91Ó°ÊÓ

along the x-axis, \( y \) along the y-axis, and \( z \) along the z-axis. These components specify the vector's direction and how far it extends in 3D space from the origin. In the plane equation problem, \( \vec{OP_{0}} = (3, -1, 2) \) represents the vector from the origin \( O(0, 0, 0) \) to the point \( P_{0}(3, -1, 2) \) on the plane.

Understanding vectors in Cartesian coordinates is fundamental for many geometrical calculations, providing a systematic way of describing positions and directions in space.

The dot product, also known as the scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is denoted by \( \vec{u} \cdot \vec{v} \) for two vectors \( \vec{u} \) and \( \vec{v} \) and is calculated as \( (u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3}) \) in three-dimensional space.

The result of a dot product can tell us if two vectors are orthogonal (perpendicular), as orthogonal vectors have a dot product of zero. This property is crucial when determining the normal vector \( \vec{n} \) to a plane, as seen in the exercise, where \( \vec{n} \cdot \vec{OP_{0}} = 0 \) was used to formulate the characteristic relationship.

A normal vector to a plane is a vector that is perpendicular to that plane. The defining property of the normal vector \( \vec{n} \) is that it should be orthogonal to any vector that lies on the plane, making it a fundamental element in describing the plane's orientation in space.

In the context of the given exercise, the normal vector \( \vec{n} = (1, 3, 0) \) was identified because it is orthogonal to vector \( \vec{OP_{0}} \) that lies on the plane. Determining the normal vector is a critical step in finding the equation of a plane, as the vector's components directly contribute to the plane's standard form equation.

The equation of a plane in three-dimensional space can be determined if you know a point \( P_{0}(x_{0}, y_{0}, z_{0}) \) that lies on the plane and a normal vector \( \vec{n} = (A, B, C) \) to the plane. The general form of the plane equation is \( A(x - x_{0}) + B(y - y_{0}) + C(z - z_{0}) = 0 \) where the normal vector's components and the known point's coordinates are substituted in.

From the exercise, using the normal vector \( \vec{n} = (1, 3, 0) \) and the point \( P_{0}(3, -1, 2) \) on the plane, the equation \( x + 3y = 6 \) defines the plane. This concise equation allows us to understand the spatial relationship of any point relative to the plane and is a direct result of the normal vector properties and the specific point through which the plane passes.