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Parametric equations provide a way to express points on a line through the use of a parameter, typically denoted by a letter like `t`. This method simplifies the representation of lines in three-dimensional space. Rather than using traditional slope-intercept or point-slope forms, parametric equations allow us to describe each coordinate of the points along the line with its own equation.
For example, consider the line given by:

• \( \frac{x-1}{-4} = \frac{y-2}{3} = \frac{z-4}{-2} = t \)
• This becomes \( x = 1 - 4t \), \( y = 2 + 3t \), \( z = 4 - 2t \).

These equations indicate how the values of \( x \), \( y \), and \( z \) change as \( t \) varies, offering a clear path along the line. Parametrization is incredibly useful, especially when determining points of intersection, as it easily sets a stage for comparing different sets of conditions given by other parametric forms.

The cross product is a vector operation used to find a vector that is perpendicular to two given vectors. This is especially useful in finding the normal vector of a plane determined by intersecting lines.

• The cross product \( \mathbf{a} \times \mathbf{b} \) of two vectors results in a new vector that is orthogonal to both \( \mathbf{a} \) and \( \mathbf{b} \).
• The magnitude of the resulting vector is determined by the formula \( |\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin\theta \), where \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \).
• For our exercise, we have \( \mathbf{d}_1 = \langle -4, 3, -2 \rangle \) and \( \mathbf{d}_2 = \langle -1, 1, 6 \rangle \) as direction vectors of two intersecting lines.
• The cross product gives the normal vector \( \mathbf{n} = \langle 20, 26, -1 \rangle \).

This `normal vector` is crucial as it helps define the plane's orientation in space. The process involves calculations using the determinant method and fully leverages linear algebra principles.

The equation of a plane in three-dimensional space can be represented in various forms. One of these is the point-normal form, which is particularly helpful when forming a plane defined by a point and a normal vector.For the plane determined by lines intersecting at point \( (1, 2, 4) \), and using the normal vector \( \langle 20, 26, -1 \rangle \), the plane’s equation is given as:

• \( 20(x - 1) + 26(y - 2) - (z - 4) = 0 \)
• After simplification, we arrive at \( 20x + 26y - z = 86 \).

In this form, \( x \), \( y \), and \( z \) represent any point on the plane. The equation is derived from setting the dot product of the vector connecting the given point to a variable point on the plane, and the vector normal to the plane, equal to zero.
This form is foundational in vector calculus and helps us understand how planes can slice through three-dimensional space with precision and clarity.

Determining where two lines intersect in space involves comparing their parametric equations to locate a common point. This process requires setting the two sets of line equations equal to one another and solving for the parameters.

• For instance, given two lines with parameters \( t \) and \( s \), equating yields three conditions based on \( x \), \( y \), and \( z \).
• This involves solving a system of equations: \( 1 - 4t = 2 - s \), \( 2 + 3t = 1 + s \), and \( 4 - 2t = -2 + 6s \).
• Through simultaneous resolution, we find \( t = 0 \) and \( s = 1 \), resulting in the intersecting point \( (1, 2, 4) \).

This method shows how critical it is to parameterize lines when finding intersections, as direct geometric approaches can be cumbersome. The intersection provides a unique point shared by both lines, helping confirm their spatial relationship and contributing to further insights, such as the geometry of the plane they determine.