Vector, Parametric, and Cartesian Equations of the Plane

A plane is described by two geometrically linearly independent vectors in different directions and a point through which the plane passes.

$$ v_1 = \begin{pmatrix} l_1 \\ m_1 \\ n_1 \end{pmatrix} \:\:\: v_2 = \begin{pmatrix} l_2 \\ m_2 \\ n_2 \end{pmatrix} $$

$$ P_0 = \begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix} $$

If we do not specify the point P₀, the two vectors v₁ and v₂ will define all planes parallel to these vectors, rather than a specific plane.

Therefore, any point on a plane can be described by the following vector equation of the plane:

$$ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix} + t_1 \cdot \begin{pmatrix} l_1 \\ m_1 \\ n_1 \end{pmatrix} + t_2 \cdot \begin{pmatrix} l_2 \\ m_2 \\ n_2 \end{pmatrix} $$

From the vector equation, we can easily derive the parametric equations of the plane:

$$ \begin{cases} x = x_0 + t_1 \cdot l_1 + t_2 \cdot l_2 \\ y = y_0 + t_1 \cdot m_1 + t_2 \cdot m_2 \\ z = z_0 + t_1 \cdot n_1 + t_2 \cdot n_2 \end{cases} $$

We can rewrite the parametric equations in the following form:

$$ \begin{cases} x - x_0 = t_1 \cdot l_1 + t_2 \cdot l_2 \\ y - y_0 = t_1 \cdot m_1 + t_2 \cdot m_2 \\ z - z_0 = t_1 \cdot n_1 + t_2 \cdot n_2 \end{cases} $$

This gives us three vectors P₀P, v₁, and v₂, which can be arranged in columns in a square matrix.

$$ \begin{pmatrix} x - x_0 & l_1 & l_2 \\ y - y_0 & m_1 & m_2 \\ z - z_0 & n_1 & n_2 \end{pmatrix} $$

By calculating the determinant of this matrix, we obtain the Cartesian equation of the plane:

$$ det \begin{pmatrix} x - x_0 & l_1 & l_2 \\ y - y_0 & m_1 & m_2 \\ z - z_0 & n_1 & n_2 \end{pmatrix} = $$

$$ (x-x_0) \cdot det \begin{pmatrix} m_1 & m_2 \\ n_1 & n_2 \end{pmatrix} + (y-y_0) \cdot det \begin{pmatrix} l_1 & l_2 \\ n_1 & n_2 \end{pmatrix} + (z-z_0) \cdot det \begin{pmatrix} l_1 & l_2 \\ m_1 & m_2 \end{pmatrix} $$

A Practical Example

Let's consider two direction vectors:

$$ v_1 = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} \:\:\: v_2 = \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} $$

and a point:

$$ P = \begin{pmatrix} 10 \\ 5 \\ 5 \end{pmatrix} $$

First, we verify if the two vectors are linearly independent.

The rank of the two vectors is two, indicating that they are linearly independent.

$$ r_k = \begin{pmatrix} 3 & 4 \\ 1 & 3 \\ 2 & 1 \end{pmatrix} = 2 $$

Note: If the rank equals the number of vectors, they are linearly independent.

Now, let's verify if point P lies on the plane.

How to Verify if a Point Belongs to the Plane

If point P is on the plane, there exist two parameters t₁ and t₂ that satisfy the following linear combination and the associated linear system:

$$ \overrightarrow{OP} = \overrightarrow{OP_0} + t_1 v_1 + t_2 v_2 $$

We substitute the direction vectors:

$$ \overrightarrow{OP} = \overrightarrow{OP_0} + t_1 \cdot \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + t_2 \cdot \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} $$

Then, we take a reference point P₀ on the plane, one of the endpoints of the vectors, for example, (3,1,2):

$$ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + t_1\cdot \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + t_2 \cdot \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} $$

Finally, we substitute the coordinates (x,y,z) with those of point (10,5,5) and check if the system has a solution:

$$ \begin{pmatrix} 10 \\ 5 \\ 5 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + t_1\cdot \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + t_2 \cdot \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} $$

The system has a solution.

To obtain point (10,5,5), we can assign the values t₁=1 and t₂=1:

$$ \begin{pmatrix} 10 \\ 5 \\ 5 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + 1 \cdot \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + 1 \cdot \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} $$

$$ \begin{pmatrix} 10 \\ 5 \\ 5 \end{pmatrix} = \begin{pmatrix} 3 + 3 + 4 \\ 1 + 1 + 3 \\ 2 + 2 + 1 \end{pmatrix} $$

$$ \begin{pmatrix} 10 \\ 5 \\ 5 \end{pmatrix} = \begin{pmatrix} 10 \\ 5 \\ 5 \end{pmatrix} $$

