91Ó°ÊÓ

Parametric equations are a set of equations that express the coordinates of the points making up a geometric object, like a line or curve, as functions of a variable, often denoted as "t". These equations are particularly useful for describing the motion along a path.

• For example, in the original exercise, the line is expressed in parametric form as:
  \(x = 2 + t, \ y = 1 + t, \ z = -\frac{1}{2} - \frac{1}{2} t\).
• Here, "t" is a parameter that varies over the real numbers.
• The direction of the line is obtained from the coefficients of "t", leading to the direction vector \(\vec{d} = \langle 1, 1, -\frac{1}{2} \rangle\).

Thus, parametric equations succinctly define each point on the line. As "t" changes, every possible point described by these equations can be traced along the line in a 3D space.

When you need to find the distance from a point to a plane, the perpendicular distance formula is a handy tool. This formula allows us to calculate the shortest distance between a point and a plane in space.
In the step-by-step solution provided, the formula is given by:
\[D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}} \]
Here's how it works:

• \((x_0, y_0, z_0)\) represents the coordinates of the point.
• \(ax + by + cz = d\) is the equation of the plane.
• The numerator \(|ax_0 + by_0 + cz_0 - d|\) calculates the absolute difference between the plane and the point based on the plane's equation.
• The denominator \(\sqrt{a^2 + b^2 + c^2}\) accounts for the normalizing factor of the plane's coefficients.

By plugging the point \((2, 1, -\frac{1}{2})\) and plane’s coefficients \(a = 1, b = 2, c = 6, d = 10\) into the formula, we compute the distance as \(\frac{9}{\sqrt{41}}\). This distance represents the shortest path from the point to the plane.

A normal vector is a vector that is perpendicular to a surface or a plane. It plays an essential role in defining and working with planes, especially in three-dimensional space.
In the problem, the plane is represented by the equation \(x + 2y + 6z = 10\). The normal vector for this plane is \(\vec{n} = \langle 1, 2, 6 \rangle\), drawn from the plane's equation.

• The coefficients \(1, 2,\) and \(6\) are the components of the normal vector.
• This vector points directly away from the plane, showing a direction that is orthogonal to any line lying flat on the plane.
• It is crucial for calculating distances and angles with respect to the plane.

In terms of planes and lines, normal vectors ensure that calculations are consistent by providing a standard reference direction. For distance calculations, like in the provided exercise, knowing the normal vector allows the application of the perpendicular distance formula, confirming the measurements are perpendicular to the surface.