hcju/proofwikiqa huggingface dataset:
Let $\vec a, \vec b, \vec c$ be $\vec{OA}, \vec{OB}, \vec{OC}$ respectively. Let the midpoint of $BC, AC, AB$ be $\vec d, \vec e, \vec f$ respectively. Then: {{begin-eqn}} {{eqn | l = \vec d       | r = \frac {\vec b + \vec c} 2 }} {{eqn | l = \vec e       | r = \frac {\vec a + \vec c} 2 }} {{eqn | l = \vec f       | r = \frac {\vec a + \vec b} 2 }} {{end-eqn}} The three medians are $\vec{AD}, \vec{BE}, \vec{CF}$ respectively: {{begin-eqn}} {{eqn | l = \vec {AD}       | r = \vec d - \vec a }} {{eqn | r = \frac {\vec b + \vec c - 2 \vec a} 2 }} {{eqn | l = \vec {BE}       | r = \vec e - \vec b }} {{eqn | r = \frac {\vec a + \vec c - 2 \vec b} 2 }} {{eqn | l = \vec {CF}       | r = \vec f - \vec c }} {{eqn | r = \frac {\vec a + \vec b - 2 \vec c} 2 }} {{end-eqn}} Their equations: {{begin-eqn}} {{eqn | n = 1       | ll= \vec {AD}:       | l = \vec r       | r = \vec a + x \paren {\frac {\vec b + \vec c - 2\vec a} 2} }} {{eqn | n = 2       | ll= \vec {BE}:       | l = \vec r       | r = \vec b + y \paren {\frac {\vec a + \vec c - 2\vec b} 2} }} {{eqn | n = 3       | ll= \vec {CF}:       | l = \vec r       | r = \vec c + z \paren {\frac {\vec a + \vec b - 2\vec c} 2} }} {{end-eqn}} It can be verified that $x = y = z = \dfrac 2 3$ produce the same point: When $x = \dfrac 2 3$, from $(1)$: :$\vec r = \vec a + \dfrac 2 3 \paren {\dfrac {\vec b + \vec c - 2\vec a} 2} = \dfrac {\vec a + \vec b + \vec c} 3$ When $y = \dfrac 2 3$, from $(2)$: :$\vec r = \vec b + \dfrac 2 3 \paren {\dfrac {\vec a + \vec c - 2\vec b} 2} = \dfrac {\vec a + \vec b + \vec c} 3$ When $z = \dfrac 2 3$, from $(3)$: :$\vec r = \vec c + \dfrac 2 3 \paren {\dfrac {\vec a + \vec b - 2\vec c} 2} = \dfrac {\vec a + \vec b + \vec c} 3$ Therefore, the three medians meet at a single point, namely $\dfrac {\vec a + \vec b + \vec c} 3$. {{qed}}
Let $z_1, z_2, z_3, z_4$ be the position vectors of the vertices of $\Box ABCD$. The midpoints of the sides of $\Box ABCD$ are, then: {{begin-eqn}} {{eqn | l = OE       | r = \frac {z_1 + z_2} 2       | c =  }} {{eqn | l = OF       | r = \frac {z_2 + z_3} 2       | c =  }} {{eqn | l = OG       | r = \frac {z_3 + z_4} 2       | c =  }} {{eqn | l = OH       | r = \frac {z_4 + z_1} 2       | c =  }} {{end-eqn}} Take two opposite sides $EF$ and $HG$: {{begin-eqn}} {{eqn | l = EF       | r = \frac {z_2 + z_3} 2 - \frac {z_1 + z_2} 2       | c =  }} {{eqn | r = \frac {z_3} 2 - \frac {z_1} 2       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = HG       | r = \frac {z_3 + z_4} 2 - \frac {z_4 + z_1} 2       | c =  }} {{eqn | r = \frac {z_3} 2 - \frac {z_1} 2       | c =  }} {{eqn | r = EF       | c =  }} {{end-eqn}} The vectors defining the opposite sides of $\Box EFGH$ are equal. The result follows by definition of parallelogram. {{qed}}