Thus, the vector equation of the plane is:

$$ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} x_0 \\ y_0 \\ z_0 \end{pmatrix} + t_1 \cdot \begin{pmatrix} l_1 \\ m_1 \\ n_1 \end{pmatrix} + t_2 \cdot \begin{pmatrix} l_2 \\ m_2 \\ n_2 \end{pmatrix} $$

$$ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 10 \\ 5 \\ 5 \end{pmatrix} + t_1 \cdot \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + t_2 \cdot \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} $$

An Alternative Verification: To verify the existence of a solution, we could transform the vector equation into three vectors: $$ \begin{pmatrix} 10 \\ 5 \\ 5 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + t_1 \cdot \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + t_2 \cdot \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} $$ $$ \begin{pmatrix} 10-3 \\ 5-1 \\ 5-2 \end{pmatrix} = t_1 \cdot \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + t_2 \cdot \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} $$ $$ \begin{pmatrix} 7 \\ 4 \\ 3 \end{pmatrix} = t_1 \cdot \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + t_2 \cdot \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} $$ Then arrange the vectors of the coefficients of the equation in a matrix: $$ \begin{pmatrix} 7 & 3 & 4 \\ 4 & 1 & 3 \\ 3 & 2 & 1 \end{pmatrix} $$ If the rank of the matrix is less than the number of vectors, which is less than three, then the three vectors are linearly dependent and the system has a solution. In this case, the rank of the matrix is two (r_k=2). Thus, the system has a solution: $$ r_k = 2 < 3 $$ The explanation is simple. The vectors v₁(3,1,2) and v₂(4,3,1) are linearly independent. For the first vector (7,1,3) to belong to the plane, it must be linearly dependent on v₁ and v₂.

Once we have the vector equation, we can easily derive the parametric equation of the plane.

We simply transform it into a system of equations:

$$ \begin{cases} x = 10 + t_1 \cdot 3 + t_2 \cdot 4 \\ y = 5 + t_1 \cdot 1 + t_2 \cdot 3 \\ z = 5 + t_1 \cdot 2 + t_2 \cdot 1 \end{cases} $$

$$ \begin{cases} x = 10 + 3t_1 + 4t_2 \\ y = 5 + t_1 + 3t_2 \\ z = 5 + 2t_1 + t_2 \end{cases} $$

Finally, to obtain the Cartesian equation of the plane, we transform the vector equation into three segments:

$$ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 10 \\ 5 \\ 5 \end{pmatrix} + t_1 \cdot \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + t_2 \cdot \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} $$

$$ \begin{pmatrix} x \\ y \\ z \end{pmatrix} - \begin{pmatrix} 10 \\ 5 \\ 5 \end{pmatrix} = t_1 \cdot \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + t_2 \cdot \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} $$

$$ \begin{pmatrix} x - 10 \\ y - 5 \\ z-5 \end{pmatrix} = t_1 \cdot \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + t_2 \cdot \begin{pmatrix} 4 \\ 3 \\ 1 \end{pmatrix} $$

Then place the three vectors in a column inside a matrix:

$$ \begin{pmatrix} x - 10 & 3 & 1 \\ y - 5 & 1 & 3 \\ z-5 & 2 & 1 \end{pmatrix} $$

The determinant of the matrix is the Cartesian equation of the plane:

We have thus found the Cartesian equation of the plane.

The Graphical Representation of the Plane in a three-dimensional diagram is as follows:

How to Convert from the Cartesian Equation to the Parametric Equation of the Plane

Let's try to reverse the process. We know the Cartesian equation of the plane and want to derive the parametric equation.

The Cartesian equation of the plane is:

$$ ax + by + cz + d = 0 $$

The vector-parametric equation of the plane is:

$$ P = P_0 + t_1 v_1 + t_2 v_2 $$

The coefficients a, b, and c of the Cartesian equation determine the normal vector of the plane, which is orthogonal to the plane:

$$ n = \begin{pmatrix} a \\ b \\ c \end{pmatrix} $$

If the normal vector is orthogonal to the plane, it is also orthogonal to any vector in the plane.

Therefore, the normal vector is orthogonal to the direction vectors of the parametric equation of the plane:

$$ \langle n , v_2 \rangle = 0 $$

$$ \langle n , v_1 \rangle = 0 $$

This relationship allows us to find two linearly independent vectors orthogonal to the normal vector and construct the vector-parametric equation of the plane.

Example

Consider the following Cartesian equation:

$$ -5x - y + 8z + 15 = 0 $$

The normal vector of the equation is:

$$ n = \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} -5 \\ -1 \\ 8 \end{pmatrix} $$

The normal vector is orthogonal to the plane.

It is therefore orthogonal to all non-null vectors in the plane:

Now we need to find two linearly independent vectors that are orthogonal to the normal vector.

To find them, we set one coordinate of the normal vector to zero, invert the other two, and change the sign of one of the latter.

