nyuuzyou/wikis huggingface dataset:
Variation of Intersection Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure or complex measure on $X$. 
Let $\cmod \mu$ be the variation of $\mu$. 
Let $F \in \Sigma$.
Let $\mu_F$ and $\cmod \mu_F$ be the intersection measures of $\mu$ and $\mu_F$ respectively. 
Let $\cmod {\mu_F}$ be the variation of $\mu_F$. 
Then:
:$\cmod \mu_F = \cmod {\mu_F}$ 
== Proof ==
Let $\map P A$ be the set of finite partitions of $A$ into $\Sigma$-measurable sets.
We have for $A \in \Sigma$:
:$\ds \map {\cmod {\mu_F} } A = \sup \set {\sum_{j \mathop = 1}^n \cmod {\map {\mu_F} {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}= \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {F \cap A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}$
If:
:$\set {A_1, A_2, \ldots, A_n} \in \map P A$
then:
:$\set {F \cap A_1, F \cap A_2, \ldots, F \cap A_n} \in \map P {A \cap F}$ 
Hence:
:$\ds \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {F \cap A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A} \le \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {F_j} } : \set {F_1, F_2, \ldots, F_n} \in \map P {A \cap F} }$
Conversely, if $\set {F_1, F_2, \ldots, F_n} \in \map P {A \cap F}$ then $\ds \set {F_1, F_2, \ldots, F_n, A \setminus F} \in \map P A$, while:
:$\ds \map \mu {\paren {A \setminus F} \cap F} = \map \mu \O = 0$
So we also have:
:$\ds \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {F_j} } : \set {F_1, F_2, \ldots, F_n} \in \map P {A \cap F} } \le \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {F \cap A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}$
Hence we have:
:$\ds \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {F_j} } : \set {F_1, F_2, \ldots, F_n} \in \map P {A \cap F} } = \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {F \cap A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}$
and so:
:$\map {\cmod \mu} {A \cap F} = \map {\cmod {\mu_F} } A$
for each $A \in \Sigma$.
So:
:$\cmod \mu_F = \cmod {\mu_F}$ 
Category:Intersection Measures
Category:Variation of Signed Measures
Category:Variation of Complex Measures
Variation of Signed Measure is Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.
Let $\size \mu$ be the variation of $\mu$. 
Then $\size \mu$ is a measure.
== Proof ==
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$. 
Then: 
:$\size \mu = \mu^+ + \mu^-$
So $\size \mu$ is a measure from Linear Combination of Measures.
Category:Measures
Category:Variation of Signed Measure
Vaughan's Identity

== Theorem ==
Let $\Lambda$ be von Mangoldt's function.
Let $\mu$ be the Möbius function.
Then for $y, z \ge 1$ and $n > z$:
:$\ds \map \Lambda n = \sum_{\substack {d \mathop \divides n \\ d \mathop \le y}} \map \mu d \map \ln {\frac n d} - \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop \le y, \, c \mathop \le z}} \map \mu d \map \Lambda c + \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y, \, c \mathop > z} } \map \mu d \map \Lambda c$
where $\divides$ denotes divisibility.
== Proof ==
By Sum Over Divisors of von Mangoldt is Logarithm:
:$\ds \ln n = \sum_{d \mathop \divides n} \map \Lambda d$
Hence:
Taking the second summation in that last line:
Again, taking the second summation in that last line:
:$\ds \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y, \, c \mathop \le z} } \map \mu d \map \Lambda c = \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \map \Lambda c - \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop \le y, \, c \mathop \le z} } \map \mu d \map \Lambda c$
Putting this together:
:$\ds \map \Lambda n = \sum_{\substack {d \mathop \divides n \\ d \mathop \le y} } \map \mu d \map \ln {\frac n d} + \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop > y, c \mathop > z} } \map \mu d \map \Lambda c + \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \map \Lambda c - \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ d \mathop \le y, c \mathop \le z} } \map \mu d \map \Lambda c$
It remains to be shown that:
:$\ds \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \map \Lambda c = 0$
The summation is expressed as:
:$\ds \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \map \Lambda c = \sum_{\substack {c \mathop \le z \\ c \mathop \divides n} } \map \Lambda c \sum_{d \mathop \divides \frac n c} \map \mu d$
Now we have $c \le z < n$, so:
:$\dfrac n c > 1$
Therefore, by Sum of Möbius Function over Divisors:
:for each $c$ the inner sum vanishes.
This shows that:
:$\ds \mathop {\sum \sum}_{\substack {d c \mathop \divides n \\ c \mathop \le z} } \map \mu d \map \Lambda c = 0$
as required.
Category:Analytic Number Theory
Vector Addition is Associative

