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A plane equation is a mathematical expression that describes all the points that lie on a flat two-dimensional surface in three-dimensional space. The most common form of the plane equation is given as \( ax + by + cz = d \), where \( a \), \( b \), and \( c \) represent the coefficients that describe the direction of the plane's normal vector:

• \( a \), \( b \), and \( c \) are the components of a vector that is perpendicular, or normal, to the plane.
• \( d \) can be thought of as the distance from the origin to the plane, along the normal vector.

In the original exercise, we used the intersection point \((2, 0, 2)\) and the normal vector \( \langle -2, -2, 0 \rangle \) to form the equation \( -2(x - 2) - 2y = 0 \). Simplifying gives us \(-2x - 2y + 4 = 0\). This shows how each point \((x, y, z)\) on the plane satisfies the equation.

Parameterization is a method for describing a line using a parameter, usually denoted by \( t \). It allows for the expression of lines in a system of equations as a set of vector functions. For a line through point \( \mathbf{r}_0 \) with a direction vector \( \mathbf{d} \), it can be given by:

• \( \mathbf{r} = \mathbf{r}_0 + t\mathbf{d} \)

Here, \( t \) is a scalar that allows the line to be traced out by varying \( t \).
The original exercise provides two parameterized lines:

• First line: \(\mathbf{r}_1 = \langle 1, 1, 0\rangle + t \langle 1, -1, 2 \rangle\)
• Second line: \( \mathbf{r}_2 = \langle 2, 0, 2 \rangle + s \langle -1, 1, 0 \rangle \)

By equating the components of the parameterized expressions, we can find specific values of \( t \) and \( s \) where the lines intersect.

The cross product is an operation on two vectors in three-dimensional space that produces another vector that is perpendicular to both inputs. The result is often used to find a vector normal to a plane, which helps in defining the plane equation:

• Given two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the cross product is \( \mathbf{a} \times \mathbf{b} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \).

In the original exercise, we use the direction vectors \( \langle 1, -1, 2 \rangle \) and \( \langle -1, 1, 0 \rangle \). Calculating their cross product yields the vector \( \langle -2, -2, 0 \rangle \), which becomes the normal vector of the plane containing the lines.

Direction vectors are vectors that define the orientation of a line in space.
In the standard parameterization of a line, the direction vector multiplies the parameter and indicates the line's path.

• The direction vector \( \mathbf{d} \) for a line parameterized by \( \mathbf{r} = \mathbf{r}_0 + t\mathbf{d} \) can be seen as the "slope" or "incline" of the line.

Using direction vectors, we can analyze relationships between lines, such as parallelism or perpendicularity. In the original example, the direction vectors of the given lines are \( \langle 1, -1, 2 \rangle \) and \( \langle -1, 1, 0 \rangle \).
These vectors were not just employed to parameterize the lines but also used to determine the normal vector of the plane through the cross product. This cross product allows us to understand how the lines behave spatially and where they intersect.