:400px Let the given intersection be at $F$. We have that $E$ is the midpoint of $AD$. Thus: {{begin-eqn}} {{eqn | l = \vec {AB} + \vec {BE}       | r = \frac {\vec {AD} } 2       | c =  }} {{eqn | ll= \implies       | l = \vec {BE}       | r = \frac {\vec {AD} } 2 - \vec {AB}       | c =  }} {{eqn | ll= \implies       | l = \vec {AF}       | r = m \paren {\frac {\vec {AD} } 2 - \vec {BE} }       | c = for some $m$ such that $0 \le m \le 1$ }} {{end-eqn}} Since $\vec {AB} + \vec {BC}$ we have $\vec {AF} = n \paren {\vec {AB} + \vec {BC} }$ where $0 \le n \le 1$. But: :$\vec {AB} + \vec {BF} = \vec {AF}$ That is: :$\vec {AB} + m \paren {\dfrac {\vec {BC} } 2 - \vec {AB} } = n \paren {\vec {AB} + \vec {BC} }$ That is: :$\paren {1 - m - n} \vec {AB} + \paren {\dfrac m 2 - n} \vec {AB} = 0$ These have a simultaneous solution because $\vec {AB}$ and $\vec {AD}$ are neither coincident nor parallel. So: :$1 - m - n = 0, \dfrac m 2 - n = 0 \implies m = 2 n$ Hence $3n = 1$ and so: :$n = \dfrac 1 3$ {{qed}}
We have: {{begin-eqn}} {{eqn | l = r_1 e^{i \theta_1} + r_2 e^{i \theta_2}       | r = r_1 \paren {\cos \theta_1 + i \sin \theta_1} + r_2 \paren {\cos \theta_2 + i \sin \theta_2}       | c = {{Defof|Polar Form of Complex Number}} }} {{eqn | r = \paren {r_1 \cos \theta_1 + r_2 \cos \theta_2} + i \paren {r_1 \sin \theta_1 + r_2 \sin \theta_2}       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = {r_3}^2       | r = r_1^2 + r_2^2 + 2 r_1 r_2 \, \map \cos {\theta_1 - \theta_2}       | c = Complex Modulus of Sum of Complex Numbers }} {{eqn | ll= \leadsto       | l = r_3       | r = \sqrt {r_1^2 + r_2^2 + 2 r_1 r_2 \, \map \cos {\theta_1 - \theta_2} }       | c =  }} {{end-eqn}} and similarly: :$\theta_3 = \map \arctan {\dfrac {r_1 \sin \theta_1 + r_2 \sin \theta_2} {r_1 \cos \theta_1 + r_2 \cos \theta_2} }$ {{qed}}
{{begin-eqn}} {{eqn | l = \sin 3 \theta       | r = \map \sin {2 \theta + \theta} }} {{eqn | r = \sin 2 \theta \cos \theta + \cos 2 \theta \sin \theta       | c = Sine of Sum }} {{eqn | r = \paren {2 \sin \theta \cos \theta} \cos \theta + \paren {\cos^2 \theta - \sin^2 \theta} \sin \theta       | c = Double Angle Formula for Sine and Double Angle Formula for Cosine }} {{eqn | r = 2 \sin \theta \cos^2 \theta + \cos^2 \theta \sin \theta - \sin^3 \theta       | c =  }} {{eqn | r = 2 \sin \theta \paren {1 - \sin^2 \theta} + \paren {1 - \sin^2 \theta} \sin \theta - \sin^3 \theta       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 2 \sin \theta - 2 \sin^3 \theta + \sin \theta - \sin^3 \theta - \sin^3 \theta       | c = multiplying out }} {{eqn | r = 3 \sin \theta - 4 \sin^3 \theta       | c = gathering terms }} {{end-eqn}} {{qed}}
We have: {{begin-eqn}} {{eqn | l = \cos 3 \theta + i \sin 3 \theta       | r = \paren {\cos \theta + i \sin \theta}^3       | c = De Moivre's Formula }} {{eqn | r = \paren {\cos \theta}^3 + \binom 3 1 \paren {\cos \theta}^2 \paren {i \sin \theta} }} {{eqn | o =        | ro=+       | r = \binom 3 2 \paren {\cos \theta} \paren {i \sin \theta}^2 + \paren {i \sin \theta}^3       | c = Binomial Theorem }} {{eqn | r = \cos^3 \theta + 3 i \cos^2 \theta \sin \theta + 3 i^2 \cos \theta \sin^2 \theta + i^3 \sin^3 \theta       | c = substituting for binomial coefficients }} {{eqn | r = \cos^3 \theta + 3 i \cos^2 \theta \sin \theta - 3 \cos \theta \sin^2 \theta - i \sin^3 \theta       | c = $i^2 = -1$ }} {{eqn | n = 1       | r = \cos^3 \theta - 3 \cos \theta \sin^2 \theta  }} {{eqn | o =        | ro=+       | r = i \paren {3 \cos^2 \theta \sin \theta - \sin^3 \theta}       | c = rearranging }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \sin 3 \theta       | r = 3 \cos^2 \theta \sin \theta - \sin^3 \theta       | c = equating imaginary parts in $(1)$ }} {{eqn | r = 3 \paren {1 - \sin^2 \theta} \sin \theta - \sin^3 \theta       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 3 \sin \theta - 4 \sin^3 \theta       | c = multiplying out and gathering terms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos 3 \theta       | r = \cos \paren {2 \theta + \theta} }} {{eqn | r = \cos 2 \theta \cos \theta - \sin 2 \theta \sin \theta       | c = Cosine of Sum }} {{eqn | r = \paren {\cos^2 \theta - \sin^2 \theta} \cos \theta - \paren {2 \sin \theta \cos \theta} \sin \theta       | c = Double Angle Formula for Cosine and Double Angle Formula for Sine }} {{eqn | r = \cos^3 \theta - \sin^2 \theta \cos \theta - 2 \sin^2 \theta \cos \theta       | c =  }} {{eqn | r = \cos^3 \theta - \paren {1 - \cos^2 \theta} \cos \theta - 2 \paren {1 - \cos^2 \theta} \cos \theta       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \cos^3 \theta - \cos \theta + \cos^3 \theta - 2 \cos \theta + 2 \cos^3 \theta       | c = multiplying out }} {{eqn | r = 4 \cos^3 \theta - 3 \cos \theta       | c = gathering terms }} {{end-eqn}} {{qed}}
We have: {{begin-eqn}} {{eqn | l = \cos 3 \theta + i \sin 3 \theta       | r = \paren {\cos \theta + i \sin \theta}^3       | c = De Moivre's Formula }} {{eqn | r = \paren {\cos \theta}^3 + \binom 3 1 \paren {\cos \theta}^2 \paren {i \sin \theta}       | c = Binomial Theorem }} {{eqn | o =        | ro=+       | r = \binom 3 2 \paren {\cos \theta} \paren {i \sin \theta}^2 + \paren {i \sin \theta}^3 }} {{eqn | r = \cos^3 \theta + 3 i \cos^2 \theta \sin \theta + 3 i^2 \cos \theta \sin^2 \theta + i^3 \sin^3 \theta       | c = substituting for binomial coefficients }} {{eqn | r = \cos^3 \theta + 3 i \cos^2 \theta \sin \theta - 3 \cos \theta \sin^2 \theta - i \sin^3 \theta       | c = $i^2 = -1$ }} {{eqn | n = 1       | r = \cos^3 \theta - 3 \cos \theta \sin^2 \theta  }} {{eqn | o =        | ro=+       | r = i \paren {3 \cos^2 \theta \sin \theta - \sin^3 \theta}       | c = rearranging }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \cos 3 \theta       | r = \cos^3 \theta - 3 \cos \theta \sin^2 \theta       | c = equating real parts in $(1)$ }} {{eqn | r = \cos^3 \theta - 3 \cos \theta \paren {1 - \cos^2 \theta}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 4 \cos^3 \theta - 3 \cos \theta       | c = multiplying out and gathering terms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sin \left({4 \theta}\right)       | r = \sin \left({3 \theta + \theta}\right) }} {{eqn | r = \sin 3 \theta \cos \theta + \cos 3 \theta \sin \theta       | c = Sine of Sum }} {{eqn | r = \left({3 \sin \theta - 4 \sin^3 \theta}\right) \cos \theta + \left({4 \cos^3 \theta - 3 \cos \theta}\right) \sin \theta       | c = Triple Angle Formula for Sine and Triple Angle Formula for Cosine }} {{eqn | r = 3 \sin \theta \cos \theta - 4 \sin^3 \theta \cos \theta + 4 \cos^3 \theta \sin \theta - 3 \cos \theta \sin \theta       | c =  }} {{eqn | r = 3 \sin \theta \cos \theta - 4 \sin^3 \theta \cos \theta + 4 \cos \theta \left({1 - \sin^2 \theta}\right) \sin \theta - 3 \cos \theta \sin \theta       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 4 \sin \theta \cos \theta - 8 \sin^3 \theta \cos \theta       | c = multiplying out and gathering terms }} {{end-eqn}} {{qed}}
We have: {{begin-eqn}} {{eqn | l = \cos 4 \theta + i \sin 4 \theta       | r = \paren {\cos \theta + i \sin \theta}^4       | c = De Moivre's Formula }} {{eqn | r = \paren {\cos \theta}^4 + \binom 4 1 \paren {\cos \theta}^3 \paren {i \sin \theta} + \binom 4 2 \paren {\cos \theta}^2 \paren {i \sin \theta}^2 }} {{eqn | o =        | ro=+       | r = \binom 4 3 \paren {\cos \theta} \paren {i \sin \theta}^3 + \paren {i \sin \theta}^4       | c = Binomial Theorem }} {{eqn | r = \cos^4 \theta + 4 i \cos^3 \theta \sin \theta - 6 \cos^2 \theta \sin^2 \theta       | c = substituting for binomial coefficients }} {{eqn | o =        | ro=-       | r = 4 i \cos \theta \sin^3 \theta + \sin^4 \theta       | c = and using $i^2 = -1$ }} {{eqn | n = 1       | r = \cos^4 \theta - 6 \cos^2 \theta \sin^2 \theta + \sin^4 \theta }} {{eqn | o =        | ro=+       | r = 4 i \cos^3 \theta \sin \theta - 4 i \cos \theta \sin^3 \theta       | c = rearranging }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \sin 4 \theta       | r = 4 \cos^3 \theta \sin \theta - 4 \cos \theta \sin^3 \theta       | c = equating imaginary parts in $(1)$ }} {{eqn | r = 4 \cos \theta \paren {1 - \sin^2 \theta} \sin \theta - 4 \cos \theta \sin^3 \theta       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 4 \cos \theta \sin \theta - 8 \sin^3 \theta \cos \theta       | c = multiplying out and gathering terms }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos 4 \theta       | r = \cos \paren {2 \theta + 2 \theta} }} {{eqn | r = \cos 2 \theta \cos 2 \theta - \sin 2 \theta \sin 2 \theta       | c = Cosine of Sum }} {{eqn | r = \paren {\cos^2 \theta - \sin^2 \theta} \paren {\cos^2 \theta - \sin^2 \theta} - \paren {2 \sin \theta \cos \theta} \paren {2 \sin \theta \cos \theta}       | c = Double Angle Formulas }} {{eqn | r = \cos^4 \theta - 2 \cos^2 \theta \sin^2 \theta + \sin^4 \theta - 4 \cos^2 \theta \sin^2 \theta       | c = multiplying out }} {{eqn | r = \cos^4 \theta - 2 \cos^2 \theta \paren {1 - \cos^2 \theta} + \paren {1 - \cos^2 \theta}^2 - 4 \cos^2 \theta \paren {1 - \cos^2 \theta}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 8 \cos^4 \theta - 8 \cos^2 \theta + 1       | c = multiplying out and gathering terms }} {{end-eqn}} {{qed}}
