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Topics This Week:
Planes
Cross products
Finding areas and volumes using vectors
Solving intersection and distance problems
Vector Representation of a Plane:
A plane in 3-space can be fixed by:
A position vector (r_0) and a direction vector (d)
Two independent vectors can be formed by subtracting two points.
Line Representation:
( \mathbf{r} = (x, y, z) ) where (s \in R) is an arbitrary parameter, represented as:
(\mathbf{r} = \mathbf{r_0} + s\mathbf{d})
Parametric Equation of a Plane:
A plane is 2-dimensional and requires two parameters.
General form: ( \mathbf{r} = \mathbf{r_0} + s\mathbf{d} + t\mathbf{e} )
Where (s, t \in R) and (\mathbf{d}) and (\mathbf{e}) are direction vectors parallel to the plane.
Ways to Define a Plane:
One point on the plane and two independent vectors parallel to the plane.
Three non-collinear points on the plane.
One point on the plane and one vector perpendicular (normal) to the plane.
One consistent linear equation in three unknowns (scalar equation):
(ax + by + cz = d)
Examples of Point and Vector Definitions:
(\mathbf{r} = \mathbf{r_0} + s\mathbf{d} + t\mathbf{e}) yields the vector parametric form of the plane.
The linear combination (s\mathbf{d} + t\mathbf{e}) results in any point on the plane, where (\mathbf{r_0}) is a specific point on it.
Suppose points A, B, and C are given by position vectors:
Take (\mathbf{r_0} = \mathbf{a}) and choose direction vectors (\mathbf{d} = \mathbf{b-a}) and (\mathbf{e} = \mathbf{c-a}).
This gives the parametric representation:
(\mathbf{r} = \mathbf{r_0} + s\mathbf{d} + t\mathbf{e})
Find the plane through the points (1, 2, 1), (5, 0, −1), and (3, −1, −1).
Use (\mathbf{r_0} = (1, 2, 1)).
Direction vectors:
(\mathbf{d} = (4, -2, -2))
(\mathbf{e} = (2, -3, -2))
Resulting vector parametric equation:
(\mathbf{r} = (1, 2, 1) + s(4, -2, -2) + t(2, -3, -2))
A plane can also be represented as:
((\mathbf{r} - \mathbf{r_0}) \cdot \mathbf{n} = 0)
Where (\mathbf{n}) is the normal vector.
This is the point-normal form.
If a point (\mathbf{x}) lies on the plane, then ((\mathbf{x}-\mathbf{r_0}) \cdot \mathbf{n} = 0).
An equation of the form (ax + by + cz = d) describes a plane in 3-space with normal vector (\mathbf{n} = (a, b, c)).
At least one of (a, b, c) must be non-zero to represent a valid plane.
Convert Scalar to Vector Parametric Form:
Given equation format, introduce parameters and rearrange.
Convert Vector Point-Normal to Scalar Form:
Expand and derive coefficients for the plane equation from the point-normal format.
Definition: The cross product of two vectors results in a vector orthogonal to both.
Formula: For vectors (\mathbf{x}) and (\mathbf{y}):
(\mathbf{x} \times \mathbf{y} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \mathbf{x}_1 & \mathbf{x}_2 & \mathbf{x}_3 \ \mathbf{y}_1 & \mathbf{y}_2 & \mathbf{y}_3 begin{matrix} \end{vmatrix} )
Direction of the Cross Product:
Determined by the right-hand rule.
Properties:
Non-commutative (i.e., (\mathbf{x} \times \mathbf{y} = - (\mathbf{y} \times \mathbf{x}))).
Area of Parallelogram: = ( ||\mathbf{x} \times \mathbf{y}|| )
Area of Triangle: = ( \frac{1}{2} ||\mathbf{x} \times \mathbf{y}|| )
Volume of Parallelepiped: = ( |\mathbf{x} \cdot (\mathbf{y} \times \mathbf{z})| )
Lines Meeting Planes:
To find the intersection of a line and a plane, ensure the plane's equation is in scalar point-normal form and solve for the parameter.
Confirm intersection by substituting back into the line's equation to find coordinates.
Finding Distances:
Distance from a point to a plane involves determining the line orthogonal to the plane and solving for intersection coordinates to compute the distance.