Learning Materials
Features
Discover
Chapter 8: Q26E (page 437)
Find an example in \({\mathbb{R}^2}\) to show that equality need not hold in the statement of Exercise 25.
It is proved that equality does not hold.
Over 30 million students worldwide already upgrade their learning with Vaia!
All the tools & learning materials you need for study success - in one app.
Get started for free
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\({\mathbb{R}^3}\).
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.
14.a. Justify each equal sign:
\(3\left( {{{\mathop{\rm r}\nolimits} _3} - {{\mathop{\rm r}\nolimits} _2}} \right) = z'\left( 1 \right) = .5x'\left( 1 \right) = \frac{3}{2}\left( {{{\mathop{\rm p}\nolimits} _3} - {{\mathop{\rm p}\nolimits} _2}} \right)\)
b. Show that \({{\mathop{\rm r}\nolimits} _2}\) is the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _2}\) to \({{\mathop{\rm p}\nolimits} _3}\).
c. Justify each equal sign: \(3\left( {{{\mathop{\rm r}\nolimits} _1} - {{\mathop{\rm r}\nolimits} _0}} \right) = z'\left( 0 \right) = .5x'\left( {.5} \right)\).
d. Use part (c) to show that \(8{{\mathop{\rm r}\nolimits} _1} = - {{\mathop{\rm p}\nolimits} _0} - {{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2} + {{\mathop{\rm p}\nolimits} _3} + 8{{\mathop{\rm r}\nolimits} _0}\).
e. Use part (d) equation (8), and part (a) to show that \({{\mathop{\rm r}\nolimits} _1}\) is the midpoint of the segment from \({{\mathop{\rm r}\nolimits} _2}\) to the midpoint of the segment from \({{\mathop{\rm p}\nolimits} _1}\) to \({{\mathop{\rm p}\nolimits} _2}\). That is, \({{\mathop{\rm r}\nolimits} _1} = \frac{1}{2}\left( {{{\mathop{\rm r}\nolimits} _2} + \frac{1}{2}\left( {{{\mathop{\rm p}\nolimits} _1} + {{\mathop{\rm p}\nolimits} _2}} \right)} \right)\).
In Exercises 7 and 8, find the barycentric coordinates of p with respect to the affinely independent set of points that precedes it.
8. \(\left( {\begin{array}{{}}0\\1\\{ - 2}\\1\end{array}} \right),\left( {\begin{array}{{}}1\\1\\0\\2\end{array}} \right),\left( {\begin{array}{{}}1\\4\\{ - 6}\\5\end{array}} \right)\), \({\mathop{\rm p}\nolimits} = \left( {\begin{array}{{}}{ - 1}\\1\\{ - 4}\\0\end{array}} \right)\)
In Exercises 5 and 6, let \({{\bf{b}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right)\), \({{\bf{b}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{1}}\\{\bf{0}}\\{ - {\bf{2}}}\end{aligned}} \right)\), and \({{\bf{b}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}\\{ - {\bf{5}}}\\{\bf{1}}\end{aligned}} \right)\) and \(S = \left\{ {{{\bf{b}}_{\bf{1}}},\,{{\bf{b}}_{\bf{2}}},\,{{\bf{b}}_{\bf{3}}}} \right\}\). Note that S is an orthogonal basis of \({\mathbb{R}^{\bf{3}}}\). Write each of the given points as an affine combination of the points in the set S, if possible. (Hint: Use Theorem 5 in section 6.2 instead of row reduction to find the weights.)
a. \({{\bf{p}}_{\bf{1}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}\\{\bf{8}}\\{\bf{4}}\end{aligned}} \right)\)
b. \({{\bf{p}}_{\bf{2}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{6}}\\{ - {\bf{3}}}\\{\bf{3}}\end{aligned}} \right)\)
c. \({{\bf{p}}_{\bf{3}}} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}\\{ - {\bf{1}}}\\{ - {\bf{5}}}\end{aligned}} \right)\)
Question: 13. Suppose \(\left\{ {{{\rm{v}}_{\rm{1}}}{\rm{,}}{{\rm{v}}_{\rm{2}}}{\rm{,}}{{\rm{v}}_{\rm{3}}}} \right\}\) is a basis for \({\mathbb{R}^3}\). Show that Span \(\left\{ {{{\rm{v}}_{\rm{2}}} - {{\rm{v}}_{\rm{1}}},{{\rm{v}}_{\rm{3}}} - {{\rm{v}}_{\rm{1}}}} \right\}\) is a plane in \({\mathbb{R}^3}\). (Hint: What can you say about \({\rm{u}}\) and \({\rm{v}}\)when Span \(\left\{ {{\rm{u,v}}} \right\}\) is a plane?)
What do you think about this solution?
We value your feedback to improve our textbook solutions.
Explanations 
Textbooks 
Exams 
GCSE 
IGCSE 
AS 
A Level 
International A Level 
University Admissions Tests 
Flashcards 
Vaia AI 
Notes 
Study Plans 
Study Sets 
Find a job 
Find a degree 
Student Deals 
Magazine 
Mobile App