$\Theta \rho ϵ\eta \Pi \alpha \tau \pi$

If $\theta =0$, then $u$ and $v$ point along the same ray, so one is a positive scalar multiple of the other. If $\theta =\pi$, then $u$ and $v$ point along opposite rays, so one is a negative scalar multiple of the other. In either case, $u$ and $v$ are collinear, so $\{u,v\}$ is linearly dependent. This contradicts the assumption that $u,v$ are linearly independent.

Write $a=(a_1,\dots ,a_n)$ and $b=(b_1,\dots ,b_n)$. By the algebraic definition of the dot product and commutativity of multiplication in $ℝ$, $$a\cdot b=\sum _{i=1}^n{a}_i{b}_i=\sum _{i=1}^n{b}_i{a}_i=b\cdot a$$

Write $a=(a_1,\dots ,a_n)$, $b=(b_1,\dots ,b_n)$, and $c=(c_1,\dots ,c_n)$. By the definition of the dot product, $$\begin{array}{cccc}& (a+b)\cdot c& =\sum _{i=1}^n(a_i+b_i)c_i& \text{<span class="tml-eqn"></span>}\\ & & =\sum _{i=1}^n(a_i{c}_i+b_i{c}_i)& \text{<span class="tml-eqn"></span>}\\ & & =\sum _{i=1}^n{a}_i{c}_i+\sum _{i=1}^n{b}_i{c}_i& \text{<span class="tml-eqn"></span>}\\ & & =a\cdot c+b\cdot c& \text{<span class="tml-eqn"></span>}\end{array}$$

By the definition of the dot product, $$\left(\lambda a\right)\cdot b=\sum _{i=1}^n\left(\lambda a_i\right)b_i=\lambda \sum _{i=1}^n{a}_i{b}_i=\lambda (a\cdot b)$$ The second equality follows from this equality and the symmetry of the dot product: $$a\cdot \left(\lambda b\right)=\left(\lambda b\right)\cdot a=\lambda (b\cdot a)=\lambda (a\cdot b)$$

By the definition of the dot product, $$a\cdot a=\sum _{i=1}^n{a}_i^2$$ Since every square $a_i^2$ is non-negative, $a\cdot a\ge 0$. Also, ${\sum }_{i=1}^n{a}_i^2=0$ exactly when each $a_i^2=0$, which happens exactly when each $a_i=0$, i.e. exactly when $a=0$.

Write $x=(x_1,x_2)$ and $y=(y_1,y_2)$. By the definition of the dot product, $$x\cdot x=x_1^2+x_2^2,y\cdot y=y_1^2+y_2^2,x\cdot y=x_1{y}_1+x_2{y}_2.$$ Substituting these expressions and multiplying out gives $$\begin{array}{cc}(x\cdot x)(y\cdot y)-{(x\cdot y)}^2& =(x_1^2+x_2^2)(y_1^2+y_2^2)-{(x_1{y}_1+x_2{y}_2)}^2\\ & =x_1^2{y}_2^2+x_2^2{y}_1^2-2x_1{x}_2{y}_1{y}_2\\ & ={(x_1{y}_2-x_2{y}_1)}^2.\end{array}$$ To show that this square is strictly positive, we must show that $x_1{y}_2-x_2{y}_1\ne 0$. Suppose instead that $x_1{y}_2-x_2{y}_1=0$, meaning $x_1{y}_2=x_2{y}_1$. Since linearly independent sets cannot contain the zero vector, $x\ne 0$, so at least one of $x_1,x_2$ is nonzero. We consider each possibility so that we divide only by a nonzero coordinate.

If $x_1\ne 0$, the assumed equality $x_1{y}_2-x_2{y}_1=0$ gives $y_2=(y_1/x_1)x_2$. Thus

$$y=(y_1,y_2)=(y_1,(y_1/x_1)x_2)=(y_1/x_1)(x_1,x_2)=(y_1/x_1)x.$$ If $x_1=0$, then $x_2\ne 0$. The assumed equality becomes $-x_2{y}_1=0$, so $y_1=0$, and therefore $$y=(0,y_2)=(y_2/x_2)(0,x_2)=(y_2/x_2)x.$$ In either case, $y$ is a scalar multiple of $x$, contradicting their linear independence. Hence $x_1{y}_2-x_2{y}_1\ne 0$, and its square is positive.