== Theorem ==
Let $\mathbf a, \mathbf b, \mathbf c$ be vectors.
Then:
:$\mathbf a + \paren {\mathbf b + \mathbf c} = \paren {\mathbf a + \mathbf b} + \mathbf c$
where $+$ denotes vector addition.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Addition and subtraction of vectors: $4$. Component and Resultant
* : Introduction: Vector Notation and Formulae
* : Chapter $\text I$: Vector Algebra: $\S 1$.
* : $\S 22$: Laws of Vector Algebra: $22.2$
* 
* 
* 
* 
Category:Vector Addition is Associative
Category:Vectors
Category:Vector Addition
Vector Addition is Associative/Proof 1

== Theorem ==
== Proof ==
:
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be positioned in space so they are end to end as in the above diagram.
Let $\mathbf v$ be a vector representing the closing side of the polygon whose other $3$ sides are represented by $\mathbf a$, $\mathbf b$ and $\mathbf c$.
By the Parallelogram Law we can add any pair of vectors, and add a third vector to their resultant.
Hence we have:
== Sources ==
* : Chapter $\text I$: Definitions. Elements of Vector Algebra: $3$. Addition and Subtraction of Vectors
Category:Vector Addition is Associative
Vector Addition is Commutative

== Theorem ==
Let $\mathbf a, \mathbf b$ be vector quantities.
Then:
:$\mathbf a + \mathbf b = \mathbf b + \mathbf a$
== Proof ==
From the Parallelogram Law:
:
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Addition and subtraction of vectors: $4$. Component and Resultant
* : Introduction: Vector Notation and Formulae
* : Chapter $\text I$: Definitions. Elements of Vector Algebra: $3$. Addition and Subtraction of Vectors
* : Chapter $\text I$: Vector Algebra: $\S 1$.
* : Chapter $1$: Vector Algebra: $1.1$ Definition of a Vector
* : $\S 22$: Laws of Vector Algebra: $22.1$
* 
* 
* 
* 
Category:Vectors
Category:Vector Algebra
Vector Addition is Continuous in Weak Topology

== Theorem ==
Let $K$ be a topological field.
Let $X$ be a topological vector space over $K$ with weak topology $w$. 
Define $s : \struct {X, w} \times \struct {X, w} \to \struct {X, w}$ by: 
:$\map s {x, y} = x + y$
for each $x, y \in X$. 
Then $s$ is continuous. 
That is, vector addition remains continuous when restricting to the weak topology.
== Proof ==
Let $X^\ast$ be the topological dual space of $X$.
From Continuity in Initial Topology, it suffices to show that for each $f \in X^\ast$ we have: 
:$f \circ s : \struct {X, w} \times \struct {X, w} \to K$ is continuous.
Define the projections $\pr_1 : \struct {X, w} \times {X, w} \to \struct {X, w}$ and $\pr_2 : \struct {X, w} \times {X, w} \to \struct {X, w}$ as the projection onto the first and second factors.
Then for each $x, y \in X$ we have: 
That is: 
:$f \circ s = f \circ \pr_1 + f \circ \pr_2$
From the definition of the product topology:
:$\pr_1 : \struct {X, w} \times \struct {X, w} \to \struct {X, w}$
and:
:$\pr_2 : \struct {X, w} \times \struct {X, w} \to \struct {X, w}$ 
are continuous.
From the definition of the weak topology, $f : \struct {X, w} \to K$ is continuous.
From Composite of Continuous Mappings is Continuous, $f \circ \pr_1 : \struct {X, w} \times \struct {X, w} \to K$ and $f \circ \pr_2 : \struct {X, w} \times \struct {X, w} \to K$ are continuous.
From Sum of Continuous Functions on Topological Ring is Continuous, $f \circ \pr_1 + f \circ \pr_2 : \struct {X, w} \times \struct {X, w} \to K$ is continuous.
So we have that $f \circ s : \struct {X, w} \to K$ is continuous for each $f \in X^\ast$. 
So from Continuity in Initial Topology, $s : \struct {X, w} \times \struct {X, w} \to \struct {X, w}$ is continuous in the weak topology. 
Category:Weak Topologies on Topological Vector Spaces
Vector Addition on Locally Convex Space is Continuous

