Ch. 9.6 Equations of Lines and Planes

The professor begins the final section of Chapter 9, focusing on the equations of lines and planes in three dimensions. He explains the analogy between lines in 2D and planes in 3D and introduces the concept of describing a line in 3D, which requires a point and a vector rather than just a point and a slope. The vector provides both the direction and the rate of change. The fixed point is denoted as \( \mathbf{p_0} = (x_0, y_0, z_0) \), and a secondary point on the line is represented as \( \mathbf{p} = (x, y, z) \). The direction vector \( \mathbf{v} \) is derived from the points \( \mathbf{p} \) and \( \mathbf{p_0} \). The professor also discusses the position vector \( \mathbf{r_0} \) from the origin to \( \mathbf{p_0} \) and the generalized position vector \( \mathbf{r} \) from the origin to \( \mathbf{p} \), with \( \mathbf{v} \) connecting these two vectors.

The professor continues by explaining how to create a line in three-dimensional space using the fixed point \( \mathbf{r_0} \) and the vector \( \mathbf{v} \). The line is parameterized by \( \mathbf{r} = \mathbf{r_0} + t\mathbf{v} \), where \( t \) represents time, and the position vector \( \mathbf{r} \) changes over time. He then transitions to discussing planes, which are two-dimensional spaces in three dimensions. To define a plane, a point on the plane and a normal vector orthogonal to the plane are needed. The normal vector ensures that any vector within the plane is perpendicular to it. The professor illustrates the process of finding the equation of a plane using the dot product of the normal vector with vectors within the plane, which should equal zero.

The professor provides a practical example to demonstrate finding the equation of a line that passes through two given points and then finding the equation of a plane that contains this line and a third point. He explains that a line in space is dimensionless and can be surrounded by an infinite number of planes unless a third point is specified. The process involves identifying vectors from the given points and using them to parameterize the line and later to find a normal vector for the plane by taking the cross product of two vectors in the plane. The resulting normal vector is then used to establish the plane's equation through the dot product with the points on the plane.

The professor delves into the detailed calculation of the plane's equation using the cross product of two vectors on the plane to find the normal vector. He calculates the cross product step by step, resulting in a normal vector \( (-5, -3, -14) \). Using this vector, he forms the dot product with the points on the plane to derive the plane's equation. The professor emphasizes the importance of understanding the process behind the formula rather than just memorizing it. He also demonstrates that any point on the plane will satisfy the derived equation, verifying the correctness of the plane's equation by substituting different points from the plane into it.

The professor concludes the lesson by summarizing the key points discussed and possibly answering any remaining questions from the class. The focus has been on understanding the equations of lines and planes in three dimensions, the importance of vectors and normal vectors, and the process of deriving these equations through practical examples. The lesson aims to provide a solid foundation for further studies in three-dimensional geometry.