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Parametrizing Plane Patches With Two Directions

Build plane-patch parametrizations from their corners, and choose parameter bounds that cover the full triangle or rectangle.

Intermediate 10 min read

What You'll Learn

  • Construct a plane parametrization from a base point and two spanning directions
  • Translate triangle vertices into a triangular parameter domain
  • Choose rectangular parameter bounds for a plane rectangle
  • Verify that a parametrization reaches the intended corners and edges
  • Distinguish the boundary edges of triangular and rectangular parameter domains
  • Check when two spanning directions produce a nondegenerate patch

From Corners to a Parametrization

A plane patch can be described by starting at one point and moving in two independent directions within the plane. The two parameters record how far to move in each direction. This construction is especially useful when a patch is specified by its corners: the corner coordinates directly provide the base point and spanning vectors, while the shape determines the parameter bounds.

The same basic formula describes both triangles and rectangles. What changes is the parameter domain. A triangular patch uses a triangular domain in the parameter plane; a rectangular patch uses a rectangular domain. Choosing the correct domain is essential: a correct formula with incorrect bounds may describe only part of the patch or a larger region than intended.

Definition: Let \(\mathbf{p}_0\) be a point in \(\mathbb{R}^3\), and let \(\mathbf{a},\mathbf{b}\) be nonparallel vectors. The map \( \mathbf{r}(s,t)=\mathbf{p}_0+s\mathbf{a}+t\mathbf{b} \) parametrizes a patch in the plane through \(\mathbf{p}_0\) spanned by \(\mathbf{a}\) and \(\mathbf{b}\), once a parameter domain for \((s,t)\) is specified. Nonparallel directions ensure that the patch has two dimensions rather than collapsing to a line or point.

Parametrizing a Triangular Patch

Suppose the triangle has vertices \(\mathbf{A},\mathbf{B},\mathbf{C}\), with \(\mathbf{A}\) chosen as the base point. The two directions from \(\mathbf{A}\) to the other vertices are \(\mathbf{B}-\mathbf{A}\) and \(\mathbf{C}-\mathbf{A}\). A parametrization is

$$ \mathbf{r}(s,t)=\mathbf{A}+s(\mathbf{B}-\mathbf{A})+t(\mathbf{C}-\mathbf{A}), \qquad s\geq 0,\quad t\geq 0,\quad s+t\leq 1. $$

The parameter domain is the triangle with parameter vertices \((0,0)\), \((1,0)\), and \((0,1)\). Substituting them gives \(\mathbf{A}\), \(\mathbf{B}\), and \(\mathbf{C}\), respectively. The inequality \(s+t\leq 1\) is what keeps the points within the triangle instead of extending beyond the edge from \(\mathbf{B}\) to \(\mathbf{C}\).

Worked Example: A Triangle Given by Three Vertices

Parametrize the triangle with vertices \(\mathbf{A}=(1,0,2)\), \(\mathbf{B}=(3,1,2)\), and \(\mathbf{C}=(0,2,5)\).

First calculate the two directions from \(\mathbf{A}\):

$$ \mathbf{B}-\mathbf{A}=(3-1,1-0,2-2)=(2,1,0), $$
$$ \mathbf{C}-\mathbf{A}=(0-1,2-0,5-2)=(-1,2,3). $$

Use these directions in the triangular parametrization:

$$ \mathbf{r}(s,t)=(1,0,2)+s(2,1,0)+t(-1,2,3), \quad s\geq0,\quad t\geq0,\quad s+t\leq1. $$

In coordinates this is \(\mathbf{r}(s,t)=(1+2s-t,\ s+2t,\ 2+3t)\). Check the three parameter vertices:

$$ \mathbf{r}(0,0)=(1,0,2)=\mathbf{A},\quad \mathbf{r}(1,0)=(3,1,2)=\mathbf{B},\quad \mathbf{r}(0,1)=(0,2,5)=\mathbf{C}. $$

The directions are independent because their cross product is nonzero:

$$ (2,1,0)\times(-1,2,3)=(3,-6,5)\ne\mathbf{0}. $$

As an additional check, the plane has normal vector \((3,-6,5)\). Its equation through \(\mathbf{A}\) is \(3(x-1)-6y+5(z-2)=0\), or \(3x-6y+5z-13=0\). Substituting \(\mathbf{B}\) gives \(9-6+10-13=0\), and substituting \(\mathbf{C}\) gives \(0-12+25-13=0\). Thus all three vertices, and hence the parametrized triangle, lie in the same plane.

