Comprehensive Study Notes on 3D Space Geometry: Scalar Products, Vector Products, Lines, Planes, and Spheres

The Scalar Product in Three-Dimensional Space

The scalar product, also known as the dot product, of two vectors $\vec{u}$ and $\vec{v}$ in three-dimensional space is a fundamental operation that results in a scalar. It can be defined geometrically and algebraically. Geometrically, the scalar product is expressed as the product of the magnitudes of the two vectors and the cosine of the angle $\theta$ between them:

$\vec{u} \cdot \vec{v} = ||\vec{u}|| \times ||\vec{v}|| \times \cos(\vec{u}, \vec{v})$

Algebraically, if the vectors are defined by their coordinates $\vec{u}(a, b, c)$ and $\vec{v}(a', b', c')$, the scalar product is the sum of the products of their corresponding components:

$\vec{u} \cdot \vec{v} = aa' + bb' + cc'$

The norm (or magnitude) of a vector $\vec{u}$ is derived from the scalar product of the vector with itself and is calculated using the square root of the sum of the squares of its components:

$||\vec{u}|| = \sqrt{a^2 + b^2 + c^2}$

Two vectors are considered orthogonal (perpendicular) if their scalar product is zero:

$\vec{u} \cdot \vec{v} = 0 \implies \vec{u} \perp \vec{v}$

This condition holds unless one of the vectors is the zero vector, where $|\vec{u}|| = 0$.

The Vector Product and Properties of Cross Products

The vector product, or cross product, of two vectors $\vec{u}$ and $\vec{v}$ results in a third vector $\vec{w}$ that is orthogonal to the plane containing $\vec{u}$ and $\vec{v}$. It is denoted as $\vec{u} \wedge \vec{v}$. A critical property of the vector product is its anticommutativity, meaning the order of the vectors matters and reverses the direction of the resulting vector:

$\vec{u} \wedge \vec{v} = -(\vec{v} \wedge \vec{u})$

If the vector product of two non-zero vectors is the zero vector, the two vectors are collinear:

$\vec{u} \wedge \vec{v} = \vec{0} \implies \vec{u} \text{ and } \vec{v} \text{ are collinear}$

The algebraic calculation of the vector product $\vec{u} \wedge \vec{v}$ for vectors $\vec{u}(a, b, c)$ and $\vec{v}(a', b', c')$ is given by the following component-wise formula:

$\vec{u} \wedge \vec{v} = (bc' - b'c)\mathbf{i} - (ac' - a'c)\mathbf{j} + (ab' - a'b)\mathbf{k}$

Where $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ are the unit vectors along the $x$, $y$, and $z$ axes respectively.

Determinants and Coplanarity

The determinant of three vectors $\vec{u}$, $\vec{v}$, and $\vec{w}$ is used to determine their spatial relationship. This relates the vector product and scalar product through the mixed product:

$(\vec{u} \wedge \vec{v}) \cdot \vec{w} = \det(\vec{u}, \vec{v}, \vec{w})$

If the determinant is equal to zero, the three vectors are coplanar, meaning they lie in the same plane or are parallel to the same plane:

$\det(\vec{u}, \vec{v}, \vec{w}) = 0 \implies \text{Vectors are coplanar}$

If the determinant is non-zero, the vectors are non-coplanar and can form a basis for three-dimensional space. To check for the collinearity of two vectors $\vec{u}$ and $\vec{v}$, one can check if the following $2 \times 2$ determinants are all zero:

$\begin{vmatrix} a & a' \\ b & b' \end{vmatrix} = 0$

$\begin{vmatrix} b & b' \\ c & c' \end{vmatrix} = 0$

$\begin{vmatrix} a & a' \\ c & c' \end{vmatrix} = 0$

Geometric Applications: Distances, Areas, and Volumes

The vector product is an essential tool for calculating distances, areas, and volumes in 3D geometry. The distance from a point $A$ to a line $\Delta$ (defined by a point and a direction vector $\vec{u}$) is calculated as:

$d(A, \Delta) = \frac{||\vec{AB} \wedge \vec{u}||}{||\vec{u}||}$

For polygons and polyhedrons, the following formulas apply:

• The area of a triangle defined by vertices $A$, $B$, and $C$ is: $\text{Area}(ABC) = \frac{1}{2} ||\vec{AB} \wedge \vec{AC}||$

• The area of a parallelogram defined by vectors $\vec{AB}$ and $\vec{AC}$ is: $\text{Area}(\text{Parallelogram}) = ||\vec{AB} \wedge \vec{AC}||$

• The volume of a tetrahedron with vertices $A$, $B$, $C$, and $D$ is: $\text{Volume}(\text{Tetrahedron}) = \frac{1}{6} |(\vec{AB} \wedge \vec{AC}) \cdot \vec{AD}|$

• The volume of a parallelepiped defined by vectors $\vec{AB}$, $\vec{AD}$, and $\vec{AE}$ is: $\text{Volume}(\text{Parallelepiped}) = |(\vec{AB} \wedge \vec{AD}) \cdot \vec{AE}|$

Equations of Lines and Planes

A line $D$ in space can be represented parametrically using a point $(x_0, y_0, z_0)$ and a direction vector $\vec{u}(a, b, c)$:

$x = x_0 + ta$

$y = y_0 + tb$

$z = z_0 + tc$

A plane $P$ can be represented parametrically using a point $A(x_A, y_A, z_A)$ and two non-collinear vectors $\vec{u}(\alpha, \beta, \gamma)$ and $\vec{v}(\alpha', \beta', \gamma')$. It can also be expressed by its Cartesian equation:

$ax + by + cz + d = 0$

In this Cartesian form, the coefficients $(a, b, c)$ represent the coordinates of a vector $\vec{n}$ that is normal (perpendicular) to the plane. A point $M(x, y, z)$ lies on the plane if the determinant formed by vectors $\vec{AM}$, $\vec{u}$, and $\vec{v}$ is zero:

$\det(\vec{AM}, \vec{u}, \vec{v}) = 0$

Relationships Between Planes and Distance to a Plane

The distance from a specific point $A(x_A, y_A, z_A)$ to a plane $P$ defined by $ax + by + cz + d = 0$ is given by the formula:

$d(A, P) = \frac{|ax_A + by_A + cz_A + d|}{\sqrt{a^2 + b^2 + c^2}}$

When considering the relationship between two planes $P$ and $Q$ with normal vectors $\vec{n_1}$ and $\vec{n_2}$:

1. If $\vec{n_1}$ and $\vec{n_2}$ are collinear, the planes are parallel ($P \parallel Q$).
2. If $\vec{n_1}$ and $\vec{n_2}$ are not collinear, the planes are secant, and their intersection is a line $\Delta$.

Geometry of Spheres

A sphere $S$ with center $I(a, b, c)$ and radius $R$ is the set of all points $M(x, y, z)$ such that the distance between $I$ and $M$ is equal to $R$. The Cartesian equation of a sphere is:

$(x - a)^2 + (y - b)^2 + (z - c)^2 = R^2$

Depending on the distance $d(I, P)$ from the center of the sphere $I$ to a plane $P$, the intersection between the sphere and the plane varies:

• If $d(I, P) > R$, the intersection is empty (the plane does not touch the sphere).
• If $d(I, P) = R$, the intersection is a single point $\text{S} \cap \text{P} = \{M\}$. In this case, the plane is tangent to the sphere at point $M$.
• If $d(I, P) < R$, the intersection is a circle $C(H, r)$ centered at point $H$ (the orthogonal projection of $I$ onto $P$) with a radius $r$ calculated as: $r = \sqrt{R^2 - d(I, P)^2}$