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \PP}$ be a locally convex space over $\GF$ equipped with the standard topolgoy.
Let $\struct {X \times X, \tau}$ be the Cartesian product $X \times X$ equipped with the product topology.
Let $+ : \struct {X \times X, \tau} \to \struct {X, \PP}$ be the vector addition defined on $X$.
Then $+ : \struct {X \times X, \tau} \to \struct {X, \PP}$ is a continuous mapping. 
== Proof ==
From the definition of the standard topolgoy, the topology on $\struct {X, \PP}$ has sub-basis: 
:$\SS = \set {\map {B_p} {x, \epsilon} : p \in \PP, \, \epsilon > 0}$
where we define: 
:$\map {B_p} {x, \epsilon} = \set {y \in X : \map p {y - x} < \epsilon}$
for each $p \in \PP$ and $\epsilon > 0$.
From Continuity Test using Sub-Basis, it suffices to show that: 
:$+^{-1} \sqbrk {\map {B_p} {x_0, \epsilon} } \in \tau$ for each $p \in \PP$, $\epsilon > 0$ and $x_0 \in X$. 
That is: 
:$\set {\tuple {x, y} \in X \times X : \map p {x + y - x_0} < \epsilon} \in \tau$
For brevity, let: 
:$U_{x_0} = \set {\tuple {x, y} \in X \times X : \map p {x + y - x_0} < \epsilon}$ 
Let $\tuple {x, y} \in U_{x_0}$ and $\tuple {x', y'} \in X \times X$. 
We then have: 
Since $\tuple {x, y} \in U_{x_0}$, we have that: 
:$\map p {x + y - x_0} < \epsilon$ 
Set: 
:$\ds \map {\epsilon'} {x, y} = \frac {\epsilon - \map p {x + y - x_0} } 2$
Then, for $\tuple {x', y'} \in \map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} }$, we have: 
so that: 
:$\map p {x + y - x_0} < \epsilon$
and so:
:$\map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} } \subseteq U_{x_0}$
for each $\tuple {x, y} \in U_{x_0}$.
From Union of Subsets is Subset: 
:$\ds \bigcup_{\tuple {x, y} \in U_{x_0} } \map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} } \subseteq U_{x_0}$ 
On the other hand, for each $\tuple {x, y} \in U_{x_0}$ we clearly have: 
:$\tuple {x, y} \in \map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} }$
So, we have: 
:$\ds U_{x_0} = \bigcup_{\tuple {x, y} \in U_{x_0} } \map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} }$ 
From Natural Basis of Product Topology of Finite Product, we have that: 
:$\set {\map {B_p} {x, \epsilon_1} \times \map {B_q} {y, \epsilon_2} : p, q \in \PP, \, x, y \in X, \, \epsilon_1, \epsilon_2 > 0}$ 
forms a basis for $\struct {X \times X, \tau}$.
So: 
:$\map {B_p} {x, \map {\epsilon'} {x, y} } \times \map {B_p} {y, \map {\epsilon'} {x, y} } \in \tau$ for each $\tuple {x, y} \in U_{x_0}$.
Since topologies are closed under set union, we have $+^{-1} \sqbrk {\map {B_p} {x_0, \epsilon} } = U_{x_0} \in \tau$, which is what we wanted.
Category:Locally Convex Spaces
Category:Vector Addition
Vector Addition on Normed Vector Space is Continuous

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}_X }$ be a normed vector space.
Let $\struct {X \times X, \norm {\, \cdot \,}_P }$ be the direct product of $X$ and $X$ with the direct product norm $\norm {\, \cdot \,}_P$. 
Let $+_{\scriptscriptstyle X} : X \times X \to X$ be the vector addition defined on $X$.
Then $+_{\scriptscriptstyle X} : X \times X \to X$ is a continuous mapping.
== Proof ==
Let $x_0, y_0 \in X$.
Let $\epsilon \in \R_{>0}$.
For $a, b \in X$, let $a-_{\scriptscriptstyle X} b$ denote the sum $a +_{\scriptscriptstyle X} \paren { -b }$, where $-b$ is the inverse vector of $b$ in $X$.
To show that $+_{\scriptscriptstyle X}$ is continuous, let $x, y \in X$ such that $\norm { x_0 -_{\scriptscriptstyle X} x }_X < \dfrac \epsilon 2$, and $\norm { y_0 -_{\scriptscriptstyle X} y }_X < \dfrac \epsilon 2$.
By definition of direct product norm, it follows that:
:$\norm { \tuple {x_0,y_0} -_{\scriptscriptstyle {X \times X} } \tuple {x,y} }_P = \map \max {\norm {x_0 -_{\scriptscriptstyle X} x}_X, \norm {y_0 -_{\scriptscriptstyle X} y}_X } < \dfrac \epsilon 2$
To show that $+_{\scriptscriptstyle X}$ is continuous at $\tuple {x_0, y_0}$, we calculate:
It follows that $+_{\scriptscriptstyle X}$ is continuous from $X \times X$ to $X$.
== Sources ==
* : $\S 5$: Normed Spaces
Category:Continuous Mappings on Normed Vector Spaces
Category:Vector Addition
Vector Augend plus Addend equals Augend implies Addend is Zero

