91Ó°ÊÓ

In geometry and physics, vector equations are used to describe lines in space. They provide a concise way of representing a line by defining a position vector, which gives a point on the line, and a direction vector that illustrates the line's direction. For instance, the vector equation \( \mathbf{x} = \mathbf{p} + s\mathbf{u} \) denotes a line through the point defined by the vector \( \mathbf{p} \). Here, \( s \) is a scalar parameter that can vary to give different points on the line, and \( \mathbf{u} \) is the direction vector ensuring the direction of the line.
For two lines to be parallel, their direction vectors are scalar multiples of one another. If represented by vector equations, two lines are said to intersect if there are parameters \( s \) and \( t \) that satisfy the equation \( \mathbf{p} + s\mathbf{u} = \mathbf{q} + t\mathbf{v} \). If neither condition is met, as in this problem, the lines are skew, meaning they do not lie in the same plane and never meet.

Planes in three-dimensional space can be represented by equations of the form \( ax + by + cz = d \). The vector form of this equation \( \mathbf{n} \cdot (\mathbf{x} - \mathbf{a}) = 0 \) uses a normal vector \( \mathbf{n} \) that is perpendicular to the plane, and \( \mathbf{a} \) is any point on the plane.
In this exercise, to find planes containing skew lines, we use cross products to determine a suitable normal vector. With the normal vector and a point from each line, we can construct the plane equations. For instance, to form a plane containing the line through point \( \mathbf{p} \) and direction \( \mathbf{u} \), the cross product of \( \mathbf{u} \) and another line's direction vector \( \mathbf{v} \) yields the normal. This results in plane equations such as \( -3x - 2y + 12z + 5 = 0 \), ensuring the lines lie in separate planes.

The cross product is a crucial operation in vector calculus, especially useful when dealing with three-dimensional problems. It takes two vectors and produces a third vector that is perpendicular to both. This is valuable for determining the normal of a plane. Given two vectors, \( \mathbf{u} = [2, -3, 1] \) and \( \mathbf{v} = [0, 6, -1] \), their cross product \( \mathbf{u} \times \mathbf{v} \) is calculated as follows:

• The determinant of the matrix \( \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -3 & 1 \ 0 & 6 & -1 \end{vmatrix} \), where \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are the unit vectors along the x, y, z axes.

This results in \( \mathbf{n} = [-3, -2, 12] \). The cross product thus helps to define a plane's orientation in space regarding given lines and is crucial for solving problems involving spatial relations between lines and planes.