Affine 3-folds and tetrahedra in 4-space

September 16, 2008

Definition.   An affine $r$-fold in ${\mathbf{R}}^n$ is a subset of ${\mathbf{R}}^n$ of the form $$\begin{array}{cc}\text{(1)}& a+t_1{v}_1+t_2{v}_2+\dots +t_r{v}_r\text{with}{t}_1,t_2,\dots ,t_r\text{varying in}\mathbf{R}\end{array}$$ where $a$ is a given point in ${\mathbf{R}}^n$ and $v_1,v_2,\dots ,v_n$ are given vectors in ${\mathbf{R}}^n$ subject to the condition that none of the vectors $v_j$ may be omitted without changing the set (1).

The following data determine a plane, i.e., an affine $2$-fold, in ${\mathbf{R}}^4$:

Example 1.   $$\begin{array}{cc}\text{(2)}& a=\left(1,-2,0,1\right);v_1=\left(1,2,1,3\right),v_2=\left(2,-1,1,0\right)\end{array}$$

Moreover, if one takes $$b=a+v_1=\left(2,0,1,4\right)\text{and}c=a+v_2=\left(3,-3,1,1\right)\text{,}$$ then $a$, $b$, and $c$ are the vertices of triangle $T=abc$ in ${\mathbf{R}}^4$, which is the set of all points

Example 2.   $$\left(1,-2,0,1\right)+t\left(1,2,1,3\right)+u\left(2,-1,1,0\right)\text{with}t,u\ge 0\text{and}t+u\le 1\text{.}$$

Equivalently, $T$ is the set of all points $$s\left(1,-2,0,1\right)+t\left(2,0,1,4\right)+u\left(3,-3,1,1\right)\text{with}s,t,u\ge 0\text{and}s+t+u=1\text{.}$$

What is the area of $T$? The idea is that the plane (2) is a normal Euclidean plane where lengths and angles may be computed using the dot product in ${\mathbf{R}}^4$. Thus the area of the triangle is half the product of the length of a base and the length of the altitude drawn to that base. If we take the side $ab$ as base, then its length is $||b-a||=||v_1||=\sqrt{15}$. For the altitude drawn from $c$ to $ab$, one observes that in the decomposition of the vector $$\stackrel{\rightharpoonup }{ac}=v_2$$ as a sum of components parallel and perpendicular to $v_1$ the perpendicular component lies over the altitude drawn from $c$ to $ab$. One finds $$\begin{array}{cccc}\text{(3)}& \hfill {\mathrm{proj}}_{v_1}\left(v_2\right)& \hfill =\hfill & \frac{1}{15}\left(1,2,1,3\right)\hfill \\ & \hfill {\mathrm{perp}}_{v_1}\left(v_2\right)& \hfill =\hfill & \frac{1}{15}\left(29,-17,14,-3\right)\hfill \end{array}$$ So the area of $T$ is given by $$A=\frac{1}{2}Bh=\frac{1}{2}\sqrt{15}\left(\frac{\sqrt{1335}}{15}\right)=\frac{1}{2}\sqrt{89}\text{.}$$

If, in addition to the point $a$ and the vectors $v_1$, $v_2$, one brings another vector $v_3=\left(-1,3,2,-1\right)$ into the picture, then one has data determining an affine $3$-fold in ${\mathbf{R}}^4$:

Example 3.   $$\begin{array}{cc}\text{(4)}& a=\left(1,-2,0,1\right);v_1=\left(1,2,1,3\right),v_2=\left(2,-1,1,0\right),v_3=\left(-1,3,2,-1\right)\end{array}$$

Just as an affine $2$-fold in ${\mathbf{R}}^3$, i.e., a plane, may be described alternatively as the set of all points satisfying a single linear equation, an affine $3$-fold in ${\mathbf{R}}^4$ — and, more generally, an affine $\left(n-1\right)$-fold in ${\mathbf{R}}^n$ — may be described as the set of all points satisfying a single linear equation in which the vector of coefficients of the coordinates is a vector that is perpendicular to the $3$-fold. To find the equation for the present example one begins by looking for a vector $u$ that is perpendicular to each of the vectors $v_1$, $v_2$, and $v_3$. The relations $u·v_1=0$, $u·v_2=0$, and $u·v_3=0$ amount to $3$ equations for the coefficients of $u$ and to make $u$ specific one may add the equation $u_4=1$. Thus, $$\begin{array}{cccc}& \hfill u_1+2u_2+u_3+3u_4& \hfill =\hfill & 0\hfill \\ & \hfill 2u_1-u_2+u_3& \hfill =\hfill & 0\hfill \\ & \hfill -u_1+3u_2+2u_3-u_4& \hfill =\hfill & 0\hfill \\ & \hfill u_4& \hfill =\hfill & 1\hfill \end{array}$$ Solution of this system of $4$ equations yields: $$u=\left(-9⁄5,-8⁄5,2,1\right)\text{,}$$ and to eliminate denominators without changing the direction of this vector, one may replace it with its scalar multiple by $-5$: $$u=\left(9,8,-10,-5\right)\text{.}$$ Therefore, this affine $3$-fold has equation of the form $$9x_1+8x_2-10x_3-5x_4=\mathrm{constant}\text{,}$$ and the constant is determined by evaluating the left side at the point $a$. Thus the equation of this “hyperplane” is: $$9x_1+8x_2-10x_3-5x_4=-12\text{.}$$

The affine $3$-fold contains, in particular, the following $4$ points: $$a=\left(1,-2,0,1\right),b=a+v_1=\left(2,0,1,4\right),c=a+v_2=\left(3,-3,1,1\right),d=a+v_3=\left(0,1,2,0\right)\text{,}$$ and these four points are the vertices of the tetrahedron $abcd$ that sits inside a copy of ${\mathbf{R}}^3$ that itself is a hyperplane in ${\mathbf{R}}^4$. What is the volume of this tetrahedron?

If we take the triangle $abc$ as “base” with area $A=\left(1⁄2\right)\sqrt{89}$, as previously calculated, principles from school geometry indicate the volume of the tetrahedron should be given by the formula $$\begin{array}{cc}\text{(5)}& V=\frac{1}{3}Ah\end{array}$$ where $h$ is the “altitude” drawn from the vertex $d$ to the base. The idea for finding the altitude is to decompose the vector $v_3=\stackrel{\rightharpoonup }{ad}$ into the sum $w^{\prime }+w^{\prime \prime }$ of two components, with $w^{\prime }$ in the base and $w^{\prime \prime }$ perpendicular to the base. Thus, $$w^{\prime }=t{v}_1+u{v}_2\text{and}{w}^{\prime \prime }=v_3-t{v}_1-u{v}_2\text{,}$$ while the condition $w^{\prime \prime }\perp \Delta abc$ gives the two conditions $w^{\prime \prime }·v_1=0$ and $w^{\prime \prime }·v_2=0$ which amount to a pair of linear equations for the two scalars $t$ and $u$. One finds $t=27⁄89$ and $u=-49⁄89$ with the result that $$w^{\prime }=\left(1⁄89\right)\left(-71,103,-22,81\right)w^{\prime \prime }=\left(1⁄89\right)\left(-18,164,200,-170\right)\text{,}$$ and, therefore, $$h=||w^{\prime \prime }||=\frac{6\sqrt{30}}{\sqrt{89}}\text{.}$$ Applying the formula (5), one sees that the volume of the tetrahedron $abcd$ is $\sqrt{30}$.