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Equation of Plane
Equation of plane represents a plane surface, in a threedimensional space. Equation of a plane can be derived through four different methods, based on the input values given. The equation of the plane can be expressed either in cartesian form or vector form.
Let us check the different methods of forming an equation of plane, the derivation of different methods, and the different forms of the equation of plane.
What Are the Equations of Plane?
The equation of a plane can be computed through different methods based on the available inputs values about the plane. The following are the four different expressions for the equation of plane.
 Equation of a plane at a perpendicular distance d from the origin and having a unit normal vector \(\hat n \) is \(\overrightarrow r. \hat n\) = d.
 The equation of a plane perpendicular to a given vector \(\overrightarrow N \), and passing through a point \(\overrightarrow a\) is \((\overrightarrow r  \overrightarrow a). \overrightarrow N = 0\)
 The equation of a plane passing through three non collinear points \(\overrightarrow a\), \(\overrightarrow b\), and \(\overrightarrow c\), is \((\overrightarrow r  \overrightarrow a)[(\overrightarrow b  \overrightarrow a) × (\overrightarrow c  \overrightarrow a)] = 0\).
 The equation of a plane passing through the intersection of two planes \(\overrightarrow r .\hat n_1 = d_1\), and \(\overrightarrow r.\hat n_2 = d_2 \), is \(\overrightarrow r(\overrightarrow n_1 + λ \overrightarrow n_2) = d_1 + λd_2\).
Derivation of Equations of Plane
Here we shall aim at understanding the proof of different methods to find the equation of plane.
Equation of a Plane in Normal Form
Let us consider a normal \(\overrightarrow ON \) to the plane. Normal is a perpendicular line drawn from the origin O to a point N in the plane, such that \(\overrightarrow ON \) is perpendicular to the pane. Let the length of the normal \(\overrightarrow ON\) be d units, such that \(\overrightarrow ON = d \hat n\). Further, we shall consider a point P in the plane, having a position vector of \(\overrightarrow r\). We now have \(\overrightarrow NP = \overrightarrow r  d. \hat n\). Also \(\overrightarrow NP\) and \(\overrightarrow ON\) are perpendicular to each other, and the dot product of these two perpendicular lines is equal to 0. Finally, we have the following expression for the dot product of these two lines as follows.
\(\overrightarrow NP . \overrightarrow ON = 0 \)
\((\overrightarrow r  d. \hat n).d \hat n = 0 \)
\((\overrightarrow r  d. \hat n)\hat n = 0 \)
\(\overrightarrow r. \hat n  d.\hat n. \hat n = 0 \)
\(\overrightarrow r. \hat n  d = 0\) (Since \(\hat n.\hat n = 1\))
\(\overrightarrow r.\hat n = d \)
Equation of a Plane Perpendicular to a given vector and through a Point
Let us consider a point A in the plane with a position vector \(\overrightarrow a\), and a vector \(\overrightarrow N\), which is perpendicular to this plane. Let us consider another point P in the plane having a position vector \(\overrightarrow r \). The line \(\overrightarrow AP \) lies in this referred plane and is perpendicular to the normal \(\overrightarrow N\). Here we have the dot product of these two lines equal to zero. \(\overrightarrow AP.\overrightarrow N = 0\). Solving this further we have the following expression.
\(\overrightarrow AP . \overrightarrow N = 0 \)
\((\overrightarrow r  \overrightarrow a). \overrightarrow N = 0 \)
Equation of a Plane Passing Through Three Non Collinear Points
Let us consider three noncollinear points A, B, C in the required plane and having the position vectors as \(\overrightarrow a\), \(\overrightarrow b\), \(\overrightarrow c\) respectively. The product \(\overrightarrow AB × \overrightarrow BC\) gives a vector which is perpendicular to this plane, and it can be referred as the normal to the plane. Here we consider a point P in the plane with the position vector \(\overrightarrow r\). The equation of a plane passing through this point P and perpendicular to \(\overrightarrow AB × \overrightarrow BC\) can be obtained from the dot product of the line \(\overrightarrow AP\), and the perpendicular \(\overrightarrow AB × \overrightarrow BC\). Finally, we have the below expression to derive the equation of the plane.
\(\overrightarrow AP . (\overrightarrow AB × \overrightarrow BC) = 0 \)
\((\overrightarrow r  \overrightarrow a) . [(\overrightarrow b  \overrightarrow a) × (\overrightarrow c  \overrightarrow a)] = 0 \)
Equation of a plane passing through the Intersection of Two Given Planes.
The given two equations of a plane are \(\overrightarrow r.\overrightarrow n_1 = d_1 \), and \(\overrightarrow r.\overrightarrow n_2 = d_2\). The position vector of any point on the line of intersection of these two planes must satisfy both the equations of the planes. If \(\overrightarrow k\) is the position vector of any point on the line of intersection of these two planes, then we have \(\overrightarrow k.\overrightarrow n_1 = d_1\), and \(\overrightarrow k.\overrightarrow n_2 = d_2\).
