91Ó°ÊÓ

Parametric equations are a way of defining a line in space using a single parameter, often denoted by letters like \( t \) or \( s \). These equations express the coordinates of any point on the line as functions of this parameter.
Let's take the line \( L \) through points \( A(5, 1, 7) \) and \( B(6, 0, 8) \):

• First, find the vector from \( A \) to \( B \), which is \( \vec{AB} = \langle 6-5, 0-1, 8-7 \rangle = \langle 1, -1, 1 \rangle \).
• The parametric equations for line \( L \) using this vector are \( x = 5 + t \), \( y = 1 - t \), and \( z = 7 + t \), where \( t \) is a parameter.

Similarly, for line \( M \) through \( C(3, 1, 3) \) and \( D(-1, 3, 3) \),

• the vector \( \vec{CD} = \langle -1-3, 3-1, 3-3 \rangle = \langle -4, 2, 0 \rangle \).
• The parametric equations are \( x = 3 - 4s \), \( y = 1 + 2s \), and \( z = 3 \), where \( s \) is the parameter.

Each point on these lines can be identified by different parameter values, which allows the lines to be easily manipulated and combined with other geometric concepts.

The cross product is a vector operation that finds a vector perpendicular to two given vectors. This is significant when dealing with geometry and vector calculus as it helps find normal vectors to surfaces and in this problem, helps us identify a direction perpendicular to both lines \( L \) and \( M \).
For lines \( L \) and \( M \), the direction vectors are \( \vec{AB} = \langle 1, -1, 1 \rangle \) and \( \vec{CD} = \langle -4, 2, 0 \rangle \), respectively. The cross product \( \vec{n} = \vec{AB} \times \vec{CD} \) is calculated by:

• \( \vec{n} = \langle ( -1 imes 0 - 1 imes 2), (1 imes 0 - 1 imes (-4)), (1 imes 2 - (-1) imes (-4)) \rangle = \langle -2, -4, -2 \rangle \).

This result, \( \vec{n} \), is a vector that signifies a direction perpendicular to both lines \( L \) and \( M \). When finding line PQ, ensuring this line's direction vector aligns with \( \vec{n} \) ensures perpendicularity.

The distance formula provides a way to calculate the distance between two points in space. It's essential for verifying that the points \( P \) and \( Q \) in this exercise are at the required distance from each other.
The distance formula is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Where \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) are the coordinates of the two points.
In this problem, point \( P \) is \( (8, -2, 10) \) and point \( Q \) is \( (3, 1, 3) \), applying the distance formula gives:
\[ d = \sqrt{(8-3)^2 + (-2-1)^2 + (10-3)^2} = \sqrt{25 + 9 + 49} = \sqrt{83} \]
Through solving, the distance verifies whether it meets the expected requirement which ensures the correctness of the calculated points.

Coordinate geometry, often referred to as analytical geometry, involves using coordinates to solve geometry problems. It brings together algebra and geometry through coordinate planes like \( xy \), \( xz \), and \( yz \). In this exercise, you're required to determine where the line through points \( P \) and \( Q \) intersects the \( xy \)-plane, where \( z = 0 \).
Given the line's parameterization resulting in \( P(8, -2, 10) \) and a direction vector \( \langle -2, -4, -2 \rangle \), the equation of line PQ can be expressed as \( (8 + k(-2), -2 + k(-4), 10 + k(-2)) \).
Substituting \( z = 0 \) gives \( 10 - 2k = 0 \). Solving for \( k \), provides the intersection coordinates.

• \( k = 5 \)
• The resulting point is \( (8 + 5(-2), -2 + 5(-4), 0) = (-2, -22, 0) \).

Consequently, coordinate geometry joins algebraic equations and geometric figures to allow such determinations.