We have: {{begin-eqn}} {{eqn | l = \cos 4 \theta + i \sin 4 \theta       | r = \paren {\cos \theta + i \sin \theta}^4       | c = De Moivre's Formula }} {{eqn | r = \paren {\cos \theta}^4 + \binom 4 1 \paren {\cos \theta}^3 \paren {i \sin \theta} + \binom 4 2 \paren {\cos \theta}^2 \paren {i \sin \theta}^2 }} {{eqn | o =        | ro= +       | r = \binom 4 3 \paren {\cos \theta} \paren {i \sin \theta}^3 + \paren {i \sin \theta}^4       | c = Binomial Theorem }} {{eqn | r = \cos^4 \theta + 4 i \cos^3 \theta \sin \theta - 6 \cos^2 \theta \sin^2 \theta       | c = substituting for binomial coefficients }} {{eqn | o =        | ro= -       | r = 4 i \cos \theta \sin^3 \theta + \sin^4 \theta       | c = and using $i^2 = -1$ }} {{eqn | n = 1       | r = \cos^4 \theta - 6 \cos^2 \theta \sin^2 \theta + \sin^4 \theta }} {{eqn | o =        | ro= +       | r = 4 i \cos^3 \theta \sin \theta - 4 i \cos \theta \sin^3 \theta       | c = rearranging }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \cos 4 \theta       | r = \cos^4 \theta - 6 \cos^2 \theta \sin^2 \theta + \sin^4 \theta       | c = equating real parts in $(1)$ }} {{eqn | r = \cos^4 \theta - 6 \cos^2 \theta \paren {1 - \cos^2 \theta} + \paren {1 - \cos^2 \theta}^2        | c = Sum of Squares of Sine and Cosine }} {{eqn | r = 8 \cos^4 \theta - 8 \cos^2 \theta + 1       | c = multiplying out and gathering terms }} {{end-eqn}} {{qed}}
Let $n = 0$. Then: {{begin-eqn}} {{eqn | l = r^0 \paren {\map \cos {0 x} + i \, \map \sin {0 x} }       | r = 1 \times \paren {\cos 0 + i \sin 0}       | c = {{Defof|Power (Algebra)|Zeroth Power}} }} {{eqn | r = 1 \paren {1 + i 0}       | c = Cosine of Zero is One and Sine of Zero is Zero }} {{eqn | r = 1       | c =  }} {{eqn | r = \paren {r \paren {\cos x + i \sin x} }^0       | c = {{Defof|Power (Algebra)|Zeroth Power}} }} {{end-eqn}} Now let $n \in \Z_{<0}$. Let $n = -m$ where $m > 0$. Thus: {{begin-eqn}} {{eqn | l = \paren {r \paren {\cos x + i \sin x} }^{-m}       | r = \frac 1 {\paren {r \paren {\cos x + i \sin x} }^m}       | c =  }} {{eqn | r = \frac 1 {r^m \paren {\map \cos {m x} + i \, \map \sin {m x} } }       | c = De Moivre's Formula: Positive Integer Index }} {{eqn | r = r^{-m} \paren {\map \cos {-m x} + i \, \map \sin {-m x} }       | c = {{Defof|Complex Division}} }} {{eqn | r = r^n \paren {\map \cos {n x} + i \, \map \sin {n x} }       | c = as $n = -m$ }} {{end-eqn}} {{qed}}
Write $p = \dfrac a b$, where $a, b \in \Z$, $b \ne 0$. Then: {{begin-eqn}} {{eqn | l = r^p \paren {\map \cos {p x} + i \, \map \sin {p x} }       | r = \paren {r^p \paren {\map \cos {p x} + i \, \map \sin {p x} } }^{\frac b b} }} {{eqn | r = \paren {r^{b p} \paren {\map \cos {b p x} + i \, \map \sin {b p x} } }^{\frac 1 b}       | c = De Moivre's Formula/Integer Index }} {{eqn | r = \paren {r^a \paren {\map \cos {a x} + i \, \map \sin {a x} } }^{\frac 1 b} }} {{eqn | r = \paren {r \paren {\cos x + i \, \sin x } }^{\frac a b}       | c = De Moivre's Formula/Integer Index }} {{eqn | r = \paren {r \paren {\cos x + i \, \sin x } }^p }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \mathbf u \cdot \mathbf v       | r = \sum_{i \mathop = 1}^n u_i v_i       | c = {{Defof|Dot Product}} }} {{eqn | r = \sum_{i \mathop = 1}^n v_i u_i       | c = Real Multiplication is Commutative }} {{eqn | r = \mathbf v \cdot \mathbf u       | c = {{Defof|Dot Product}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \mathbf u \cdot \mathbf v       | r = \norm {\mathbf u} \norm {\mathbf v} \cos \angle \mathbf u, \mathbf v       | c = {{Defof|Dot Product}} }} {{eqn | r = \norm {\mathbf v} \norm {\mathbf u} \cos \angle \mathbf u, \mathbf v       | c = Real Multiplication is Commutative }} {{eqn | r = \norm {\mathbf v} \norm {\mathbf u} \cos \angle \mathbf v, \mathbf u       | c = Cosine Function is Even }} {{eqn | r = \mathbf v \cdot \mathbf u       | c = {{Defof|Dot Product}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \mathbf u \cdot \mathbf u       | r = \sum_{i \mathop = 1}^n u_i^2       | c = Definition of Dot Product }} {{eqn | o = \ge       | r = 0       | c = as $u_i \in \R$ it follows that $u_i^2 \ge 0$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \mathbf u \cdot \mathbf u       | r = \left\Vert{\mathbf u}\right\Vert^2       | c = Dot Product of Vector with Itself }} {{eqn | o = \ge       | r = 0       | c = Square of Real Number is Non-Negative }} {{end-eqn}} {{qed}}
Let: : $z_1 := x_1 + i y_1, z_2 = x_2 + i y_2$ Then: {{begin-eqn}} {{eqn | l = z_1 \times z_2       | r = x_1 y_2 - y_1 x_2       | c = {{Defof|Vector Cross Product|subdef = Complex|index = 1|Complex Cross Product}} }} {{eqn | r = -\left({x_2 y_1 - y_2 x_1}\right)       | c = Real Addition is Commutative and Real Multiplication is Commutative }} {{eqn | r = -\left({z_2 \times z_1}\right)       | c = {{Defof|Vector Cross Product|subdef = Complex|index = 1|Complex Cross Product}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = z       | r = r e^{i \theta}       | c =  }} {{eqn | r = r \paren {\cos \theta + i \sin \theta}       | c = Euler's Formula }} {{eqn | ll= \leadsto       | l = \overline z       | r = r \paren {\cos \theta - i \sin \theta}       | c = Polar Form of Complex Conjugate }} {{eqn | r = r e^{-i \theta}       | c = Corollary to Euler's Formula }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = z_1 \circ z_2       | r = \map \Re {\overline {z_1} z_2}       | c = {{Defof|Dot Product|subdef = Complex|index = 3}} }} {{eqn | r = \map \Re {r_1 e^{-i \theta_1} r_2 e^{i \theta_2} }       | c = Exponential Form of Complex Conjugate }} {{eqn | r = \map \Re {r_1 r_2 e^{i \paren {\theta_2 - \theta_1} } }       | c = Product of Complex Numbers in Exponential Form }} {{eqn | r = \map \Re {r_1 r_2 \paren {\map \cos {\theta_2 - \theta_1} + i \, \map \sin {\theta_2 - \theta_1} } }       | c = {{Defof|Polar Form of Complex Number}} }} {{eqn | r = r_1 r_2 \map \cos {\theta_2 - \theta_1}       | c = {{Defof|Real Part}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = z_1 \times z_2       | r = \map \Im {\overline {z_1} z_2}       | c = {{Defof|Vector Cross Product|subdef = Complex|index = 3|Complex Cross Product}} }} {{eqn | r = \map \Im {r_1 e^{-i \theta_1} r_2 e^{i \theta_2} }       | c = Exponential Form of Complex Conjugate }} {{eqn | r = \map \Im {r_1 r_2 e^{i \paren {\theta_2 - \theta_1} } }       | c = Product of Complex Numbers in Exponential Form }} {{eqn | r = \map \Im {r_1 r_2 \paren {\map \cos {\theta_2 - \theta_1} + i \, \map \sin {\theta_2 - \theta_1} } }       | c = {{Defof|Polar Form of Complex Number}} }} {{eqn | r = r_1 r_2 \map \sin {\theta_2 - \theta_1}       | c = {{Defof|Imaginary Part}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \left({\mathbf u + \mathbf v}\right) \cdot \mathbf w       | r = \sum_{i \mathop = 1}^n \left({u_i + v_i}\right) w_i       | c = {{Defof|Vector Sum}} and {{Defof|Dot Product}} }} {{eqn | r = \sum_{i \mathop = 1}^n \left({u_i w_i + v_i w_i}\right)       | c = Real Multiplication Distributes over Real Addition }} {{eqn | r = \sum_{i \mathop = 1}^n u_i w_i + \sum_{i \mathop = 1}^n v_i w_i       | c = }} {{eqn | r = \mathbf u \cdot \mathbf w + \mathbf v \cdot \mathbf w       | c = {{Defof|Dot Product}} }} {{end-eqn}} {{qed}}
Let the vectors $\mathbf u$, $\mathbf v$ and $\mathbf w$ be embedded in a Cartesian $3$-space. It is noted that $\mathbf u$, $\mathbf v$ and $\mathbf w$ are not necessarily coplanar. :520px Let instances of $\mathbf u$ and $\mathbf w$ be selected so their initial points are at some point $O$. Let an instance of $\mathbf v$ be selected so its initial point is positioned at the terminal point $U$ of $\mathbf u$. Let the terminal point $\mathbf v$ be $V$. Let $UA$ be dropped perpendicular to $\mathbf w$. Let $VB$ be dropped perpendicular to $\mathbf w$. Let another instance of $\mathbf w$ be selected so that its initial point is at $U$. Let $VC$ be dropped perpendicular to this second instance of $\mathbf w$. Let $CB$ be dropped from $C$ to the first instance of $\mathbf w$. We have that: :$UA \parallel CB$ and: :$UC \parallel AB$ Thus $\Box ABCU$ is a parallelogram. Hence: :$AB = UC$ Then we have that: {{begin-eqn}} {{eqn | l = OA       | r = OU \cos \angle \mathbf u, \mathbf w       | c =  }} {{eqn | r = \norm {\mathbf u} \cos \angle \mathbf u, \mathbf w       | c =  }} {{eqn | l = AB       | r = UV \cos \angle \mathbf v, \mathbf w       | c =  }} {{eqn | r = \norm {\mathbf v} \cos \angle \mathbf v, \mathbf w       | c =  }} {{eqn | l = OB       | r = OV \cos \angle \paren {\mathbf u + \mathbf v}, \mathbf w       | c =  }} {{eqn | r = \norm {\paren {\mathbf u + \mathbf v} } \cos \angle \paren {\mathbf u + \mathbf v}, \mathbf w       | c =  }} {{eqn | r = \norm {\mathbf u} \cos \angle \mathbf u, \mathbf w + \norm {\mathbf v} \cos \angle \mathbf v, \mathbf w       | c =  }} {{eqn | ll= \leadsto       | l = \norm {\paren {\mathbf u + \mathbf v} } \norm {\mathbf w} \cos \angle \paren {\mathbf u + \mathbf v}, \mathbf w       | r = \norm {\mathbf u} \norm {\mathbf w} \cos \angle \mathbf u, \mathbf w + \norm {\mathbf v} \norm {\mathbf w} \cos \angle \mathbf v, \mathbf w       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\mathbf u + \mathbf v} \cdot \mathbf w       | r = \mathbf u \cdot \mathbf w + \mathbf v \cdot \mathbf w       | c = {{Defof|Dot Product|index = 2}} }} {{end-eqn}} Hence the result. {{qed}}