== Theorem ==
Let $V$ be a vector space over a field $F$.
Let $\mathbf a, \mathbf b \in V$.
Let $\mathbf a + \mathbf b = \mathbf a$.
Then:
:$\mathbf b = \bszero$
where $\bszero$ is the zero vector of $V$.
== Proof ==
== Sources ==
* : Chapter $7$: Vector Spaces: $\S 32$. Definition of a Vector Space: Theorem $64 \ \text{(i)}$
Category:Vector Spaces
Vector Cross Product/Examples/Couple Exerted by Force

== Example of Vector Cross Product ==
Let $\mathbf F$ be a force acting at a point $P$ on a body $B$ axially about an axis of rotation $R$ such that the distance from $P$ to $R$ is represented by the displacement vector $\mathbf d$.
Then the couple exerted on $B$ by $\mathbf F$ is defined as:
:$\mathbf T = \mathbf F \times \mathbf d = \norm {\mathbf F} \norm {\mathbf d} \mathbf {\hat t} \sin \theta$
where:
:$\times$ denotes vector cross product
:$\mathbf {\hat t}$ denotes the unit vector perpendicular to both $\mathbf F$ and $\mathbf d$ according to the right-hand rule
:$\theta$ is the angle between the directions of $\mathbf F$ and $\mathbf d$.
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $1$. General
Category:Examples of Vector Cross Product
Vector Cross Product Distributes over Addition

== Theorem ==
The vector cross product is distributive over addition.
That is, in general:
:$\mathbf a \times \paren {\mathbf b + \mathbf c} = \paren {\mathbf a \times \mathbf b} + \paren {\mathbf a \times \mathbf c}$
for $\mathbf a, \mathbf b, \mathbf c \in \R^3$.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $\S 22$: Cross or Vector Product: $22.14$
Category:Vector Cross Product
Category:Vector Addition
Category:Examples of Distributive Operations
Category:Vector Cross Product Distributes over Addition
Vector Cross Product Distributes over Addition/Proof 1

== Theorem ==
== Proof ==
Let:
:$\mathbf a = \begin {bmatrix} a_x \\ a_y \\a_z \end {bmatrix}$, $\mathbf b = \begin {bmatrix} b_x \\ b_y \\ b_z \end {bmatrix}$, $\mathbf c = \begin {bmatrix} c_x \\ c_y \\ c_z \end {bmatrix}$
be vectors in $\R^3$.
Then:
Category:Vector Cross Product Distributes over Addition
Vector Cross Product Distributes over Addition/Proof 2

== Theorem ==
== Proof ==
We draw a triangular prism whose parallel edges are in the direction of $\mathbf a$ and with its end faces as triangles with sides $\mathbf b$, $\mathbf c$ and $\mathbf b + \mathbf c$.
:
From Magnitude of Vector Cross Product equals Area of Parallelogram Contained by Vectors, the vector areas of these triangular end faces are $\dfrac {\mathbf b \times \mathbf c} 2$ and $\dfrac {\mathbf c \times \mathbf b} 2$.
The remaining vector areas are $\mathbf b \times \mathbf a$, $\mathbf c \times \mathbf a$ and $\mathbf a \times \paren {\mathbf b + \mathbf c}$.
From Total Vector Area of Polyhedron is Zero:
:$\paren {\mathbf b \times \mathbf a} + \paren {\mathbf c \times \mathbf a} + \paren {\mathbf a \times \paren {\mathbf b + \mathbf c} } + \dfrac {\mathbf b \times \mathbf c} 2 + \dfrac {\mathbf c \times \mathbf b} 2 = 0$
from which we get:
:$\paren {-\mathbf b \times \mathbf a} + \paren {-\mathbf c \times \mathbf a} = \paren {\mathbf a \times \paren {\mathbf b + \mathbf c} }$
The result follows from Vector Cross Product is Anticommutative:
:$\mathbf a \times \mathbf b + \mathbf a \times \mathbf c = \mathbf a \times \paren {\mathbf b + \mathbf c}$
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $6$. Application to Vector Products
Category:Vector Cross Product Distributes over Addition
Vector Cross Product Distributes over Addition/Proof 3

