91Ó°ÊÓ

In vector geometry, understanding parametric equations provides a way to describe a line in space. They express the coordinates of the points that make up the line as functions of a parameter, typically denoted by \( t \), which can be any real number.
For a line defined by the vector equation \( \mathbf{r} = (1+t) \mathbf{i} + 3t \mathbf{j} + (1-t) \mathbf{k} \), parametric equations let us break this down into individual coordinate components:

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

By adjusting the parameter \( t \), you can explore all possible points along the line without directly computing them one by one. This capability is vital in analyzing geometric problems, especially when trying to intersect the line with other geometric surfaces such as planes.

The equation of a plane in three-dimensional space is often represented in the form \( ax + by + cz = d \). This notation captures the plane's orientation and position by way of its normal vector and intercepts. In the specific problem you're tackling, the equation \( x + y + cz = d \) simplifies the scenario somewhat, with fixed coefficients for \( x \) and \( y \).
Substituting the parametric equations of the line into the plane equation helps verify if a line lies within the plane. By ensuring that substituting the coordinates satisfies the equation, if it holds true for all \( t \), we confirm the line exists entirely within the plane.
To reach this conclusion, adjustments might be needed in parameters like \( c \) and \( d \). This approach demonstrates the elegance and power of linear equations in revealing spatial relationships.

Analyzing the intersection of a line and a plane involves a few key areas of focus. To find out if a line fully lies within a plane, you can use substitution and simplification. In this exercise, after substituting the parametric equations of the line into the plane's equation, simplifying gives:

• \( 1 + 4t + c - ct = d \)

For the line completely to lie in the plane, the coefficient of \( t \) must cancel out, resulting in \( 4 - c = 0 \), and therefore, \( c = 4 \). The constant terms should also equate to zero for the equation to hold, leading to \( 1 + c = d \). Solving gives \( d = 5 \).
Thus, when \( c = 4 \) and \( d = 5 \), substituting back we find \( c + d = 9 \), confirming not only alignment but complete coincidence, where the line isn't just piercing through but effectively becoming part of the plane's points. This method is a foundational tool in both geometry and linear algebra applications.