91Ó°ÊÓ

3D geometry involves the study of shapes and figures in three-dimensional space. This space is characterized by three coordinates: x, y, and z, which determine the position of points. For instance, a point P is represented as \( (x, y, z) \). In this exercise, points P and \( P_0 \) lie in \(\mathbb{R}^3\). Distance measurements, lines, curves, and shapes are crucial applications of 3D geometry.

Understanding how to navigate and manipulate these dimensions is vital for solving problems like finding the shortest distance from a point to a line. The shortest distance from a point to a line involves finding the perpendicular segment from the point to the line. We often represent lines in 3D using parametric equations for ease of calculation and visualization.

Parametric equations define a line in terms of parameters, usually denoted as t. For a line passing through a point \(P_0 = (1, 1, 1)\) with a direction vector \(\vec{d} = [3, 0, 1]\), its parametric equation is given by:

\[\vec{r}(t) = \vec{P_0} + t \vec{d} = \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} + t \begin{bmatrix} 3 \ 0 \ 1 \end{bmatrix} = \begin{bmatrix} 1 + 3t \ 1 \ 1 + t \end{bmatrix}\]

Parametric equations are useful because they allow us to find exact coordinates on the line by varying the parameter t. For instance, when solving for the point on the line that is closest to a given point, these equations simplify the computation by expressing all coordinates as functions of t.

Vectors are fundamental in 3D geometry for representing points, directions, and distances. Operations on vectors (such as addition, subtraction, and scalar multiplication) help manipulate these quantities.

In this problem, we define the vector \(\vec{PQ} \) from point P to any point Q on the line L. Then, \(\vec{PQ} = \begin{bmatrix} (1 + 3t) - 0 \ 1 - 2 \ (1 + t) - 1 \end{bmatrix} = \begin{bmatrix} 1 + 3t \ -1 \ t \end{bmatrix} \).

Understanding these vector manipulations is critical. It helps translate geometric problems into algebraic computations, which can be solved step-by-step to find distances, directions, and specific points.

The dot product is a scalar representation of two vectors' 'directional alignment'. Mathematically, the dot product of vectors \(\vec{a} = [a_1, a_2, a_3]\) and \(\vec{b} = [b_1, b_2, b_3]\) is given by:

\[\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3\]

For the shortest distance problem, \(\vec{PQ} \) must be perpendicular to the direction vector \(\vec{d}\). This condition is found using the dot product:

\[\vec{PQ} \cdot \vec{d} = 0 \rightarrow (1 + 3t) \cdot 3 + t \cdot 1 = 0 \rightarrow 3 + 9t + t = 0 \rightarrow 10t + 3 = 0 \rightarrow t = -\frac{3}{10}\]

The dot product is essential for verifying orthogonality, simplifying complex 3D problems into solvable equations.

Calculating the shortest distance from a point to a line in 3D involves finding the magnitude of the vector \(\vec{PQ} \), where Q is the closest point on the line L.
First, solve for Q by substituting the calculated t value back into the parametric equations:

\[Q = \begin{bmatrix} 1 + 3(-\frac{3}{10}) \ 1 \ 1 + (-\frac{3}{10}) \end{bmatrix} = \begin{bmatrix} 1 - \frac{9}{10} \ 1 \ 1 - \frac{3}{10} \end{bmatrix} = \begin{bmatrix} \frac{1}{10} \ 1 \ \frac{7}{10} \end{bmatrix}\]

Then, calculate the distance as the magnitude of vector \(\vec{PQ}\):

\[\vec{PQ} = \begin{bmatrix} \frac{1}{10} - 0 \ 1 - 2 \ \frac{7}{10} - 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{10} \ -1 \ -\frac{3}{10} \end{bmatrix}\]

The shortest distance is:

\[\|\vec{PQ}\| = \sqrt{\left( \frac{1}{10} \right)^2 + (-1)^2 + \left( -\frac{3}{10} \right)^2} = \sqrt{\frac{1}{100} + 1 + \frac{9}{100}} = \sqrt{\frac{110}{100}} = \sqrt{1.1}\]

Distance calculations in 3D require careful vector operations and an understanding of fundamental geometric principles.