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Chapter 8: Q-8.2-18E (page 437)

Let\(T\)be a tetrahedron in “standard” position, with three edges along the three positive coordinate axes in\({\mathbb{R}^3}\), and suppose the vertices are\(a{{\bf{e}}_1}\),\(b{{\bf{e}}_2}\),\(c{{\bf{e}}_{\bf{3}}}\), and 0, where\(\left[ {\begin{array}{*{20}{c}}{{{\bf{e}}_1}}&{{{\bf{e}}_2}}&{{{\bf{e}}_3}}\end{array}} \right] = {I_3}\). Find formulas for the barycentric coordinates of an arbitrary point\({\bf{p}}\)in\({\mathbb{R}^3}\).

Short Answer

Expert verified

The barycentric coordinates are \(\frac{x}{a},\frac{y}{b},\frac{z}{c},1 - \left( {\frac{x}{a} + \frac{y}{b} + \frac{z}{c}} \right)\).

Step by step solution

01

State the barycentric coordinates of a point

As the edges are along three coordinate axes in\({\mathbb{R}^3}\), let the set of vertices be\(\left\{ {\left[ {\begin{array}{*{20}{c}}a\\0\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}0\\b\\0\end{array}} \right],\left[ {\begin{array}{*{20}{c}}0\\0\\c\end{array}} \right],\left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]} \right\}\).

Let a point\(p = \left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right]\).

The point can also be written as shown below:

\(\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{x}{a}\left[ {\begin{array}{*{20}{c}}a\\0\\0\end{array}} \right] + \frac{y}{b}\left[ {\begin{array}{*{20}{c}}0\\b\\0\end{array}} \right] + \frac{z}{c}\left[ {\begin{array}{*{20}{c}}0\\0\\c\end{array}} \right] + \alpha \left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\)

Here, \(\alpha \) is a constant.

02

State the formulas for barycentric coordinates

The sum of weights must be 1. So,

\(\begin{array}{c}\frac{x}{a} + \frac{y}{b} + \frac{z}{c} + \alpha = 1\\\alpha = 1 - \left( {\frac{x}{a} + \frac{y}{b} + \frac{z}{c}} \right)\end{array}\)

It becomes as shown below:

\(\left[ {\begin{array}{*{20}{c}}x\\y\\z\end{array}} \right] = \frac{x}{a}\left[ {\begin{array}{*{20}{c}}a\\0\\0\end{array}} \right] + \frac{y}{b}\left[ {\begin{array}{*{20}{c}}0\\b\\0\end{array}} \right] + \frac{z}{c}\left[ {\begin{array}{*{20}{c}}0\\0\\c\end{array}} \right] + 1 - \left( {\frac{x}{a} + \frac{y}{b} + \frac{z}{c}} \right)\left[ {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right]\)

Therefore, the barycentric coordinates are \(\frac{x}{a},\frac{y}{b},\frac{z}{c},1 - \left( {\frac{x}{a} + \frac{y}{b} + \frac{z}{c}} \right)\).

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Most popular questions from this chapter

Let\({v_1} = \left[ {\begin{array}{*{20}{c}}1\\3\\{ - 6}\end{array}} \right]\),\({v_{\bf{2}}} = \left[ {\begin{array}{*{20}{c}}{\bf{7}}\\3\\{ - {\bf{5}}}\end{array}} \right]\), \({v_{\bf{3}}} = \left[ {\begin{array}{*{20}{c}}{\bf{3}}\\{\bf{9}}\\{ - {\bf{2}}}\end{array}} \right]\), \({\bf{a}} = \left[ {\begin{array}{*{20}{c}}{\bf{0}}\\{\bf{0}}\\{\bf{9}}\end{array}} \right]\), \({\bf{b}} = \left[ {\begin{array}{*{20}{c}}{1.4}\\{{\bf{1}}.{\bf{5}}}\\{ - {\bf{3}}.{\bf{1}}}\end{array}} \right]\), and \({\bf{x}}\left( t \right) = {\bf{a}} + t{\bf{b}}\)for \(t \ge {\bf{0}}\).Find the point where the ray\({\bf{x}}\left( t \right)\)intersects the plane that contains the triangle with vertices\({v_1}\),\({v_{\bf{2}}}\), and\({v_{\bf{3}}}\). Is this point inside the triangle?

