀ + cz₀

Thus, the Cartesian form of the equation of a plane is:

ax + by + cz = d

where ( d = ax₀ + by₀ + cz₀) is a constant.

Read More about Coordinate Axes and Planes in 3D Space.

Intercept Form of Equation of Plane

To derive the equation of a plane in intercept form, we start with the general form of the equation of a plane:

ax + by + cz + d = 0

where ( a ), ( b ), and ( c ) are the coefficients representing the direction of the plane’s normal vector, and ( d ) is the distance from the origin along the normal vector.

To obtain the intercept form, we divide the entire equation by ( -d ) to isolate ( x ), ( y ), and ( z ):

[Tex]\frac{x}{-\frac{d}{a}} + \frac{y}{-\frac{d}{b}} + \frac{z}{-\frac{d}{c}} = 1 [/Tex]

Simplifying each term, we get:

[Tex]\frac{x}{\frac{-d}{a}} + \frac{y}{\frac{-d}{b}} + \frac{z}{\frac{-d}{c}} = 1[/Tex]

⇒ [Tex]\frac{x}{\frac{a}{-d}} + \frac{y}{\frac{b}{-d}} + \frac{z}{\frac{c}{-d}} = 1[/Tex]

⇒ [Tex]\frac{x}{\frac{a}{d}} + \frac{y}{\frac{b}{d}} + \frac{z}{\frac{c}{d}} = 1 [/Tex]

Finally, we rewrite the equation in the intercept form:

[Tex]\frac{x}{\frac{d}{a}} + \frac{y}{\frac{d}{b}} + \frac{z}{\frac{d}{c}} = 1 [/Tex]

Hence, the equation of a plane in intercept form is:

[Tex]\frac{x}{\frac{d}{a}} + \frac{y}{\frac{d}{b}} + \frac{z}{\frac{d}{c}} = 1 [/Tex]

Example of Planes in Intercept Form

In intercept form, the equation of a plane is given by its intercepts on the three coordinate axes. Here are some examples:

XY-Plane: ax​ + by ​+ cz ​= 1

• This equation represents a plane intersecting the x-axis at a, the y-axis at b, and the z-axis at c.

YZ-Plane: ax​=0

• This equation represents a plane parallel to the yz-plane and intersecting the x-axis at a.

XZ-Plane: by​=0

• This equation represents a plane parallel to the xz-plane and intersecting the y-axis at b.

Plane with Intercepts: x/2 + y/3 + z/4 = 1

• This equation represents a plane intersecting the x-axis at 2, the y-axis at 3, and the z-axis at 4.

Equation of a Plane in Parametric Form

The vector form of the equation of a plane:

[Tex]\vec{r} = \vec{r_0} + s\vec{v} + t\vec{w}[/Tex]

Where ( [Tex]\vec{r}[/Tex] ) is a point on the plane, ( [Tex]\vec{r_0}[/Tex] ) is a known point on the plane, and ( [Tex]\vec{v}[/Tex]) and ( [Tex]\vec{w} [/Tex]) are two non-parallel vectors lying on the plane.

We can rewrite this equation as:

• x = x₀ + sv₁ + tw₁
• y = y₀ + sv₂ + tw₂
• z = z₀ + sv₃ + tw₃

Where (v₁, v₂, v₃) and (w₁, w₂, w₃) are the components of vectors ( [Tex]\vec{v}[/Tex]) and ( [Tex]\vec{w}[/Tex] ) respectively.

These equations represent the parametric form of the equation of a plane. The parameters ( s ) and ( t ) allow us to generate any point (x, y, z) lying on the plane by varying ( s ) and ( t ) over the real numbers.

Read More,

• Points, Lines, Planes
• Equation of a Line in 3D
• Real Life Examples of a Plane in Geometry
• What is a Plane in Geometry?

Solved Examples

Example 1: Given three non-collinear points P₁​(x₁​,y₁​,z₁​), P₂​(x₂​,y₂​,z₂​), and P₃​(x₃​,y₃​,z₃​), find the equation of the plane passing through these points.

Solution:

Given three points P₁(1, 2, 3), P₂(4, 5, 6), and P₃(7, 8, 9).

1. Find two vectors on the plane:

[Tex]\vec{v_1}[/Tex] = ❬ 4 – 1, 5 – 2, 6 – 3 ❭ = ❬ 3, 3, 3 }

[Tex]\vec{v_2}[/Tex] = ❬ 7 – 1, 8 – 2, 9 – 3 ❭ = ❬ 6, 6, 6 }

2. Compute the cross product of these vectors to find the normal vector:

[Tex]\vec{n} = \vec{v_1} \times \vec{v_2} [/Tex]

= [Tex]\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 3 & 3 \\ 6 & 6 & 6 \end{vmatrix}[/Tex]

= [Tex]\hat{i}(3*6 – 3*6) – \hat{j}(3*6 – 3*6) + \hat{k}(3*6 – 3*6)[/Tex]

= ❬ 0,0,0 ❭

3. Choose one of the given points, say P₁(1, 2, 3), and use the point-normal form of the equation of the plane:

0(x – 1) + 0(y – 2) + 0(z – 3) = 0

⇒ 0 = 0

The equation simplifies to (0 = 0), which is always true. This means that the points (P₁), (P₂), and (P₃) are collinear, and there’s no unique plane passing through all three points.

