Cross Products

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Recall

Definition. A vector $\vec{AB}$ is a geometric object or quantity that has both 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 of the vector $\vec{A}$, also known as length or norm, is given by the square root of the sum of its components squared, $|\vec{A}|~ or~ ||\vec{A}|| = \sqrt{a_1^2+a_2^2+a_3^2}$, 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 sum of two vectors is the sum of their components, $\vec{A}+ \vec{B} = (a_1+b_1)\vec{i}+(a_2+b_2)\vec{j}+(a_3+b_3)\vec{k}$ = < (a₁+b₁), (a₂+b₂), (a₃+b₃) >.

• The subtraction of two vectors is similar to addition and is also done component-wise, it is given by simply subtracting their corresponding components (x, y, and z in 3D), $\vec{A} - \vec{B} = (a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k}$ = < (a₁-b₁), (a₂-b₂), (a₃-b₃) >.

• Scalar multiplication is the multiplication of a vector by a scalar, a real number, changing its magnitude without altering its direction. It is effectively multiplying each component of the vector by the scalar value, $c\vec{A} = (ca_1)\vec{i}+(ca_2)\vec{j}+(ca_3)\vec{k} = < ca_1, ca_2, ca_3>$.

• 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 the sum of the products of their corresponding components: $\vec{A}·\vec{B} = \sum a_ib_i = a_1b_1 + a_2b_2 + a_3b_3.$, e.g. $\vec{A}·\vec{B} = \sum a_ib_i = ⟨2, 2, -1⟩·⟨5, -3, 2⟩ = a_1b_1 + a_2b_2 + a_3b_3 = 2·5+2·(-3)+(-1)·2 = 10-6-2 = 2.$

It is the product of their magnitudes multiplied by the cosine of the angle between them, $\vec{A}·\vec{B}=||\vec{A}||·||\vec{B}||·cos(θ).$

Determinant in space

Let $\vec{A}, \vec{B},~\text{and} \vec{C}$ be three vectors in ℝ³ with components,

$\vec{A} = (\begin{smallmatrix}a_1\\a_2\\a_3\end{smallmatrix}), \vec{B} = (\begin{smallmatrix}b_1\\b_2\\b_3\end{smallmatrix}), \vec{C} = (\begin{smallmatrix}c_1\\c_2\\c_3\end{smallmatrix})$.

The determinant of the matrix formed by these three vectors, denoted as $det(\vec{A}, \vec{B}, \vec{C})$, is given by

$det(\vec{A}, \vec{B}, \vec{C}) = |\begin{smallmatrix}a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{smallmatrix}| = a_1det(\begin{smallmatrix}b_2 & b_3\\ c_2 & c_3\end{smallmatrix})-a_2det(\begin{smallmatrix}b_1 & b_3\\ c_1 & c_3\end{smallmatrix})+a_3det(\begin{smallmatrix}b_1 & b_2\\ c_1 & c_2\end{smallmatrix})$

Where the 2 × 2 determinants (also called minors) are calculated as follows:

$det(\begin{smallmatrix}b_2 & b_3\\ c_2 & c_3\end{smallmatrix}) = b_2c_3-b_3c_2, det(\begin{smallmatrix}b_1 & b_3\\ c_1 & c_3\end{smallmatrix}) = b_1c_3 -b_3c_1, det(\begin{smallmatrix}b_1 & b_2\\ c_1 & c_2\end{smallmatrix})=b_1c_2-b_2c_1$

In conclusion, $det(\vec{A}, \vec{B}, \vec{C}) = a_1(b_2c_3-b_3c_2) -a_2(b_1c_3 -b_3c_1) + a_3(b_1c_2-b_2c_1)$

Geometric Interpretation

Geometrically, the determinant $det(\vec{A}, \vec{B}, \vec{C})$ represents the volume of the parallelepiped spanned or formed by the vectors $\vec{A}, \vec{B},and~ \vec{C}$ (Figure ii). The sign of the determinant indicates the orientation of the three vectors. If the determinant is positive, the vectors form a right-handed system (if when you extend the fingers of your right hand along the direction of vector $\vec{A}$ and close them (middle finger) in the direction of $\vec{B}$, the thumb points roughly in the direction of $\vec{C}$); if negative, they form a left-handed system, e.g., $\vec{i}, \vec{j}, \vec{k}$ form a right handed system.

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 results in a vector that is perpendicular to both of the input vectors 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 in the direction of $\vec{A}$, second finger points to $\vec{B}$), your thumb points in the direction of $\vec{A} \times \vec{B}$ (Figure iii).