Using norm squared is the dot product, dot product distributes into vector addition from the right, dot product distributes into vector addition from the left, and symmetry of the dot product, we get $$\begin{array}{cccc}& {\Vert x+y\Vert }^2& =(x+y)\cdot (x+y)& \text{<span class="tml-eqn"></span>}\\ & & =x\cdot x+x\cdot y+y\cdot x+y\cdot y& \text{<span class="tml-eqn"></span>}\\ & & ={\Vert x\Vert }^2+2x\cdot y+{\Vert y\Vert }^2& \text{<span class="tml-eqn"></span>}\end{array}$$ Replacing $y$ by $-y$, and using compatibility with scalar multiplication, gives $${\Vert x-y\Vert }^2={\Vert x\Vert }^2-2x\cdot y+{\Vert y\Vert }^2$$

TODO: Add a Euclidean geometry proof here.

The vectors $a$, $b$, and $a-b$ determine a triangle with side lengths $\Vert a\Vert$, $\Vert b\Vert$, and $\Vert a-b\Vert$. By the Euclidean law of cosines applied to this triangle, $${\Vert a-b\Vert }^2={\Vert a\Vert }^2+{\Vert b\Vert }^2-2\Vert a\Vert \Vert b\Vert \cos \left(\theta \right)$$ On the other hand, by the dot product expansion of norm squared, $${\Vert a-b\Vert }^2={\Vert a\Vert }^2-2a\cdot b+{\Vert b\Vert }^2$$ Equating the two expressions and subtracting ${\Vert a\Vert }^2+{\Vert b\Vert }^2$ from both sides gives $$-2a\cdot b=-2\Vert a\Vert \Vert b\Vert \cos \left(\theta \right)$$ Dividing by $-2$, we get $$a\cdot b=\Vert a\Vert \Vert b\Vert \cos \left(\theta \right)$$

Since $x=\lambda \omega$ and $\omega$ is a unit vector, $$\Vert x\Vert =\Vert \lambda \omega \Vert =\left|\lambda \right|\Vert \omega \Vert =\left|\lambda \right|.$$ Also, $$x\cdot \omega =\left(\lambda \omega \right)\cdot \omega =\lambda (\omega \cdot \omega )=\lambda ,$$ because $\omega \cdot \omega ={\Vert \omega \Vert }^2=1$. Therefore $\Vert x\Vert =|x\cdot \omega |$.

Since $x=\lambda n$, $$|x\cdot \omega |=|\left(\lambda n\right)\cdot \omega |=\left|\lambda \right||n\cdot \omega |.$$ Since $n$ is a unit vector, $$\Vert x\Vert =\Vert \lambda n\Vert =\left|\lambda \right|\Vert n\Vert =\left|\lambda \right|.$$ Therefore $|x\cdot \omega |=\Vert x\Vert |n\cdot \omega |$.

Write $a=(a_1,a_2,a_3)$ and $b=(b_1,b_2,b_3)$. By the definition of the dot product and the component formula for the cross product, $$\begin{array}{cccc}& a\cdot (a\times b)& =a_1(a_2{b}_3-a_3{b}_2)+a_2(a_3{b}_1-a_1{b}_3)+a_3(a_1{b}_2-a_2{b}_1)& \text{<span class="tml-eqn"></span>}\\ & & =a_1{a}_2{b}_3-a_1{a}_3{b}_2+a_2{a}_3{b}_1-a_1{a}_2{b}_3+a_1{a}_3{b}_2-a_2{a}_3{b}_1& \text{<span class="tml-eqn"></span>}\\ & & =0& \text{<span class="tml-eqn"></span>}\end{array}$$ The same calculation gives $$\begin{array}{cccc}& b\cdot (a\times b)& =b_1(a_2{b}_3-a_3{b}_2)+b_2(a_3{b}_1-a_1{b}_3)+b_3(a_1{b}_2-a_2{b}_1)& \text{<span class="tml-eqn"></span>}\\ & & =a_2{b}_1{b}_3-a_3{b}_1{b}_2+a_3{b}_1{b}_2-a_1{b}_2{b}_3+a_1{b}_2{b}_3-a_2{b}_1{b}_3& \text{<span class="tml-eqn"></span>}\\ & & =0& \text{<span class="tml-eqn"></span>}\end{array}$$

Write the vectors in coordinates. The component formula for the cross product shows that each component of $a\times (b+c)$ distributes over the coordinate sums in $b+c$. Therefore $a\times (b+c)=a\times b+a\times c$.

Write the vectors in coordinates. The component formula for the cross product shows that each component of $a\times \left(\lambda b\right)$ has a common factor $\lambda$. Factoring it out gives $a\times \left(\lambda b\right)=\lambda (a\times b)$.