== Theorem ==
== Proof ==
Let $\mathbf b'$ and $\mathbf c'$ be the projections of $\mathbf b$ and $\mathbf c$ onto the plane perpendicular to $\mathbf a$.
Then $\mathbf b' + \mathbf c'$ is the projection of $\mathbf b + \mathbf c$ onto that plane.
We have:
Because:
:$\mathbf a$ is perpendicular to $\mathbf b'$ and $\mathbf c'$
:$\mathbf a \times \mathbf b'$ lies in the plane perpendicular to $\mathbf a$
it follows that:
:$\mathbf a \times \mathbf b'$ is $\norm {\mathbf a}$ times the length of $\mathbf b'$ and is perpendicular to $\mathbf b'$.
Similarly:
:$\mathbf a \times \mathbf c'$ is $\norm {\mathbf a}$ times the length of $\mathbf c'$ and is perpendicular to $\mathbf c'$.
Hence:
:$\mathbf a \times \mathbf b' + \mathbf a \times \mathbf c'$ also lies in the plane perpendicular to $\mathbf a$.
:$\mathbf a \times \mathbf b' + \mathbf a \times \mathbf c'$ is $\norm {\mathbf a}$ times the length of $\mathbf b' + \mathbf c'$ and is perpendicular to $\mathbf b' + \mathbf c'$.
In other words:
:$\mathbf a \times \mathbf b + \mathbf a \times \mathbf c = \mathbf a \times \paren {\mathbf b + \mathbf c}$
== Sources ==
* : Chapter $\text I$: Vector Algebra: $\S 3$
Category:Vector Cross Product Distributes over Addition
Vector Cross Product Operator is Bilinear

== Theorem ==
Let $\mathbf u$, $\mathbf v$ and $\mathbf w$ be vectors in a vector space $\mathbf V$ of $3$ dimensions:
where $\left({\mathbf i, \mathbf j, \mathbf k}\right)$ is the standard ordered basis of $\mathbf V$.
Let $c$ be a real number.
Then:
: $\left({c \mathbf u + \mathbf v}\right) \times \mathbf w = c \left({ \mathbf u \times \mathbf w}\right) + \mathbf v \times \mathbf w$
== Proof ==
Category:Vector Cross Product
Vector Cross Product is Anticommutative

== Theorem ==
The vector cross product is anticommutative:
:$\forall \mathbf a, \mathbf b \in \R^3: \mathbf a \times \mathbf b = -\paren {\mathbf b \times \mathbf a}$
=== Complex Plane ===
The same result holds for complex numbers:
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Sources ==
* : Chapter $\text I$: Vector Algebra: $\S 3$: $(6)$
* : $\S 22$: Cross or Vector Product: $22.13$
* 
* 
Category:Vector Cross Product is Anticommutative
Category:Vector Cross Product
Category:Examples of Anticommutativity
Vector Cross Product is Anticommutative/Complex

== Theorem ==
The complex cross product is anticommutative:
:$\forall z_1, z_2 \in \C: z_1 \times z_2 = -\paren {z_2 \times z_1}$
== Proof ==
Let:
: $z_1 := x_1 + i y_1, z_2 = x_2 + i y_2$
Then:
== Examples ==
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The Dot and Cross Product: $111 \ \text{(b)}$
Category:Vector Cross Product is Anticommutative
Category:Complex Cross Product
Vector Cross Product is Anticommutative/Proof 4

== Theorem ==
== Proof ==
From the definition of the vector cross product:
Hence we have that:
:$\mathbf a \times \mathbf b$ is a vector whose direction is specified according to the right-hand rule
while:
:$\mathbf b \times \mathbf a$ is a vector whose direction, also specified according to the right-hand rule, is exactly the opposite of that for $\mathbf a \times \mathbf b$.
That is:
:$\mathbf a \times \mathbf b = -\mathbf b \times \mathbf a$
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $4$. The Vector Product: $(2.13)$
Category:Vector Cross Product is Anticommutative
Vector Cross Product is Orthogonal to Factors