Question: In Exercises 15-20, write a formula for a linear functional f and specify a number d, so that \(\left( {f:d} \right)\) the hyperplane H described in the exercise.

Let H be the plane in \({\mathbb{R}^{\bf{3}}}\) spanned by the rows of \(B = \left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{3}}&{\bf{5}}\\{\bf{0}}&{\bf{2}}&{\bf{4}}\end{array}} \right)\). That is, \(H = {\bf{Row}}\,B\). (Hint: How is H is related to Nul B?see section 6.1.)

Question: 24. Repeat Exercise 23 for \({v_1} = \left( \begin{array}{l}1\\2\end{array} \right)\), \({v_2} = \left( \begin{array}{l}5\\1\end{array} \right)\), \({v_3} = \left( \begin{array}{l}4\\4\end{array} \right)\) and \(p = \left( \begin{array}{l}2\\3\end{array} \right)\).

Question: In Exercise 3, determine whether each set is open or closed or neither open nor closed.

3. a. \(\left\{ {\left( {x,y} \right):y > {\bf{0}}} \right\}\)

b. \(\left\{ {\left( {x,y} \right):x = {\bf{2}}\,\,\,and\,\,{\bf{1}} \le y \le {\bf{3}}} \right\}\)

c. \(\left\{ {\left( {x,y} \right):x = {\bf{2}}\,\,\,and\,\,{\bf{1}} < y < {\bf{3}}} \right\}\)

d. \(\left\{ {\left( {x,y} \right):xy = {\bf{1}}\,\,\,and\,\,x > {\bf{0}}} \right\}\)

e. \(\left\{ {\left( {x,y} \right):xy \ge {\bf{1}}\,\,\,and\,\,x > {\bf{0}}} \right\}\)

In Exercises 13-15 concern the subdivision of a Bezier curve shown in Figure 7. Let \({\mathop{\rm x}\nolimits} \left( t \right)\) be the Bezier curve, with control points \({{\mathop{\rm p}\nolimits} _0},...,{{\mathop{\rm p}\nolimits} _3}\), and let \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) be the subdividing Bezier curves as in the text, with control points \({{\mathop{\rm q}\nolimits} _0},...,{{\mathop{\rm q}\nolimits} _3}\) and \({{\mathop{\rm r}\nolimits} _0},...,{{\mathop{\rm r}\nolimits} _3}\), respectively.

15. Sometimes only one-half of a Bezier curve needs further subdividing. For example, subdivision of the “left” side is accomplished with parts (a) and (c) of Exercise 13 and equation (8). When both halves of the curve \({\mathop{\rm x}\nolimits} \left( t \right)\) are divided, it is possible to organize calculations efficiently to calculate both left and right control points concurrently, without using equation (8) directly.

a. Show that the tangent vector \(y'\left( 1 \right)\) and \(z'\left( 0 \right)\) are equal.

b. Use part (a) to show that \({{\mathop{\rm q}\nolimits} _3}\) (which equals \({{\mathop{\rm r}\nolimits} _0}\)) is the midpoint of the segment from \({{\mathop{\rm q}\nolimits} _2}\) to \({{\mathop{\rm r}\nolimits} _1}\).

c. Using part (b) and the results of Exercises 13 and 14, write an algorithm that computes the control points for both \({\mathop{\rm y}\nolimits} \left( t \right)\) and \({\mathop{\rm z}\nolimits} \left( t \right)\) in an efficient manner. The only operations needed are sums and division by 2.
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