Example 2: Given a point P₀​(x₀​,y₀​,z₀​) on the plane and a normal vector [Tex]\vec{n}[/Tex] =⟨a,b,c⟩, find the equation of the plane passing through P₀​ with the given normal vector.

Solution:

Given a point P₀(1, -2, 3) on the plane and a normal vector ( [Tex]\vec{n}[/Tex] = ❬ 2, -1, 4 ❭), let’s find the equation of the plane passing through ( P₀) with the given normal vector.

Using the point-normal form of the equation of a plane:

[Tex]\vec{n} \cdot (\vec{r} – \vec{r_0}) = 0[/Tex]

Substitute the given values:

❬ 2, -1, 4 ❭ · (❬ x, y, z ❭ – ❬ 1, -2, 3 ❭) = 0 \]

⇒ 2(x – 1) – 1(y + 2) + 4(z – 3) = 0

⇒ 2x – 2 – y – 2 + 4z – 12 = 0

⇒ 2x – y + 4z – 16 = 0

Hence, the equation of the plane passing through the point P₀(1, -2, 3) with the normal vector ( [Tex]\vec{n}[/Tex] = ❬ 2, -1, 4 ❭) is:

2x – y + 4z – 16 = 0

Example 3: Given two planes with their equations ax + by + cz + d₁​=0 and ex + fy + gz + d₂​=0, find the equation of the line of intersection between these two planes.

Solution:

Given two planes with equations (2x + y – z + 4 = 0) and (3x – y + 2z – 6 = 0), let’s find the equation of the line of intersection between these two planes.

To find the line of intersection, we will solve the system of equations formed by equating the two planes:

• 2x + y – z + 4 = 0
• 3x – y + 2z – 6 = 0

1. Multiply the first equation by 3 and the second equation by 2 to make the coefficients of (y) equal and opposite:

• 6x + 3y – 3z + 12 = 0
• 6x – 2y + 4z – 12 = 0

2. Subtract the second equation from the first:

(6x + 3y – 3z + 12) – (6x – 2y + 4z – 12) = 0

⇒ 6x + 3y – 3z + 12 – 6x + 2y – 4z + 12 = 0

⇒ 5y – 7z + 24 = 0

3. Rearrange the equation to isolate (y):

5y = 7z – 24

⇒ y = 7/5z – 24/5

4. Express (z) in terms of a parameter (t):

z = t

5. Substitute (z = t) into the equation for (y):

y = 7/5t – 24/5

6. Express (x) in terms of (t) using one of the original equations.

2x + y – z + 4 = 0

⇒ 2x + (7/5t – 24/5) – t + 4 = 0

⇒ 2x + 7/5t – t – 24/5 + 4 = 0

⇒ 2x + 7/5t – 5/5t – 24/5 + 20/5 = 0

⇒ [Tex]2x + \frac{2}{5}t – \frac{24}{5} = 0[/Tex]

⇒ [Tex]2x + \frac{2}{5}t = \frac{24}{5}[/Tex]

⇒ 2x = [Tex]\frac{24}{5} – \frac{2}{5}t[/Tex]

⇒ 2x = [Tex]\frac{24 – 2t}{5}[/Tex]

⇒ x = [Tex]\frac{24 – 2t}{10}[/Tex]

⇒ x = [Tex]\frac{12 – t}{5}[/Tex]

Hence, the parametric equations for the line of intersection between the two planes are:

• x = [Tex]\frac{12 – t}{5}[/Tex]
• y = 7/5t – 24/5
• z = t

Practice Questions: Equation of Plane

Q1: Find the equation of the plane passing through the point (2, -1, 3) and perpendicular to the line with direction vector (❬ 1, 2, -3❭).

Q2: Determine the equation of the plane passing through the point (-1, 4, 2) and parallel to the plane with equation (2x – y + 3z = 5).

Q3: Find the equation of the plane passing through the points (-3, 1, 0), (1, 2, 3), and (2, -1, 4).

Q4: Determine the equation of the plane passing through the line of intersection of the planes (x – y + 2z = 3) and (2x + y – z = 1), and parallel to the plane (3x + 2y + 4z = 5).

Q5: Find the equation of the plane passing through the origin and containing the lines given by the parametric equations: (x = 2t – 1), (y = 3t + 2), (z = t + 4), and (x = t + 1), (y = 2t – 3), (z = 3t + 5).

FAQs: Equation of Plane

What is Plane?

A plane, in mathematics and geometry, refers to a flat, two-dimensional surface that extends infinitely in all directions. 

What are the Different forms of Equations of Plane?

There are various forms of the equation of a plane, each serving different purposes. These forms include the general form, normal form, intercept form, and vector form.

What are the General Equation of Plane?

The general equation of a plane takes the form ax + by + cz + d =0, where a, b, and c are coefficients representing the plane’s normal vector, and d is the distance from the origin along the normal vector.

What is Intercept form of Equation of Plane?

Intercept form of the equation of a plane is expressed as x/a​ + y/b + z/c ​= 1, where a, b, and c are the intercepts of the plane on the x, y, and z axes respectively.

Write Equation of Plane in Vector Form.

Equation of plane in vector form is expressed as ([Tex]\vec{r} = \vec{r_0} + s\vec{v} + t\vec{w}[/Tex]), where ([Tex]\vec{r_0}[/Tex]) is the known point, ([Tex]\vec{v}[/Tex]) and ([Tex]\vec{w}[/Tex]) are the non-parallel vectors, and (s) and (t) are scalar parameters.