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be vectors in the real Euclidean space $\R^3$.
Let $\mathbf a \times \mathbf b$ denote the vector cross product.
Then:
:$(1): \quad$ $\mathbf a$ and $\mathbf a \times \mathbf b$ are orthogonal.
:$(2): \quad$ $\mathbf b$ and $\mathbf a \times \mathbf b$ are orthogonal.
== Proof ==
Let $\mathbf a = \begin {bmatrix} a_1 \\ a_2 \\ a_3 \end {bmatrix}$, and $\mathbf b = \begin {bmatrix} b_1 \\ b_2 \\ b_3 \end {bmatrix}$.
Then the dot product of $\mathbf a$ and $\mathbf a \times \mathbf b$ is:
Since the dot product is equal to zero, the vectors $\mathbf a$ and $\mathbf a \times \mathbf b$ are orthogonal by definition.
Similarly, $\mathbf b$ and $\mathbf a \times \mathbf b$ are orthogonal:
== Sources ==
* : $\S 7.5$
* 
Category:Vector Cross Product
Vector Cross Product is not Associative

== Theorem ==
The vector cross product is not associative.
That is, in general:
:$\mathbf a \times \paren {\mathbf b \times \mathbf c} \ne \paren {\mathbf a \times \mathbf b} \times \mathbf c$
for $\mathbf a, \mathbf b, \mathbf c \in \R^3$.
== Proof ==
Proof by Counterexample:
Let:
be vectors in the real Euclidean space $\R^3$.
== Sources ==
* : Chapter $\text {II}$: The Products of Vectors: $7$. Products of Three Vectors
* : Chapter $\text I$: Vector Algebra: $\S 5$
Category:Vector Cross Product
Vector Cross Product of Vector Cross Products

== Theorem ==
Let $\mathbf a, \mathbf b, \mathbf c, \mathbf d$ be vectors in a vector space $\mathbf V$ of $3$ dimensions:
Let $\mathbf a \times \mathbf b$ denote the vector cross product of $\mathbf a$ with $\mathbf b$.
Let $\sqbrk {\mathbf a, \mathbf b, \mathbf c}$ denote the scalar triple product of $\mathbf a$, $\mathbf b$ and $\mathbf c$.
Then:
=== Corollary ===
== Proof ==
Then:
 
== Sources ==
* : Chapter $\text I$: Vector Algebra: $\S 6$: $(15)$
* : $\S 22$: Miscellaneous Formulas involving Dot and Cross Products: $22.21$
Category:Vector Cross Product of Vector Cross Products
Category:Vector Cross Product
Vector Cross Product of Vector Cross Products/Corollary

== Theorem ==
Let $\mathbf a, \mathbf b, \mathbf c, \mathbf d$ be vectors in a vector space $\mathbf V$ of $3$ dimensions:
Let $\sqbrk {\mathbf a, \mathbf b, \mathbf c}$ denote the scalar triple product of $\mathbf a$, $\mathbf b$ and $\mathbf c$.
Then:
:$\sqbrk {\mathbf a, \mathbf b, \mathbf c} \mathbf d = \sqbrk {\mathbf b, \mathbf c, \mathbf d} \mathbf a + \sqbrk {\mathbf a, \mathbf d, \mathbf c} \mathbf b + \sqbrk {\mathbf a, \mathbf b, \mathbf d} \mathbf c$
== Proof ==
From Vector Cross Product of Vector Cross Products:
 
== Sources ==
* : Chapter $\text I$: Vector Algebra: $\S 6$: $(16)$
Category:Vector Cross Product of Vector Cross Products
Category:Scalar Triple Product
Vector Cross Product satisfies Jacobi Identity

== Theorem ==
Let $\mathbf a, \mathbf b, \mathbf c$ be vectors in $3$ dimensional Euclidean space.
Let $\times$ denotes the cross product.
Then:
:$\mathbf a \times \paren {\mathbf b \times \mathbf c} + \mathbf b \times \paren {\mathbf c \times \mathbf a} + \mathbf c \times \paren {\mathbf a \times \mathbf b} = \mathbf 0$
That is, the cross product operation satisfies the Jacobi identity.
== Proof ==
Category:Algebra
Category:Vector Cross Product
Vector Equation of Plane/Also presented as

== Equation of Plane: Also presented as == 
The vector form of the equation of the plane can also be presented in the form:
:$\mathbf r \cdot \mathbf n = p$
where:
:$\mathbf r$ is the position vector of an arbitrary point on $P$
:$\mathbf n$ is the unit normal vector to $P$
:$p$ is the length of the normal vector to $P$ through the origin.
== Proof ==
== Sources ==
* 
* 
Category:Vector Equation of Plane