The cross product $\vec{A}x\vec{B}$ can be computed using the following formula,

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

Where the 2×2 determinants (minors) are calculated as follows:

$|\begin{smallmatrix}a_2 & a_3\\ b_2 & b_3\end{smallmatrix}| = a_2b_3-a_3b_2, |\begin{smallmatrix}a_1 & a_3\\ b_1 & b_3\end{smallmatrix}| = a_1b_3-a_3b_1, |\begin{smallmatrix}a_1 & a_2\\ b_1 & b_2\end{smallmatrix}| = a_1b_2-a_2b_1.$

In conclusion, $\vec{A}x\vec{B} = (a_2b_3-a_3b_2)\vec{i}-(a_1b_3-a_3b_1)\vec{j}+(a_1b_2-a_2b_1)\vec{k}$

Examples

• $\vec{i}x\vec{j} = det(\begin{smallmatrix}i & j & k\\ 1 & 0 & 0\\ 0 & 1 & 0\end{smallmatrix}) = (0·0-0·1)\vec{i}-(1·0-0·0)\vec{j}+(1·1-0·0)\vec{k} = \vec{k}$

• $⟨1,0,2⟩x⟨−1,−2,4⟩ = det(\begin{smallmatrix}i & j & k\\ 1 & 0 & 2 \\ -1 & -2 & 4\end{smallmatrix})=4\vec{i}-6\vec{j}+-2\vec{k}=⟨4, -6, -2⟩$.

• $⟨−2,1,3⟩x⟨5,2,−1⟩ = det(\begin{smallmatrix}i & j & k\\ -2 & 1 & 3 \\ 5 & 2 & -1\end{smallmatrix})=-7\vec{i}+13\vec{j}+-9\vec{k}=⟨-7, 13, -9⟩$.

Another way of calculating the volume of a parallelepiped spanned by the vectors $\vec{A}, \vec{B},and~ \vec{C}$ (Figure v) is using the base times heigh formula. This can be expressed as:

Volume = $|\vec{B}x\vec{C}|(\vec{A}·\vec{n})$ where $\vec{n}$ is a unit vector perpendicular to the parallelogram formed by $\vec{B} ~\text{and}~ \vec{C}$:

$\vec{n} = \frac{\vec{B}x\vec{C}}{|\vec{B}x\vec{C}|}$

Volume = $|\vec{B}x\vec{C}|(\vec{A}·\frac{\vec{B}x\vec{C}}{|\vec{B}x\vec{C}|}) = \vec{A}·(\vec{B}x\vec{C}) = det(\vec{A},\vec{B},\vec{C})$  

Verification

To verify this, let’s look at the determinant form:

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

Expanding $\vec{A}·(\vec{B}x\vec{C})$

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

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

To find the equation of a plane given three points P₁, P₂, and P₃ in the plane, we need to determine a condition that ensures a fourth point P lies in the same plane.

Consider the vectors $\vec{P_1P}, \vec{P_1P_2},and~ \vec{P_1P_3}$. For point P to be in the planed defined by P₁, P₂, and P₃, the parallelepiped spanned by these vectors must be flat. This condition is mathematically represented by the determinant being zero: (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., A normal vector is perpendicular (orthogonal) to every vector that lies in the plane.

An alternative way to find the equation of the plane is by using a normal vector. For point P to be in the plane, the vector $\vec{P_1P}$ must be perpendicular to the normal vector $\vec{N}$, $\vec{P_1P}⊥\vec{N}$ where $\vec{N}$ is a normal vector to our plane. This can be expressed as $\vec{P_1P}·\vec{N}=0$ [🚀]

Finding the Normal Vector

A normal vector $\vec{N}$ to the plane can be found using the cross product of the vectors $\vec{P_1P_2}$ and $\vec{P_1P_3}$: $\vec{N}=\vec{P_1P_2}x\vec{P_1P_3}$

Thus, the condition that the point P lies in the plane can be rewritten as:

$\vec{P_1P}·\vec{N}=0 [🚀] ⇒ \vec{P_1P}·(\vec{P_1P_2}x\vec{P_1P_3})=0$ (In the accompanying figure vi, the vectors and the plane can be visualized to better understand the geometric relationships and the conditions derived). This shows that the dot product of $\vec{P_1P}$ with the normal vector $\vec{N}$ must be zero for P to lie in the plane.

Solved exercises

• 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.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.