Find a transformation from u,v space to x,y,z space that takes the triangle U = [(0,0),(1,0),(0,1)] to the triangle T=[(1,0),-2),(-1,2,0),(1,1,2)]

Question
Upper Level Math
asked 2021-01-02
Find a transformation from u,v space to x,y,z space that takes the triangle \(\displaystyle{U}={\left[\begin{array}{cc} {0}&{0}\\{1}&{0}\\{0}&{1}\end{array}\right]}\) to the triangle
\(\displaystyle{T}={\left[{\left({1},{0}\right)},-{2}\right)},{\left(-{1},{2},{0}\right)},{\left({1},{1},{2}\right)}{]}\)

Answers (1)

2021-01-03
Step 1
The triangle \(\displaystyle{U}={\left[\begin{array}{cc} {0}&{0}\\{1}&{0}\\{0}&{1}\end{array}\right]}\) and the triangle
\(\displaystyle{T}={\left[\begin{array}{ccc} {1}&{0}&-{2}\\-{1}&{2}&{0}\\{1}&{1}&{2}\end{array}\right]}\)
Step 2
Consider \(\displaystyle{T}{\left({u},{v}\right)}={\left({a}_{{1}}{u}+{b}_{{1}}{v}+{c}_{{1}},{a}_{{2}}{u}+{b}_{{2}}{v}+{c}_{{2}},{a}_{{3}}{u}+{b}_{{3}}{v}={c}_{{3}}\right)}\) .Obtain the transformation from u, v to x, y, z as
\(\displaystyle{T}{\left({0},{0}\right)}={\left({a}_{{1}}{\left({0}\right)}+{b}_{{1}}{\left({0}\right)}+{c}_{{1}},{a}_{{2}}{\left({0}\right)}+{b}_{{2}}{\left({0}\right)}+{c}_{{2}},{a}_{{3}}{\left({0}\right)}+{b}_{{3}}{\left({0}\right)}+{c}_{{3}}\right)}\)
\(\displaystyle{T}{\left({0},{0}\right)}={\left({c}_{{1}},{c}_{{2}},{c}_{{3}}\right)}\)
\(\displaystyle{T}{\left({0},{0}\right)}={\left({1},{0},-{2}\right)}\)
\(\displaystyle\Rightarrow{c}_{{1}}={1},{c}_{{2}}={0},{c}_{{3}}=-{2}\)
Step 3
Further evaluate as,
\(\displaystyle{T}{\left({1},{0}\right)}={\left({a}_{{1}}{\left({1}\right)}\right)}+{b}_{{1}}{\left({0}\right)}+{c}_{{1}},{a}_{{2}}{\left({1}\right)}+{b}_{{2}}{\left({0}\right)}+{c}_{{2}},{a}_{{3}}{\left({1}\right)}+{b}_{{2}}{\left({0}\right)}+{c}_{{3}}{)}\)
\(\displaystyle{T}{\left({0},{0}\right)}={\left({a}_{{1}}+{c}_{{1}},{a}_{{2}}+{c}_{{2}},{a}_{{3}}+{c}_{{3}}\right)}\)
\(\displaystyle{T}{\left({0},{0}\right)}={\left(-{1},{2},{0}\right)}\)
\(\displaystyle\Rightarrow{a}_{{1}}={c}_{{1}}=-{1},{a}_{{2}}+{c}_{{2}}={2},{a}_{{3}}+{c}_{{3}}={0}\)
\(\displaystyle\Rightarrow{a}_{{1}}=-{2},{a}_{{2}}={2},{a}_{{3}}={2}\)
Step 4
Also evaluate,
\(\displaystyle{T}{\left({0},{1}\right)}={\left({a}_{{1}}{\left({0}\right)}+{b}_{{1}}{\left({1}\right)}+{c}_{{1}},{a}_{{2}}{\left({0}\right)}+{b}_{{2}}{\left({1}\right)}={c}_{{2}},{a}_{{3}}{\left({0}\right)}+{b}_{{3}}{\left({1}\right)}={c}_{{3}}\right)}\)
\(\displaystyle{T}{\left({0},{0}\right)}={\left({b}_{{1}}+{c}_{{1}},{b}_{{2}}+{c}_{{2}},{b}_{{3}}+{c}_{{3}}\right)}\)
\(\displaystyle{T}{\left({0},{0}\right)}={\left({1},{1},{2}\right)}\)
\(\displaystyle\Rightarrow{b}_{{1}}+{c}_{{1}}={1},{b}_{{2}}+{c}_{{2}}={1},{b}_{{3}}+{c}_{{3}}={2}\)
\(\displaystyle\Rightarrow{b}_{{1}}={0},{b}_{{2}}={1},{b}_{{3}}={4}\)
Step 5
Therefore, \(\displaystyle{T}{\left({u},{v}\right)}={\left(-{2}{u}+{1},{2}{u}+{v},{2}{u}+{4}{v}-{2}\right)}\)
0

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