91Ó°ÊÓ

A direction vector is essential when working with lines in three-dimensional space. It helps us define the orientation or direction in which the line extends. When looking at the parametric form of a line's equation, like

• \( x = 4t + 3 \)
• \( y = 2t - 1 \)
• \( z = t + 3 \)

the direction vector emerges naturally from the coefficients of \( t \). Here, these coefficients form the vector \((4, 2, 1)\). This direction vector tells us that for every one unit increase in \( t \), \( x \) increases by 4, \( y \) increases by 2, and \( z \) increases by 1.

Understanding the direction vector is crucial because it's used in many significant calculations, like finding if two lines are parallel, determining the shortest distance between a point and a line, and computing vector projections.

The position vector represents a specific point in three-dimensional space relative to the origin. In the exercise, we deal with two main vectors: the position vector of a point on the line and the position vector of the given point \( M \). For a line described by parametric equations, the position vector of a point when \( t = 0 \) is straightforward to find. In this example, substituting \( t = 0 \) into the parametric equations gives the point \( (3, -1, 3) \). This serves as the position vector \( \mathbf{r_0} \).

The position vector for point \( M \) is noted directly as \( \mathbf{r_1} = (1, 1, 3) \). These vectors not only help us locate points but are also vital for calculations like evaluating cross products or determining distances using vector mathematics.

The cross product is a vector operation that outputs a vector perpendicular to two given vectors. It is especially useful in geometry to find vectors orthogonal to a plane defined by two vectors. In our scenario, the cross product helps identify a vector perpendicular to both the direction vector of the line and the vector connecting point \( M \) with a point on the line.

By calculating \( \mathbf{n} \times (\mathbf{r_0} - \mathbf{r_1}) \), where \( \mathbf{n} \) is the direction vector \( (4, 2, 1) \), and \( \mathbf{r_0} - \mathbf{r_1} = (2, -2, 0) \), we achieve this task. The determinant form used:

• \[ \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \ 4 & 2 & 1 \ 2 & -2 & 0\end{vmatrix} = (-2\mathbf{i} - 2\mathbf{j} - 12\mathbf{k}) \]

The resulting vector helps confirm orthogonality and contributes to the distance calculation between the point and the line.

The distance formula in three-dimensional space is an indispensable tool when evaluating distances between geometric elements like a point and a line. It involves using vectors and their magnitudes. For instance, the distance \( d \) between a point \( M(x_1, y_1, z_1) \) and a line can be found using the formula:

• \[ d = \frac{| \mathbf{n} \cdot (\mathbf{r_0} - \mathbf{r_1}) |}{\| \mathbf{n} \|} \]

Here, \( \mathbf{n} \) is the direction vector of the line, \( \mathbf{r_0} \) is a position vector on the line, and \( \mathbf{r_1} \) is the position vector of the point \( M \). Ultimately, you compute magnitudes and employ the dot and cross products to find the minimum distance. The given solution calculates the distance as 0, illustrating that the point \( M \) lies on the line, a unique scenario of the typical use of this formula.