Why the Triangular Bounds Cover Exactly the Triangle

The bounds are not an arbitrary convention. For any point in the parameter domain, define \(\lambda_0=1-s-t\), \(\lambda_1=s\), and \(\lambda_2=t\). The inequalities defining the domain imply that each coefficient is nonnegative and that their sum is one. The parametrization can therefore be rewritten as a weighted combination of the three vertices:

$$ \mathbf{r}(s,t) =(1-s-t)\mathbf{A}+s\mathbf{B}+t\mathbf{C} =\lambda_0\mathbf{A}+\lambda_1\mathbf{B}+\lambda_2\mathbf{C}, \qquad \lambda_0+\lambda_1+\lambda_2=1. $$

Conversely, any point expressed as a combination of the vertices with nonnegative coefficients summing to one has this form: set \(s=\lambda_1\) and \(t=\lambda_2\). Then \(s\geq0\), \(t\geq0\), and \(s+t=1-\lambda_0\leq1\). This proves that the parameter domain maps onto exactly the filled triangle, including its edges and vertices.

If the two directions are independent, each point of the plane has at most one pair of parameters in this formula. Indeed, if two parameter pairs give the same point, subtracting their expressions gives \((s_1-s_2)\mathbf{a}+(t_1-t_2)\mathbf{b}=\mathbf{0}\). Independence forces both coefficient differences to be zero. Thus the parametrization does not fold distinct parameter points onto the same point.

Boundary of the triangular domain: The triangle's three edges come from \(s=0\), \(t=0\), and \(s+t=1\). In particular, the edge from \(\mathbf{B}\) to \(\mathbf{C}\) is \(s+t=1\); it is not obtained by fixing one parameter alone.

Parametrizing a Rectangular Patch

For a rectangle, choose a corner \(\mathbf{P}_0\) and let \(\mathbf{a}\) and \(\mathbf{b}\) be the two edge vectors from that corner. If these are the full edge vectors, the parametrization is

$$ \mathbf{r}(s,t)=\mathbf{P}_0+s\mathbf{a}+t\mathbf{b}, \qquad 0\leq s\leq1,\quad 0\leq t\leq1. $$

The four parameter corners map to the four patch corners. For the patch to be a geometric rectangle, the edge vectors must be perpendicular as well as nonzero. If they are independent but not perpendicular, the same formula and bounds describe a parallelogram, not a rectangle.

Worked Example: A Rectangle in a Tilted Plane

Parametrize the rectangle with one corner \(\mathbf{P}_0=(1,-1,0)\) and edge vectors \(\mathbf{a}=(2,0,2)\) and \(\mathbf{b}=(1,2,-1)\).

First verify that the edges meet at a right angle:

$$ \mathbf{a}\cdot\mathbf{b}=2(1)+0(2)+2(-1)=2-2=0. $$

Both vectors are nonzero, so they span a rectangle in their plane. The parametrization with unit-square bounds is

$$ \mathbf{r}(s,t)=(1,-1,0)+s(2,0,2)+t(1,2,-1), \qquad 0\leq s\leq1,\quad 0\leq t\leq1. $$

Its four corners are

$$ \mathbf{r}(0,0)=(1,-1,0), $$
$$ \mathbf{r}(1,0)=(1,-1,0)+(2,0,2)=(3,-1,2), $$
$$ \mathbf{r}(0,1)=(1,-1,0)+(1,2,-1)=(2,1,-1), $$
$$ \mathbf{r}(1,1)=(1,-1,0)+(2,0,2)+(1,2,-1)=(4,1,1). $$

The normal vector is \(\mathbf{a}\times\mathbf{b}=(-4,4,4)\), so a simpler normal is \((-1,1,1)\). The plane equation through \(\mathbf{P}_0\) is \(-x+y+z+2=0\). Substitution at the four corners gives, respectively, \(-1-1+0+2=0\), \(-3-1+2+2=0\), \(-2+1-1+2=0\), and \(-4+1+1+2=0\). The corners lie in the plane, and the parametrization covers the rectangle between them.

Bounds and Boundary Edges

In the rectangular domain \(0\leq s,t\leq1\), fixing one parameter at either endpoint and allowing the other to vary traces an edge. For example, \(s=0\) traces the edge from \(\mathbf{P}_0\) to \(\mathbf{P}_0+\mathbf{b}\), and \(t=1\) traces the edge from \(\mathbf{P}_0+\mathbf{b}\) to \(\mathbf{P}_0+\mathbf{a}+\mathbf{b}\). The four sides correspond to \(s=0\), \(s=1\), \(t=0\), and \(t=1\).