For any real values of a constant λ, we have \(\overrightarrow k.(\overrightarrow n_1 + λ.\overrightarrow n_2) = d_1 + λ.d_2\). Here \(\overrightarrow k\) is arbitrary and can be replaced with r to obtain the required equation of the plane.
Thus the equation of the plane passing through the line of intersection of the two planes \(\overrightarrow r.\overrightarrow n_1 = d_1\), and \(\overrightarrow r.\overrightarrow n_2 = d_2\) is \(\overrightarrow k.(\overrightarrow n_1 + λ.\overrightarrow n_2) = d_1 + λ.d_2\). Further this equation can be solved for λ, to obtain the required equation of the Plane.
Cartesian Form of Equation of Plane
The equation of a plane in vector form can easily be transformed into cartesian form by presenting the values of each of the vectors in the equation.
Equation of Plane in Normal Form
The vector form of equation of a plane is \(\overrightarrow r.\hat n = d \). Here let us substitute \(\overrightarrow r = x\hat i + y\hat j + z\hat k \), and the unit normal vector \(\hat n = l\hat i + m\hat j + n\hat k \).
\((x\hat i + y\hat j + z\hat k )( l\hat i + m\hat j + n\hat k) = d\)
lx + my + nk = d
lx + my + nk = d is the required cartesian form of equation of a line.
Equation of a Plane Perpendicular to a given vector and through a Point
The vector form of equation of a plane is \((\overrightarrow r  \overrightarrow a). \overrightarrow N = 0 \). Here we take \(r = x\hat i + y\hat j + z\hat k\), \(a = x_1 \hat i + y_1 \hat j + z_1 \hat k \), and \(N = A\hat i + B\hat j + C\hat k\) respectively. Substituting these in the vector form of the equation of the line we have the following expression.
\([(x\hat i + y\hat j + z\hat k)  (x_1 \hat i + y_1 \hat j + z_1 \hat k )].( A\hat i + B\hat j + C\hat k) = 0\)
\([(x  x_1)\hat i + (y  y_1)\hat j + (z  z_1)\hat k] ( A\hat i + B\hat j + C\hat k) = 0\)
\(A(x  x_1) + B(y  y_1) + C(z  z_1) = 0\)
Equation of a Plane Passing Through Three NonCollinear Points
The equation of plane passing through three noncollinear points A, B, C, having the position vectors as \(\overrightarrow a\), \(\overrightarrow b\), \(\overrightarrow c\) is \(\overrightarrow AP\), and the perpendicular \(\overrightarrow AB × \overrightarrow BC\). Here we take \(r = x\hat i + y\hat j + z\hat k\), and the points as A\((x_1, y_1, z_1)\), B\((x_2, y_2. z_2)\), and C\((x_3, y_3, z_3)\).
AP = \((x  x_1) \hat i + (y  y_1)\hat j + (z  z_1)\hat k \)
AB = \((x_2  x_1) \hat i + (y_2  y_1)\hat j + (z_2  z_1)\hat k \)
AC = AB = \((x_3  x_1) \hat i + (y_3  y_1)\hat j + (z_3  z_1)\hat k \)
Substituting these in the above equation of the plane we have the following cartesian form of equation of plane.
\( \begin{bmatrix}x  x_1&y  y_1 & z  z_1\\x_2  x_1&y_2  y_1 & z_2  z_1\\x_3  x_1&y_3  y_1&z_3  z_1\end{bmatrix}=0\)
Equation of a plane passing through the Intersection of Two Given Planes.
The equation of a plane passing through the intersection of two planes \(\overrightarrow r.\overrightarrow n_1 = d_1 \), and \(\overrightarrow r.\overrightarrow n_2 = d_2\), is \(\overrightarrow r.(\overrightarrow n_1 + λ.\overrightarrow n_2) = d_1 + λ.d_2\). To convert this equation of plane in cartesian form let us take \(\overrightarrow n = A_1\hat i + B_1\hat j + C_1\hat k \), \(\overrightarrow n = A_2\hat i + B_2\hat j + C_2\hat k \), and \(\overrightarrow r = x\hat i + y\hat j + z\hat k \).
Substituting these vectors in the above equation of a plane, we have the following expression.
\(x(A_1 + λ A_2) + y(B_1 + λB_2) + z(C_1 + λ C_2) = d_1 + λd_2\)
\((A_1x + B_1y + C_1z  d_1) + λ(A_2x + B_2y + C_2z  d_2) = 0\)
Related Topics
The following topics would help in a better understanding of the equation of a plane.
Examples on Equation of Plane

Example 1: Find the vector equation of plane passing through a point (2, 1, 3), and having the direction ratios of its normal as (5, 2, 4).
Solution:
The coordinates of the point (2, 1, 3) can be represented as a position vector \(\overrightarrow a = 2\hat i 1 \hat j + 3 \hat k\). And the normal having the direction ratios (5, 2, 4) can be represented as a vector \(\overrightarrow N = 5\hat i + 2\hat j + 4 \hat k\).