Write $a=(a_1,a_2,a_3)$ and $b=(b_1,b_2,b_3)$. By the component formula for the cross product, $$b\times a=(b_2{a}_3-b_3{a}_2,b_3{a}_1-b_1{a}_3,b_1{a}_2-b_2{a}_1).$$ Multiplying by $-1$ gives $$-(b\times a)=(a_2{b}_3-a_3{b}_2,a_3{b}_1-a_1{b}_3,a_1{b}_2-a_2{b}_1)=a\times b.$$

By expanding the cross product over vector addition and pulling scalar factors out of the first and second inputs, $$u^{\prime }\times v^{\prime }=u\times v-(v\cdot \omega )(u\times \omega )-(u\cdot \omega )(\omega \times v)+(u\cdot \omega )(v\cdot \omega )(\omega \times \omega ).$$ Since $\omega \times \omega =0$, the last term is $0$. Also $u\times \omega$ and $\omega \times v$ are orthogonal to their factors, so both are orthogonal to $\omega$. Taking the dot product with $\omega$ leaves $$(u^{\prime }\times v^{\prime })\cdot \omega =(u\times v)\cdot \omega .$$

Let $n=u\times v$. Let $\theta$ be the angle between $u$ and $v$. Since $u,v\in {ℝ}^3\setminus \left\{0\right\}$, we have $\Vert u\Vert >0$ and $\Vert v\Vert >0$ by nonzero vectors have positive length. Since $u$ and $v$ are linearly independent, the angle satisfies $\theta \ne 0$ and $\theta \ne \pi$ by linearly independent nonzero vectors have non-extreme angle. Also $\theta \in [0,\pi ]$ by the definition of angle, so $\sin \left(\theta \right)\ne 0$ by sine is zero on $[0,\pi ]$ only at the endpoints. Therefore, by the definition of the cross product, $$\Vert u\times v\Vert =\Vert u\Vert \Vert v\Vert \sin \left(\theta \right)\ne 0$$ and so $n\ne 0$.

First suppose that $x\in \mathrm{span}\left(\{u,v\}\right)$. By the definition of span, $x=su+tv$ for some $s,t\in ℝ$. Since $n=u\times v$ is orthogonal to both $u$ and $v$ by the cross product is orthogonal to its factors proposition,

$$x\cdot n=(su+tv)\cdot n=s(u\cdot n)+t(v\cdot n)=0$$ Thus $$\mathrm{span}\left(\{u,v\}\right)\subseteq \{x\in {ℝ}^3:x\cdot n=0\}$$ For the reverse containment, note that $u,v,n$ are linearly independent. To see this, suppose that $$au+bv+cn=0$$ Taking the dot product of both sides with $n$, we get $$a(u\cdot n)+b(v\cdot n)+c(n\cdot n)=0$$ Since $u\cdot n=0$, $v\cdot n=0$, and $n\cdot n={\Vert n\Vert }^2\ne 0$ by norm squared is the dot product, it follows that $c=0$. Therefore $au+bv=0$, and since $u,v$ are linearly independent, we get $a=b=0$. Thus $u,v,n$ are linearly independent.

Since $\dim \left({ℝ}^3\right)=3$ by the dimension of ${ℝ}^n$, the three linearly independent vectors $u,v,n$ form a basis of ${ℝ}^3$.

Now suppose that $x\cdot n=0$. Since $u,v,n$ form a basis, we can write $x=su+tv+rn$. Then

$$0=x\cdot n=s(u\cdot n)+t(v\cdot n)+r(n\cdot n)=r{\Vert n\Vert }^2$$ Since ${\Vert n\Vert }^2\ne 0$, we have $r=0$. Thus $x=su+tv$, so $x\in \mathrm{span}\left(\{u,v\}\right)$ by the definition of span.

Let $q=(x,y,z)$. By the definition of the dot product, we have $$\begin{array}{cccc}& (q-p)\cdot n=0& ⟺(x-p_1,y-p_2,z-p_3)\cdot (a,b,c)=0& \text{<span class="tml-eqn"></span>}\\ & & ⟺a(x-p_1)+b(y-p_2)+c(z-p_3)=0& \text{<span class="tml-eqn"></span>}\\ & & ⟺ax+by+cz=a{p}_1+b{p}_2+c{p}_3& \text{<span class="tml-eqn"></span>}\\ & & ⟺ax+by+cz=p\cdot n& \text{<span class="tml-eqn"></span>}\\ & & ⟺ax+by+cz=d& \text{<span class="tml-eqn"></span>}\end{array}$$ Therefore the coordinate equation is exactly the translated normal form written out component by component.