For example, setting the x-coordinate of the normal vector to zero:

$$ \begin{pmatrix} 0 \\ -1 \\ 8 \end{pmatrix} $$

Inverting the other two coordinates:

$$ \begin{pmatrix} 0 \\ 8 \\ -1 \end{pmatrix} $$

Then changing the sign of one of the latter:

We find the first directional vector of the plane equation:

$$ v_1 = \begin{pmatrix} 0 \\ 8 \\ -1 \end{pmatrix} $$

Using the same method, we find the second directional vector:

For example, setting the y-coordinate to zero:

$$ \begin{pmatrix} -5 \\ 0 \\ 8 \end{pmatrix} $$

Inverting the remaining coordinates:

$$ \begin{pmatrix} 8 \\ 0 \\ -5 \end{pmatrix} $$

Then changing the sign of one of the latter:

$$ v_2 = \begin{pmatrix} 8 \\ 0 \\ 5 \end{pmatrix} $$

We have found two vectors orthogonal to the normal vector:

Verify that they are linearly independent vectors:

$$ r_k \begin{pmatrix} 0 & 8 \\ 8 & 0 \\ -1 & 5 \end{pmatrix} = 2 $$

The rank of the matrix with the vectors in the column is equal to 2, which is the number of vectors.

So the two vectors are linearly independent.

Next, verify that they are also orthogonal to the normal vector:

$$ \langle n , v_1 \rangle = \langle \begin{pmatrix} -5 \\ -1 \\ 8 \end{pmatrix} , \begin{pmatrix} 0 \\ 8 \\ -1 \end{pmatrix} \rangle = 0 $$

$$ \langle n , v_2 \rangle = \langle \begin{pmatrix} -5 \\ -1 \\ 8 \end{pmatrix} , \begin{pmatrix} 8 \\ 0 \\ 5 \end{pmatrix} \rangle = 0 $$

The dot product is zero.

So the two vectors v₁ and v₂ are orthogonal vectors to the normal vector.

We can use them as directional vectors of the plane.

Once linear independence and orthogonality are verified, we can substitute directional vectors v₁ and v₂ into the vector equation of the plane:

$$ P = P_0 + t_1 v_1 + t_2 v_2 $$

$$ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = P_0 + t_1 \begin{pmatrix} 0 \\ 8 \\ -1 \end{pmatrix} + t_2 \begin{pmatrix} 8 \\ 0 \\ 5 \end{pmatrix} $$

This is the parametric equation of the plane and all parallel planes.

To restrict the equation to the specific plane of interest, we just need to add any point P₀ on the plane.

For example, set x=0 and z=0 in the Cartesian equation to determine the y-coordinate:

$$ -5x -y +8z + 15 = 0 \\ -5(0) -y +8(0) + 15 = 0 \\ y = 15$$

We thus find a point on the plane P(x,y,z) = (0,15,0).

Substitute this into the vector equation in place of P₀.

We have thus found the vector equation of the plane:

$$ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 15 \\ 0 \end{pmatrix} + t_1 \begin{pmatrix} 0 \\ 8 \\ -1 \end{pmatrix} + t_2 \begin{pmatrix} 8 \\ 0 \\ 5 \end{pmatrix} $$

Once we have the vector equation, it is easy to derive the parametric equation of the plane.

Simply transform the vectors into a system of equations:

$$ \begin{cases} x = 8t_2 \\ y = 15 + 8 t_1 \\ z = -t_1 + 5t_2 \end{cases} $$

How to Find the Cartesian Equation from the Normal Vector

Consider the following Cartesian equation:

$$ -5x - y + 8z + 15 = 0 $$

The normal vector of the equation is:

$$ n = \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} -5 \\ -1 \\ 8 \end{pmatrix} $$

The normal vector is orthogonal to the plane.

Identify any point on the plane.

Set the y and z coordinates to zero to calculate the value of x:

$$ -5x -y +8z + 15 = 0 $$

$$ -5x -(0) +8(0) + 15 = 0 $$

$$ x = 15 $$

We find a point P₁ on the plane:

$$ P_1 = \begin{pmatrix} 15 \\ 0 \\ 0 \end{pmatrix} $$

Consider a generic point on the plane:

$$ P = \begin{pmatrix} x \\ y \\ z \end{pmatrix} $$

Calculate the segment between the two points:

$$ P_1P = \begin{pmatrix} x \\ y \\ z \end{pmatrix} - \begin{pmatrix} 15 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} x - 15 \\ y \\ z \end{pmatrix} $$

We thus find a generic vector v of the plane.

We already know that any vector in the plane must be orthogonal to the normal vector.

Therefore, the dot product <v,n> is zero:

$$ \langle P_1P , n \rangle = 0 $$

Perform the calculations of the dot product:

$$ \langle \begin{pmatrix} x - 15 \\ y \\ z \end{pmatrix} , \begin{pmatrix} -5 \\ -1 \\ 8 \end{pmatrix} \rangle = -5(x-15)-y+8z = 0 $$

$$ -5x+75-y+8z = 0 $$

We thus find the Cartesian equation of the plane.