91Ó°ÊÓ

Understanding direction vectors is crucial when dealing with lines in space. A direction vector indicates the direction in which a line extends. In the example given, we have two lines: \(L1\) and \(L2\). To find the direction vector of these lines, we look at the coefficients of the parameters \(t\) and \(s\) in the parametric equations of the lines.

For \(L1: x = t, y = 3 - 3t, z = -2 - t\), the parameter is \(t\), and the corresponding direction vector \(\mathbf{d}_1\) is \(\langle 1, -3, -1 \rangle\). Similarly, in line \(L2: x = 1 + s, y = 4 + s, z = -1 + s\), \(s\) is the parameter, and its direction vector \(\mathbf{d}_2\) is \(\langle 1, 1, 1 \rangle\).

Direction vectors are fundamental because they help us determine how lines relate to each other, such as being parallel or intersecting. In this case, both lines intersect at a single point, which we can further explore using these vectors.

The cross product is an operation on two vectors in three-dimensional space that results in another vector, which is perpendicular (or normal) to the plane formed by the original two vectors. Here, we have direction vectors \( \mathbf{d}_1 = \langle 1, -3, -1 \rangle \) and \( \mathbf{d}_2 = \langle 1, 1, 1 \rangle \).

To find the plane that contains both lines \(L1\) and \(L2\), we calculate the cross product \( \mathbf{d}_1 \times \mathbf{d}_2 \), resulting in the vector \( \mathbf{n} = \langle -2, -2, 4 \rangle \).

The resulting vector \(\mathbf{n}\) is normal to the plane formed by the intersecting lines. Calculating the cross product is a key step in finding the equation of a plane, as the normal vector is used in its formula.

The equation of a plane in three-dimensional space can be expressed in the standard form: \( ax + by + cz = d \), where \( \langle a, b, c \rangle \) is the normal vector to the plane and \((x, y, z)\) are the coordinates. Once we have the normal vector and a point on the plane, we can formulate the equation of the plane.

In this exercise, the normal vector \( \mathbf{n} = \langle -2, -2, 4 \rangle \) is obtained from the cross product, and the point \((1, 1, -3)\) is where the lines intersect. By substituting these into the formula, we establish the initial plane equation: \(-2(x - 1) - 2(y - 1) + 4(z + 3) = 0\).

Simplifying gives us \(-2x - 2y + 4z = -12\). For ease, this is further simplified by dividing by \(-2\), resulting in \(x + y - 2z = 6\). This final equation represents the plane encompassing the intersecting lines, hinging on the critical understanding of both the normal vector and its relationship to the plane's form.