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The vector form of a line provides a powerful tool for describing lines in three-dimensional space. It is expressed as \( \vec{r} = \vec{a} + p \vec{b} \). Here's what each component means:

• \( \vec{a} \) is a point on the line, essentially anchoring it in space.
• \( \vec{b} \) is the direction vector, indicating the direction in which the line proceeds.
• \( p \) is the scalar parameter that can be any real number (\( p \in \mathbf{R} \)).

By adjusting \( p \), you can find any point along the line. For example, if \( p = 0 \), you obtain the point \( \vec{a} \). Increasing \( p \) moves you along the direction of \( \vec{b} \).
The line \( L: \vec{r} = (-1,3,4) + p(6,1,-2) \) means:
- The line passes through the point \((-1, 3, 4)\).- It moves in the direction of \((6, 1, -2)\).This framework allows for a clear and concise depiction of lines, easing intersection calculations with other geometric entities such as planes.

Planes in three-dimensional space are commonly described using the equation \( Ax + By + Cz + D = 0 \). This is known as the general form of the equation of a plane. Here's a breakdown of what this represents:

• \( A, B, \) and \( C \) are coefficients that define the orientation of the plane.
• \( D \) is the constant term, impacting the position of the plane relative to the origin.
• \((x, y, z) \) are the coordinates of any point on the plane.

When a line intersects a plane, their equations must be satisfied simultaneously. In Exercise a, we have the plane \( \pi: x + 2y - z + 29 = 0 \), and for Exercise b, \( \pi: 2x + 7y + z + 15 = 0 \).
To find the intersection, we substitute the expressions for \( x, y, \) and \( z \) from the line into the plane equation. Solving the resulting equation gives us the parameter value (such as \( p \) or \( t \)) at which the line intersects the plane.

Parametrization is a method of expressing complex geometric objects or paths using a single parameter. This technique is especially useful for delineating the entirety of a geometric figure or curve with a simple variable.
In exercise b, the line \( L: \frac{x-1}{4} = \frac{y+2}{-1} = z-3 \) is parameterized as:

• \( x = 4t + 1 \)
• \( y = -t - 2 \)
• \( z = t + 3 \)

The parameter \( t \) controls the lines' equations, letting you calculate any point on the line by substituting different values of \( t \). With \( t = -3 \), for example, we find that the line intersects the plane at the point \((-11, 1, 0)\).
Using parametrization, not only can you describe linear intersections, but you can also explore complex contours or trajectories in multi-dimensional spaces, making it a vital tool in geometric calculations.