91Ó°ÊÓ

Parametric equations play a crucial role in representing lines and curves using parameters, often denoted by variables like \( t \). For the intersection of planes, such equations describe how each coordinate on the line changes as the parameter varies. In our exercise, the intersection of the planes \( x + y + z = 1 \) and \( x + 2y + 2z = 1 \) results in a line. The cross product of the normals from these planes provides a direction vector, \( \langle 0, -1, 1 \rangle \). To fully describe the line, we also need a point that lies on it, such as \((1, 0, 0)\). With these, the parametric equations can be written as:
\[ x = 1, \ y = -t, \ z = t \]
These equations tell us exactly what each coordinate is at any point along the line, simply by changing \( t \).
Parametric equations are essential because they offer a straightforward way to handle lines and surfaces in three-dimensional space. By expressing each coordinate in terms of a single parameter, it simplifies calculations and visualizations for further geometric and algebraic problems.

The cross product is a method of finding a vector perpendicular to two given vectors in three-dimensional space. It's incredibly useful in determining the direction vector of the line of intersection between two planes. In the original solution, the normal vectors of the planes are extracted from the plane equations: \( \mathbf{n_1} = \langle 1, 1, 1 \rangle \) and \( \mathbf{n_2} = \langle 1, 2, 2 \rangle \).

The cross product formula is:
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \]
Applying it here:
\[ \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 1 & 2 & 2 \end{vmatrix} = \langle 0, -1, 1 \rangle \]
This new vector is the direction vector of the intersection line, ensuring it complies with both planes' constraints. The cross product not only gives us this necessary direction but also highlights the perpendicular relationship inherent between vectors and planes.

The angle between two planes in space is an important geometric parameter that can be derived from the normal vectors of the planes. In this regard, the dot product can help us find the cosine of the angle between these normals. For the planes \( x + y + z = 1 \) and \( x + 2y + 2z = 1 \), the normal vectors are \( \mathbf{n_1} = \langle 1, 1, 1 \rangle \) and \( \mathbf{n_2} = \langle 1, 2, 2 \rangle \).

The formula to find the angle \( \theta \) between two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is:
\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \]
Let's apply it here:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = 5 \]
The magnitudes are \( \|\mathbf{n_1}\| = \sqrt{3} \) and \( \|\mathbf{n_2}\| = 3 \).
Thus,
\[ \cos \theta = \frac{5}{3\sqrt{3}} \]
Using the inverse cosine function, the angle can be found in radians. Understanding the geometric relationship between two intersecting planes is enhanced by this calculation, offering insights into their spatial arrangement.