The equation of a line passing through a point and having a normal is \((\overrightarrow r  \overrightarrow a). \overrightarrow N = 0\).
\((\overrightarrow r  (2\hat i 1 \hat j + 3 \hat k)). (5\hat i + 2\hat j + 4 \hat k) = 0\).
Therefore, \((\overrightarrow r  (2\hat i 1 \hat j + 3 \hat k)). (5\hat i + 2\hat j + 4 \hat k) = 0\) is the required vector equation of plane.

Example 2: Find the vector equation of plane passing through the points A(2, 5, 3), B(3, 3, 5), C(4, 2, 3).
Solution:
The three given points in the plane A(2, 5, 3), B(3, 3, 5), C(4, 2, 3), can be represented as position vectors \(\overrightarrow a = 2\hat i +5 \hat j  3 \hat k\), \(\overrightarrow b = 3\hat i +3 \hat j 5 \hat k\), and \(\overrightarrow c = 4\hat i 2 \hat j + 3 \hat k\).
The required equation of plane passing through the three given points is \((\overrightarrow r  \overrightarrow a)[(\overrightarrow b  \overrightarrow a) × (\overrightarrow c  \overrightarrow a)] = 0\)
\((\overrightarrow r  (2\hat i +5 \hat j  3 \hat k))[((3\hat i +3 \hat j 5 \hat k)  (2\hat i +5 \hat j  3 \hat k)) × ((4\hat i 2 \hat j + 3 \hat k)  (2\hat i +5 \hat j 3 \hat k))] = 0\)
\((\overrightarrow r  (2\hat i +5 \hat j  3 \hat k))[(\hat i  2\hat j 2 \hat k) × (2 \hat i 7\hat j + 6\hat k)] = 0\)
FAQs on Equation of Plane
What Are the Equations of Plane?
The different equations of plane are as follows.
 The equation of plane having a unit normal vector and at a distance from the origin is \(\overrightarrow r. \hat n\) = d.
 The equation of a plane passing through a point and having a normal is \((\overrightarrow r  \overrightarrow a). \overrightarrow N = 0\)
 The equation of plane passing through three non collinear points is \((\overrightarrow r  \overrightarrow a)[(\overrightarrow b  \overrightarrow a) × (\overrightarrow c  \overrightarrow a)] = 0\).
 The equation of plane passing through the intersection of two planes is \(\overrightarrow r(\overrightarrow n_1 + λ \overrightarrow n_2) = d_1 + λd_2\).
How To You Find the Equation of Plane From the Given Points?
The equation of the plane passing through three given points A, B, C can be calculated by taking the position vectors of these points as \(\overrightarrow a\), \(\overrightarrow b\), \(\overrightarrow c\) respectively. The product \(\overrightarrow AB × \overrightarrow BC\) gives a vector which is perpendicular to this plane, and it can be referred as the normal to the plane. Here we consider a point P in the plane with the position vector \(\overrightarrow r\). The equation of a plane passing through this point P and perpendicular to this normal is \((\overrightarrow r  \overrightarrow a) . [(\overrightarrow b  \overrightarrow a) × (\overrightarrow c  \overrightarrow a)] = 0 \).
What Are the Different Forms of Equation of Plane?
The equation of a plane is generally represented in two forms. The vector form of the equation of a plane, and the cartesian form of the equation of a plane.
What Is Intercept Form of Equation of Plane
The intercept form of equation of plane is of the form x/a + y/b + z/c = 1. Here a, b, c are the xintercept, yintercept, and zintercepts respectively. Further this plane cuts the xaxis at the point (a, 0, 0), yaxis at the point (0, b, 0), and the zaxis at the point(0, 0, c).
How Do You Find Equation of Plane in Vector Form?
The equation of a plane in vector form is obtained by representing the normal and the points in the plane in vector form. The following are the vector form of equations of the plane.
 \(\overrightarrow r. \hat n\) = d.
 \((\overrightarrow r  \overrightarrow a). \overrightarrow N = 0\)
 \((\overrightarrow r  \overrightarrow a)[(\overrightarrow b  \overrightarrow a) × (\overrightarrow c  \overrightarrow a)] = 0\).
 \(\overrightarrow r(\overrightarrow n_1 + λ \overrightarrow n_2) = d_1 + λd_2\).
How Do You Find Equation of Plane in Cartesian Form?
The equation of plane in cartesian form is obtained by representing the normal and the points as coordinates in a cartesian plane. The cartesian form of equations of a plane are as follows.
 \(lx + my + nk = d \)

\(A(x  x_1) + B(y  y_1) + C(z  z_1) = 0\)
 \((A_1x + B_1y + C_1z  d_1) + λ(A_2x + B_2y + C_2z  d_2) = 0\)

\( \begin{bmatrix}x  x_1&y  y_1 & z  z_1\\x_2  x_1&y_2  y_1 & z_2  z_1\\x_3  x_1&y_3  y_1&z_3  z_1\end{bmatrix}=0\)
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