Plane passing through three points (cartesian form)

Plane Passing Through Three Points (Cartesian Form)

In three-dimensional geometry, a plane can be uniquely determined by three non-collinear points. Non-collinear means that the points do not all lie on the same line. If you have three such points, you can find the equation of the plane in Cartesian form. The Cartesian form of a plane's equation is typically written as:

$$ Ax + By + Cz + D = 0 $$

where (A), (B), (C), and (D) are real numbers, and (x), (y), and (z) are the coordinates of any point on the plane.

Determining the Equation of a Plane

To find the equation of a plane that passes through three points (P_1(x_1, y_1, z_1)), (P_2(x_2, y_2, z_2)), and (P_3(x_3, y_3, z_3)), follow these steps:

  1. Find two vectors that lie in the plane by subtracting the coordinates of the points:

    • (\vec{AB} = \vec{P_2} - \vec{P_1} = \langle x_2 - x_1, y_2 - y_1, z_2 - z_1 \rangle)
    • (\vec{AC} = \vec{P_3} - \vec{P_1} = \langle x_3 - x_1, y_3 - y_1, z_3 - z_1 \rangle)

  2. Compute the cross product of these two vectors to find a normal vector (\vec{n}) to the plane:

    • (\vec{n} = \vec{AB} \times \vec{AC})

  3. The components of the normal vector (\vec{n} = \langle A, B, C \rangle) give us the coefficients for the plane's equation.

  4. Substitute the coordinates of one of the points into the plane's equation to solve for (D).

  5. Write the final equation of the plane.

Formula for the Cross Product

The cross product of two vectors (\vec{a} = \langle a_1, a_2, a_3 \rangle) and (\vec{b} = \langle b_1, b_2, b_3 \rangle) is given by:

$$ \vec{a} \times \vec{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle $$

Example

Let's find the equation of the plane passing through the points (P_1(1, 2, 3)), (P_2(4, 5, 6)), and (P_3(7, 8, 9)).

  1. Calculate the vectors (\vec{AB}) and (\vec{AC}):

    • (\vec{AB} = \langle 4 - 1, 5 - 2, 6 - 3 \rangle = \langle 3, 3, 3 \rangle)
    • (\vec{AC} = \langle 7 - 1, 8 - 2, 9 - 3 \rangle = \langle 6, 6, 6 \rangle)

  2. Compute the cross product (\vec{n} = \vec{AB} \times \vec{AC}):

    • (\vec{n} = \langle 3 \cdot 6 - 3 \cdot 6, 3 \cdot 6 - 3 \cdot 6, 3 \cdot 6 - 3 \cdot 6 \rangle = \langle 0, 0, 0 \rangle)

  3. In this case, the cross product is the zero vector, which means that our initial points were collinear, and no unique plane exists that passes through all three. This is an exceptional case.

Let's try another set of points: (P_1(1, 1, 1)), (P_2(2, 3, 4)), and (P_3(0, -1, 2)).

  1. Calculate the vectors (\vec{AB}) and (\vec{AC}):

    • (\vec{AB} = \langle 2 - 1, 3 - 1, 4 - 1 \rangle = \langle 1, 2, 3 \rangle)
    • (\vec{AC} = \langle 0 - 1, -1 - 1, 2 - 1 \rangle = \langle -1, -2, 1 \rangle)

  2. Compute the cross product (\vec{n} = \vec{AB} \times \vec{AC}):

    • (\vec{n} = \langle 2 \cdot 1 - 3 \cdot (-2), 3 \cdot (-1) - 1 \cdot 1, 1 \cdot (-2) - 2 \cdot (-1) \rangle)
    • (\vec{n} = \langle 2 + 6, -3 - 1, -2 + 2 \rangle = \langle 8, -4, 0 \rangle)

  3. The normal vector components give us (A = 8), (B = -4), and (C = 0).

  4. Substitute the coordinates of (P_1) into the plane's equation to find (D):

    • (8 \cdot 1 - 4 \cdot 1 + 0 \cdot 1 + D = 0)
    • (D = -4)

  5. The final equation of the plane is:

    • (8x - 4y + 0z - 4 = 0)
    • Simplified: (2x - y - 1 = 0)

Differences and Important Points

Aspect Description
Determination A plane is uniquely determined by three non-collinear points.
Equation Form The general form of a plane's equation in Cartesian coordinates is (Ax + By + Cz + D = 0).
Normal Vector The coefficients (A), (B), and (C) in the plane's equation are the components of a normal vector to the plane.
Cross Product The cross product of two vectors in the plane yields a normal vector.
Collinear Points If the given points are collinear, the cross product will be the zero vector, indicating that no unique plane exists.
Substitution Substituting the coordinates of one of the points into the plane's equation allows us to solve for (D).

Understanding the concept of a plane passing through three points is crucial for solving problems in 3D geometry, especially in fields such as computer graphics, physics, and engineering.