Find an equation of the plane. The plane through the origin and the points (2, -4, 6) and (5, 1, 3).

Find an equation of the plane. The plane through the origin and the points (2, -4, 6) and (5, 1, 3).

Finding the Equation of a Plane

In this article, we will learn how to find the equation of a plane that passes through the origin and two given points.

Step 1: Understand the problem

We are given the following information:

  • The plane passes through the origin (0, 0, 0)
  • The plane also passes through the points (2, -4, 6) and (5, 1, 3)

To find the equation of the plane, we need to determine the normal vector of the plane, which is perpendicular to the plane.

Step 2: Find the normal vector

The normal vector of the plane can be found by taking the cross product of two vectors that lie on the plane.

Let's define two vectors:

  • Vector 1: from the origin to the point (2, -4, 6)
  • Vector 2: from the origin to the point (5, 1, 3)

Vector 1 = (2, -4, 6) Vector 2 = (5, 1, 3)

The normal vector is the cross product of these two vectors: Normal vector = Vector 1 × Vector 2 = (2, -4, 6) × (5, 1, 3) = (-27, -15, 17)

Step 3: Write the equation of the plane

The equation of a plane in 3D space can be written in the form: Ax + By + Cz + D = 0

where (A, B, C) is the normal vector, and D is a constant.

Since the plane passes through the origin, we know that the constant D is equal to 0.

Therefore, the equation of the plane is: -27x - 15y + 17z = 0

This is the equation of the plane that passes through the origin and the points (2, -4, 6) and (5, 1, 3).

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