Let $Q=\{p+su+tv:s,t\in ℝ\}$. By the translated plane as a span definition, $x\in Q$ exactly when $x-p\in \mathrm{span}\left(\{u,v\}\right)$. By the span plane has cross product normal proposition, this happens exactly when $(x-p)\cdot (u\times v)=0$, which is exactly the normal equation $$(x-p)\cdot (u\times v)=0$$

We know that $ax+by+c=0$ defines a line, but we need to verify that $p,q$ are both on that line, but since $p\cdot L=0$ and $q\cdot L=0$ we know that they are on the line, as needed.

By the dot product expansion of norm squared, $$\begin{array}{cccc}& {\Vert x+y\Vert }^2+{\Vert x-y\Vert }^2& =({\Vert x\Vert }^2+2x\cdot y+{\Vert y\Vert }^2)+({\Vert x\Vert }^2-2x\cdot y+{\Vert y\Vert }^2)& \text{<span class="tml-eqn"></span>}\\ & & =2{\Vert x\Vert }^2+2{\Vert y\Vert }^2& \text{<span class="tml-eqn"></span>}\end{array}$$

By the dot product expansion of norm squared and the geometric formula for the dot product, $$\begin{array}{cccc}& {\Vert x+y\Vert }^2& ={\Vert x\Vert }^2+2x\cdot y+{\Vert y\Vert }^2& \text{<span class="tml-eqn"></span>}\\ & & ={\Vert x\Vert }^2+2\Vert x\Vert \Vert y\Vert \cos \left(\theta \right)+{\Vert y\Vert }^2& \text{<span class="tml-eqn"></span>}\end{array}$$

The vectors $x=c-a$ and $y=b-a$ point from $a$ toward the other two vertices, while $z=p-a$ points from $a$ toward $p$. We first express $z$ in terms of these two edge vectors; its coefficients will give the barycentric coordinates.

By assumption, $x$ and $y$ are linearly independent and therefore form a basis of ${ℝ}^2$. Thus there are unique $u,v\in ℝ$ with $z=ux+vy$. Rearranging gives

$$p=a+u(c-a)+v(b-a)=(1-u-v)a+vb+uc.$$ Hence $(1-u-v,v,u)$ are the barycentric coordinates. We now want to compute $u$ and $v$. To get two scalar equations in these two unknowns, take the dot product of both sides of $z=ux+vy$ with a fixed vector. We choose $x$ once and $y$ once. This is what it means here to “multiply” a vector equation by $x$ or $y$: each dot product gives a scalar equation. $$\begin{array}{cc}x\cdot z& =u(x\cdot x)+v(x\cdot y),\\ y\cdot z& =u(x\cdot y)+v(y\cdot y).\end{array}$$ We solve these equations by making the coefficients of one unknown equal and then subtracting, so that unknown cancels. The positive dot product expression proposition tells us that $D>0$, because $x$ and $y$ are linearly independent. Thus we may divide by $D$ after each cancellation.

First, isolate $v$. Multiply the first scalar equation by $x\cdot y$ and the second by $x\cdot x$. Both $u$-coefficients are then $(x\cdot x)(x\cdot y)$. Subtract the scaled first equation from the scaled second:

$$\begin{array}{cc}(x\cdot x)(y\cdot z)-(x\cdot y)(x\cdot z)& =(x\cdot x)[u(x\cdot y)+v(y\cdot y)]\\ & -(x\cdot y)[u(x\cdot x)+v(x\cdot y)]\\ & =v[(x\cdot x)(y\cdot y)-(x\cdot y{)}^2]\\ & =vD.\end{array}$$ Dividing by $D$ gives $$v=\frac{(x\cdot x)(y\cdot z)-(x\cdot y)(x\cdot z)}{D}.$$

Similarly, to isolate $u$, multiply the first scalar equation by $y\cdot y$ and the second by $x\cdot y$. Both $v$-coefficients are then $(x\cdot y)(y\cdot y)$. Subtract the scaled second equation from the scaled first:

$$\begin{array}{cc}(y\cdot y)(x\cdot z)-(x\cdot y)(y\cdot z)& =(y\cdot y)[u(x\cdot x)+v(x\cdot y)]\\ & -(x\cdot y)[u(x\cdot y)+v(y\cdot y)]\\ & =u[(x\cdot x)(y\cdot y)-(x\cdot y{)}^2]\\ & =uD.\end{array}$$ Thus $$u=\frac{(y\cdot y)(x\cdot z)-(x\cdot y)(y\cdot z)}{D},$$ as claimed.