From Dot Product Operator is Bilinear: :$\paren {c \mathbf u + \mathbf v} \cdot \mathbf w = c \paren {\mathbf u \cdot \mathbf w} + \paren {\mathbf v \cdot \mathbf w}$ Setting $c = 1$ yields the result. {{qed}}
{{begin-eqn}} {{eqn | l = \paren {c \mathbf u + \mathbf v} \cdot \mathbf w       | r = c \sum_{i \mathop = 1}^n \paren {u_i + v_i} w_i       | c = {{Defof|Dot Product|index = 1}} }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {c u_i + v_i} w_i       | c = Real Multiplication Distributes over Real Addition }} {{eqn | r = \sum_{i \mathop = 1}^n \paren {c u_i w_i + v_i w_i}       | c = Real Multiplication Distributes over Real Addition }} {{eqn | r = \sum_{i \mathop = 1}^n c u_i w_i + \sum_{i \mathop = 1}^n v_i w_i       | c = Real Multiplication is Commutative }} {{eqn | r = c \sum_{i \mathop = 1}^n u_i w_i + \sum_{i \mathop = 1}^n v_i w_i       | c = Real Multiplication Distributes over Real Addition }} {{eqn | r = c \paren {\mathbf u \cdot \mathbf w} + \mathbf v \cdot \mathbf w       | c = {{Defof|Dot Product|index = 1}} }} {{end-eqn}} {{qed}} Category:Dot Product 36k7dqpm7b0l2qlcb912iz5izgz60iw
{{begin-eqn}} {{eqn | l = \left({c \mathbf u}\right) \cdot \mathbf v       | r = \sum_{i \mathop = 1}^n \left({c u_i}\right) v_i       | c = {{Defof|Dot Product}} }} {{eqn | r = \sum_{i \mathop = 1}^n c \left({ u_i v_i }\right)       | c = Real Multiplication is Associative }} {{eqn | r = c \sum_{i \mathop = 1}^n u_i v_i       | c = Real Multiplication Distributes over Real Addition }} {{eqn | r = c \left({\mathbf u \cdot \mathbf v}\right)       | c = {{Defof|Dot Product}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \left({c \mathbf u}\right) \cdot \mathbf v       | r = \left\Vert{c \mathbf u}\right\Vert \left\Vert{\mathbf v}\right\Vert \cos \left({\angle c \mathbf u}\right), \mathbf v       | c = {{Defof|Dot Product}} }} {{eqn | r = \sqrt{ \sum_{i \mathop = 1}^n \left({ c u_i }\right)^2} \left\Vert{ \mathbf v }\right\Vert \cos \angle c \mathbf u, \mathbf v       | c = Definition of Vector Length in $\R^n$ }} {{eqn | r = \sqrt{c^2 \sum_{i \mathop = 1}^n u_i^2} \left\Vert{\mathbf v}\right\Vert \cos \angle \mathbf u, \mathbf v       | c = $c \mathbf u$ and $\mathbf u$ are in the same direction }} {{eqn | r = c \sqrt{\sum_{i \mathop = 1}^n u_i^2} \left\Vert{\mathbf v}\right\Vert \cos \angle \mathbf u, \mathbf v       | c = }} {{eqn | r = c \left\Vert{\mathbf u}\right\Vert \left\Vert{\mathbf v}\right\Vert \cos \angle \mathbf u, \mathbf v       | c = Definition of Vector Length in $\R^n$ }} {{eqn | r = c \left({\mathbf u \cdot \mathbf v}\right)       | c = {{Defof|Dot Product}} }} {{end-eqn}} {{qed}}
From Dot Product Operator is Bilinear: :$\left({c \mathbf u + \mathbf v}\right) \cdot \mathbf w = c \left({\mathbf u \cdot \mathbf w}\right) + \left({\mathbf v \cdot \mathbf w}\right)$ Setting $\mathbf v = 0$ and renaming $\mathbf w$ yields the result. {{qed}}