Cross Products

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Recall

Definition. A vector $\vec{AB}$ is a geometric object that has magnitude (or length) and direction. Vectors in an n-dimensional Euclidean space can be represented as coordinates vectors in a Cartesian coordinate system.

Definition. The magnitude or length of the vector $\vec{A}$ is given by $|\vec{A}|~ or~ ||\vec{A}|| = \sqrt{a_1^2+a_2^2+a_3^2}$, e.g., $||< 3, 2, 1 >|| = \sqrt{3^2+2^2+1^2}=\sqrt{14}$, $||< 3, -4, 5 >|| = \sqrt{3^2+(-4)^2+5^2}=\sqrt{50}=5\sqrt{2}$, or $||< 1, 0, 0 >|| = \sqrt{1^2+0^2+0^2}=\sqrt{1}=1$.

The dot or scalar product is a fundamental operation between two vectors. It produces a scalar quantity that represents the projection of one vector onto another. The dot product is defined as follows: $\vec{A}·\vec{B} = \sum a_ib_i = a_1b_1 + a_2b_2 + a_3b_3,$ e.g. $\vec{A}·\vec{B} = \sum a_ib_i = ⟨2, 2, -1⟩·⟨5, -3, 2⟩ = a_1b_1 + a_2b_2 + a_3b_3 = 2·5+2·(-3)+(-1)·2 = 10-6-2 = 2.$

Determinant in space

$det(\vec{A}, \vec{B}, \vec{C}) = [\begin{smallmatrix}a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{smallmatrix}] = a_1[\begin{smallmatrix}b_2 & b_3\\ c_2 & c_3\end{smallmatrix}]-a_2[\begin{smallmatrix}b_1 & b_3\\ c_1 & c_3\end{smallmatrix}]+a_3[\begin{smallmatrix}b_1 & b_2\\ c_1 & c_2\end{smallmatrix}]$

Theorem. Geometrically, $det(\vec{A}, \vec{B}, \vec{C})$ = ± the volume of a parallelepiped spanned by the vectors $\vec{A}, \vec{B},and~ \vec{C}$(Figure ii).

Cross Product

Definition. The cross product, denoted by $\vec{A}x\vec{B}$, is a binary operation on two vectors in three-dimensional space. It is a vector that is perpendicular to both of the input vectors (perpendicular to the parallelogram) and has a magnitude equal to the area of the parallelogram formed by the two input vectors.

The direction of the resulting vector is determined by the right-hand rule: if you curl the fingers of your right hand from $\vec{A}$ to $\vec{B}$ (first finger points $\vec{A}$, second finger points to $\vec{B}$), your thumb points in the direction of $\vec{A} \times \vec{B}$ (Figure iii).  

$\vec{A}x\vec{B} = [\begin{smallmatrix}i & j & k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\end{smallmatrix}] =[\begin{smallmatrix}a_2 & a_3\\ b_2 & b_3\end{smallmatrix}]\vec{i}-[\begin{smallmatrix}a_1 & a_3\\ b_1 & b_3\end{smallmatrix}]\vec{j}+[\begin{smallmatrix}a_1 & a_2\\ b_1 & b_2\end{smallmatrix}]\vec{k}$

Examples:

• $\vec{i}x\vec{j} = \vec{k}$, $\vec{i}x\vec{j} = [\begin{smallmatrix}i & j & k\\ 1 & 0 & 0 \\ 0 & 1 & 0\end{smallmatrix}] =0\vec{i}-0\vec{j}+1\vec{k}=\vec{k}$.

Another way of seeing the volume of a parallelepiped spanned by the vectors $\vec{A}, \vec{B},and~ \vec{C}$ (Figure v)= base x height = $|\vec{B}x\vec{C}|(\vec{A}·\vec{n})$ where n is a unit vector perpendicular to the parallelogram formed by $\vec{B},and~ \vec{C}$ = $|\vec{B}x\vec{C}|(\vec{A}·\frac{\vec{B}x\vec{C}}{|\vec{B}x\vec{C}|}) = \vec{A}·(\vec{B}x\vec{C}) = det(\vec{A},\vec{B},\vec{C})$  

Let’s check it out,

$det(\vec{A}, \vec{B}, \vec{C}) = [\begin{smallmatrix}a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{smallmatrix}] = a_1[\begin{smallmatrix}b_2 & b_3\\ c_2 & c_3\end{smallmatrix}]-a_2[\begin{smallmatrix}b_1 & b_3\\ c_1 & c_3\end{smallmatrix}]+a_3[\begin{smallmatrix}b_1 & b_2\\ c_1 & c_2\end{smallmatrix}]$

