91Ó°ÊÓ

Parametric equations are a fundamental concept in vector calculus. They allow us to represent curves and lines in a way that relates each point on the curve to a parameter, usually denoted by symbols like \( t \). For instance, the parametric form of a line involves three equations for \( x \), \( y \), and \( z \) coordinates, each depending on the same parameter.

For the line \(L\) given in our problem, the equations are: \( x = -2 - 2t \), \( y = 4t \), and \( z = 7 + t \).

Key points to remember:

• Each equation depends on the parameter \( t \)
• By changing \( t \), you can find different points on the line
• This form is very useful when dealing with vector operations

Understanding how to convert between different forms of line equations and using parametric forms makes complex vector problems more manageable.

In vector calculus, the cross product is a critical operation to find a vector orthogonal to two other vectors. The result of the cross product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) provides a new vector perpendicular to both.

For two vectors \( \mathbf{v} = (-3 - 2t, 4t + 1, 8 + t) \) and direction vector \( \mathbf{d} = (-2, 4, 1) \), their cross product is calculated and can be depicted by a determinant matrix:
\[ \mathbf{v} \times \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -3 - 2t & 4t + 1 & 8 + t \ -2 & 4 & 1 \end{vmatrix} \]
The calculation involves:

• Combining components of vectors in a determinant form
• Expanding the determinant to get the vector components
• Resultant cross product gives \( (-31, -13, -10) \)

The magnitude of the cross product helps in determining the area of the parallelogram formed by the vectors and is pivotal in finding the distance.

Euclidean distance is the straight-line distance between two points in three-dimensional space. It's calculated using the Pythagorean theorem. In our context, understanding this helps find the shortest distance from a point to a line.

The formula for Euclidean distance between two points \( P1 = (x_1, y_1, z_1) \) and \( P2 = (x_2, y_2, z_2) \) is:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Steps involved in our problem involve:

• Finding vector \( \mathbf{v} \) from the given point to a point on the line
• Using the cross product and the magnitude of the direction vector
• Employing the formula \( d = \frac{| \mathbf{v} \times \mathbf{d} |}{| \mathbf{d} |} \)

By dividing the magnitude of the cross product by the magnitude of the direction vector, we directly find the shortest distance to the line as approximately 7.65 units.