Parameter bounds can also be scaled to match the chosen directions. If \(\mathbf{a}\) and \(\mathbf{b}\) are unit directions rather than full edge vectors, and the side lengths are \(L\) and \(M\), use \(0\leq s\leq L\) and \(0\leq t\leq M\). More generally, non-unit intervals simply change which portions of the two directions are included. State the bounds explicitly so the intended patch is unambiguous.

Worked Example: Using Lengths as Parameter Bounds

Consider the rectangle starting at \(\mathbf{P}_0=(0,1,0)\), with perpendicular unit edge directions \(\mathbf{u}=(1,0,1)/\sqrt{2}\) and \(\mathbf{v}=(0,1,0)\), and side lengths \(2\sqrt{2}\) and \(3\). Since \(\mathbf{u}\cdot\mathbf{v}=0\), these directions are perpendicular. Using distance along each unit direction as the parameter gives

$$ \mathbf{r}(s,t)=(0,1,0)+s\left(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}\right)+t(0,1,0), \qquad 0\leq s\leq2\sqrt{2},\quad 0\leq t\leq3. $$

At the parameter corners, the points are \(\mathbf{r}(0,0)=(0,1,0)\), \(\mathbf{r}(2\sqrt{2},0)=(2,1,2)\), \(\mathbf{r}(0,3)=(0,4,0)\), and \(\mathbf{r}(2\sqrt{2},3)=(2,4,2)\). For example, the second point follows because \(2\sqrt{2}\mathbf{u}=(2,0,2)\). These are exactly the starting corner, the endpoint in the first direction, the endpoint in the second direction, and the opposite corner. The bound lengths therefore give the full rectangle, not merely a unit-sized portion of it.

A Common Parameter-Domain Mistake

A frequent error is to use rectangular boundary reasoning for a triangular domain. For the triangle \(s\geq0,\ t\geq0,\ s+t\leq1\), someone might claim that fixing \(s=1\) and varying \(t\) traces an edge. But the bounds require \(1+t\leq1\), so \(t\leq0\); together with \(t\geq0\), this forces \(t=0\). The only allowed point is \((1,0)\), a vertex, not an edge.

The correction is to read the triangle's three boundary conditions directly: \(s=0\), \(t=0\), and \(s+t=1\). Along the third edge, \(t=1-s\) with \(0\leq s\leq1\), and the image is

$$ \mathbf{r}(s,1-s) =\mathbf{A}+s(\mathbf{B}-\mathbf{A})+(1-s)(\mathbf{C}-\mathbf{A}), \qquad 0\leq s\leq1. $$

At \(s=0\) this is \(\mathbf{C}\), and at \(s=1\) it is \(\mathbf{B}\), so this condition traces the full edge between those vertices.

Why These Parametrizations Are Useful

A parametrization converts a geometric patch into a formula together with a precise domain. The formula records where the patch sits in space, while the parameter bounds record which portion of the plane is included. Keeping these roles separate makes it easier to check corners, edges, and the shape of the domain before using the parametrization in later calculations.

Before accepting a proposed parametrization, check that the base point is the intended corner, that the two directions reach the other relevant corners, and that the parameter domain has the correct shape. For a nondegenerate triangle or rectangle, the directions must be independent. For a rectangle, also verify that its edge directions are perpendicular. Finally, substitute the parameter-domain corners and confirm that they give the stated patch corners.

Check Your Understanding

Use the definitions and checks in this tutorial to answer the following questions.

  1. Given triangle vertices \(\mathbf{A},\mathbf{B},\mathbf{C}\), write a parametrization based at \(\mathbf{A}\) and state its parameter bounds.
  2. Which three parameter-domain equations describe the edges of the triangular domain \(s\geq0,\ t\geq0,\ s+t\leq1\)?
  3. For a rectangle with full edge vectors \(\mathbf{a}\) and \(\mathbf{b}\), what bounds give the full patch, and what additional condition makes it a rectangle rather than a parallelogram?
  4. Why must the two spanning directions be independent to produce a two-dimensional patch?
  5. If \(s\) and \(t\) measure distances along perpendicular unit directions and the side lengths are \(L\) and \(M\), what parameter bounds cover the rectangle?

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