$\vec{A}·(\vec{B}x\vec{C}) =\vec{A}·([\begin{smallmatrix}b_2 & b_3\\ c_2 & c_3\end{smallmatrix}]\vec{i}-[\begin{smallmatrix}b_1 & b_3\\ c_1 & c_3\end{smallmatrix}]\vec{j}+[\begin{smallmatrix}b_1 & b_2\\ c_1 & c_2\end{smallmatrix}]\vec{k}) = a_1[\begin{smallmatrix}b_2 & b_3\\ c_2 & c_3\end{smallmatrix}]-a_2[\begin{smallmatrix}b_1 & b_3\\ c_1 & c_3\end{smallmatrix}]+a_3[\begin{smallmatrix}b_1 & b_2\\ c_1 & c_2\end{smallmatrix}]$

Properties of the Cross Product

Let $\vec{u}, \vec{v}$, and $\vec{w}$ be vectors in space, and let c be a scalar, the following statements hold true (Credits: The Cross Product -Mathematics Libre Texts),

1. Anticommutative property: $\vec{u}x\vec{v}=-(\vec{v}x\vec{u})$
2. Distributive property: $\vec{u}x(\vec{v}+\vec{w})=(\vec{u}x\vec{v})+(\vec{u}x\vec{w})$
3. Multiplication by a constant: $c(\vec{u}x\vec{v})=(c\vec{u})x\vec{v}=\vec{u}x(c\vec{v})$
4. Cross product of the zero vector: $\vec{u}x\vec{0} = \vec{0}x\vec{u} = \vec{0}$
5. Cross product of a vector with itself: $\vec{u}x\vec{u}=\vec{0}$
6. Triple scalar product: $\vec{u}·(\vec{v}x\vec{w})=(\vec{u}x\vec{v})·\vec{w}$
7. Triple cross product: $\vec{u}x(\vec{v}x\vec{w})=(\vec{u}·\vec{w})\vec{v}-(\vec{u}·\vec{v})\vec{w}$

Find the Equation of a Plane Given Three Points

We want to find the equation of a plane given three points P₁, P₂, and P₃ in the plane. Let P be a fourth point, the condition that P is in the same plane is that the parallelepiped spanned by the vectors $\vec{P_1P}, \vec{P_1P_2},and~ \vec{P_1P_3}$ is flat (Figure vi) ↭ $det(\vec{P_1P}, \vec{P_1P_2},~ \vec{P_1P_3})$ = 0

Definition. A normal vector to a plane is a vector that is perpendicular to the plane, hence is perpendicular (orthogonal) to every vector that lies in the plane.

An alternative to the previously found solution is as follows, P is in the plane ↭ $\vec{P_1P}⊥\vec{N}$ where N is a normal vector to our plane ↭ $\vec{P_1P}·\vec{N}=0$

Such a normal vector could be found using the following formula, $\vec{N}=\vec{P_1P_2}x\vec{P_1P_3}$ ⇒[$\vec{P_1P}·\vec{N}=0$] $\vec{P_1P}·(\vec{P_1P_2}x\vec{P_1P_3})=0$ (Figure vi).

• Exercise. Determine the equation of the plane that contains the points P₁ = ⟨1, −2, 0⟩, P₂ = ⟨3, 1, 4⟩, and P₃ = ⟨0, -1, 2⟩.

Recall that if P₁ has coordinates (x₁, y₁, z₁) and P₂ has coordinates (x₂, y₂, z₂), then the componentes of $\vec{P_1P_2}$ are ⟨x₂-x₁, y₂-y₁, z₂-z₁⟩, i.e., we subtract the coordinates of P₁ from the coordinates of P₂.

$\vec{P_1P_2} = ⟨2, 3, 4⟩, \vec{P_1P_3} = ⟨-1, 1, 2⟩$

$\vec{N}=\vec{P_1P_2}x\vec{P_1P_3} = [\begin{smallmatrix}i & j & k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\end{smallmatrix}] = [\begin{smallmatrix}i & j & k\\ 2 & 3 & 4\\ -1 & 1 & 2\end{smallmatrix}] = 2\vec{i}-8\vec{j}+5\vec{k}$

P is in the plane ↭ $\vec{P_1P}⊥\vec{N}$ where N is a normal vector to our plane ↭ $\vec{P_1P}·\vec{N}=⟨x-1, y+2, z-0⟩·2\vec{i}-8\vec{j}+5\vec{k} = (x-1)2-8(y+2)+5z = 0 ↭ 2x -2 -8y -16+5z = 0 ↭ 2x -8y +5z = 18$, and this is the equation of the plane.

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
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