Question

Find an equation of the plane. The plane passes through \((1,2,3),(3,2,1),\) and (-1,-2,2) .

Short answer

The equation of the plane that passes through the points \((1,2,3),(3,2,1),\) and \((-1,-2,2)\) is \( 4x + 6y - 8z = 14 \).

Step-by-step solution

Determine Two Vectors from the Given Points

Firstly, calculate two vectors on the plane using the coordinates of the given points. Vector \( \mathbf{AB} = (B_x - A_x, B_y - A_y, B_z - A_z) = (3 - 1, 2 - 2, 1 - 3) = (2, 0, -2) \) and Vector \( \mathbf{AC} = (C_x - A_x, C_y - A_y, C_z - A_z) = (-1 - 1, -2 - 2, 2 - 3) = (-2, -4, -1) \).

Find the Normal Vector

The normal vector to the plane is given by the cross product between two vectors on the plane. Using the cross product formula, we get \( \mathbf{N} = \mathbf{AB} \times \mathbf{AC} = (0* -1 -((-2)* -2), -2* -1 - (2* -4), 2* -4 - (0* -2)) = (4, 6, -8) \).

Find the Plane Equation

Plugging the coefficients from the normal vector and the coordinates of a point on the plane (it could be any of the given points) into the plane equation, we get \( 4(x - 1) + 6(y - 2) - 8(z - 3) = 0 \). Simplifying this equation, we obtain \( 4x + 6y - 8z = 14 \).

Key concepts

Vector Mathematics

Vector mathematics is fundamental to understanding geometric and physical problems. In the context of plane equations, vectors represent quantities with both magnitude and direction. They are denoted by an arrow over a letter and written in coordinate form, such as \( \mathbf{AB} \), which is the vector from point A to point B.

When finding a plane's equation, vectors are crucial for determining the direction of the plane, and their coordinates are used to express the position of points within a reference frame. By subtracting the coordinates of the initial point from the terminal point, we obtain the direction vector, which is an essential first step for calculating further properties such as the normal vector of the plane.

Cross Product

The cross product, also known as the vector product, is an operation on two vectors that produces a third vector perpendicular to both. It's denoted by the symbol '×' and is crucial in finding the normal vector of a plane.

The formula for the cross product of vectors \( \mathbf{A} \) and \( \mathbf{B} \) is \( \mathbf{A} \times \mathbf{B} \), and the result is a vector not in the same plane as \( \mathbf{A} \) and \( \mathbf{B} \). In our exercise, we applied this concept to find the normal vector \( \mathbf{N} \) by crossing the vectors we found on the plane.

Normal Vector

The normal vector is a key concept in defining a plane. It is a vector that is perpendicular to every line lying in the plane, essentially defining the plane's orientation in three-dimensional space.

Once the two vectors lying on the plane are determined, as seen in the solution steps \( \mathbf{AB} \) and \( \mathbf{AC} \), the normal vector \( \mathbf{N} \) can be obtained by computing their cross product. The coordinates of this normal vector are then used as the coefficients in the scalar equation of the plane, as they describe the plane's perpendicular direction in the three-dimensional Euclidean space.

Coordinate Geometry

Coordinate geometry, or analytic geometry, provides a connection between algebra and geometry through the use of an ordered pair of numbers—the coordinates. It allows us to describe geometric shapes in a numerical way, which is particularly useful for equations of lines, circles, and planes in two and three dimensions.

In the context of the plane's equation, after finding the normal vector, we use the coordinates of a known point on the plane—such as point A in our exercise—to anchor the plane in space. The final step develops the equation based on both the normal vector and this anchor point, linking coordinate geometry with vector mathematics to provide a complete description of the plane.