91Ó°ÊÓ

In plane geometry, direction vectors help us understand the orientation of a line in three-dimensional space. They indicate the direction in which a line progresses. When you have a parametric equation of a line, the coefficients of the parameter (usually 't' or 's') are what we call the direction vectors. For instance, consider the line L1 described by the parametric equations:

• \( x=t \)
• \( y=3-3t \)
• \( z=-2-t \)

The direction vector for L1 is \( \mathbf{d_1} = (1, -3, -1) \). This vector is derived from the coefficients of the parameter 't' appearing in the equations. Similarly, for another line L2, given by

• \( x=1+s \)
• \( y=4+s \)
• \( z=-1+s \)

the direction vector is \( \mathbf{d_2} = (1, 1, 1) \). Using these vectors, we can analyze how two lines relate to each other in space.

The cross product is a vital operation in vector mathematics that allows us to find a vector perpendicular to two given vectors. This perpendicular vector is essential for determining the normal vector of a plane. The normal vector can help define the orientation of a plane in 3D space. To find the cross product of the two direction vectors from our lines L1 and L2,

• \( \mathbf{d_1} = (1, -3, -1) \)
• \( \mathbf{d_2} = (1, 1, 1) \)

we use the determinant method:\[\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & -3 & -1 \ 1 & 1 & 1 \end{vmatrix} = \mathbf{i}(-3 \cdot 1 - (-1 \cdot 1)) - \mathbf{j}(1 \cdot 1 - (-1 \cdot 1)) + \mathbf{k}(1 \cdot 1 - 1 \cdot -3)\]This results in the normal vector \( \mathbf{n} = (2, -2, 4) \). This vector is crucial because it is used to define the plane that contains both lines L1 and L2.

Parametric equations are a useful way of expressing a line or curve where the coordinates are defined as functions of a parameter, usually represented by symbols such as 't' or 's'. Consider the parametric equations of line L1:

• \( x = t \)
• \( y = 3 - 3t \)
• \( z = -2 - t \)

In these equations, 't' serves as the parameter, representing a variable that traces the position along the line as it changes. Similarly, line L2 is described as:

• \( x = 1 + s \)
• \( y = 4 + s \)
• \( z = -1 + s \)

The parameter 's' changes to trace a different path. Parametric equations simplify finding intersection points, calculating direction vectors, and constructing related geometric entities like planes.

In mathematics, a plane can be defined by a point and a normal vector. These two components pinpoint both the location and orientation of the plane in 3D space. To find the equation of a plane containing lines L1 and L2, we first need a point on the plane and a normal vector to the plane. We've already calculated the point of intersection

• \((0, 3, -2)\)

and the normal vector \(\mathbf{n} = (2, -2, 4)\) using the cross product of the direction vectors. The general equation for a plane is:\[\mathbf{n} \cdot (\mathbf{r} - \mathbf{r_0}) = 0\]Substituting our point and normal vector yields:\[ 2(x - 0) - 2(y - 3) + 4(z + 2) = 0 \]Upon simplification, we derive the plane equation:\[ x - y + 2z + 7 = 0 \]This equation represents a plane capturing all points lying along both lines.