Question

Show that the transformation in Exercise 7 is equivalent to a rotation about the origin followed by a translation by p. Find p.

Short answer

The transformation is equivalent to a rotation about the origin. And the value of p is \({\bf{p}} = \left[ {\begin{array}{*{20}{c}}{3 + 4\sqrt 3 }\\{4 - 3\sqrt 3 }\end{array}} \right]\).

Step-by-step solution

Obtain the translation and rotation matrices

As p is in\({\mathbb{R}^{\bf{2}}}\), thetranslation matrix to\({\bf{p}} = \left( { - 6, - 8} \right)\)is

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 6}\\0&1&{ - 8}\\0&0&1\end{array}} \right]\).

As p is in\({\mathbb{R}^{\bf{2}}}\), thetranslation matrix to\({\bf{p}} = \left( {6,8} \right)\)is

\(\left[ {\begin{array}{*{20}{c}}1&0&6\\0&1&8\\0&0&1\end{array}} \right]\).

Obtain therotation matrix through\(60^\circ \), as shown below:

\(\begin{array}{c}\left[ {\begin{array}{*{20}{c}}{\cos \varphi }&{ - \sin \varphi }&0\\{\sin \varphi }&{\cos \varphi }&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\cos \left( {60^\circ } \right)}&{ - \sin \left( {60^\circ } \right)}&0\\{\sin \left( {60^\circ } \right)}&{\cos \left( {60^\circ } \right)}&0\\0&0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1/2}&{ - \sqrt 3 /2}&0\\{\sqrt 3 /2}&{1/2}&0\\0&0&1\end{array}} \right]\end{array}\)

Obtain the image of the homogeneous coordinate

Obtain the image of thehomogeneous coordinate rotating about\(60^\circ \)and translated by\({\bf{p}} = \left( { - 6, - 8} \right)\).

\(\left[ {\begin{array}{*{20}{c}}1&0&{ - 6}\\0&1&{ - 8}\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\\1\end{array}} \right]\)

By rotating about the origin, it becomes

\(\left[ {\begin{array}{*{20}{c}}{1/2}&{ - \sqrt 3 /2}&0\\{\sqrt 3 /2}&{1/2}&0\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&{ - 6}\\0&1&{ - 8}\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\\1\end{array}} \right]\).

Translation at\({\bf{p}} = \left( {6,8} \right)\)is

\(\left[ {\begin{array}{*{20}{c}}1&0&6\\0&1&8\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{1/2}&{ - \sqrt 3 /2}&0\\{\sqrt 3 /2}&{1/2}&0\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&{ - 6}\\0&1&{ - 8}\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}x\\y\\1\end{array}} \right]\).

Obtain the product of matrices

Compute the product of\(\left[ {\begin{array}{*{20}{c}}1&0&6\\0&1&8\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{1/2}&{ - \sqrt 3 /2}&0\\{\sqrt 3 /2}&{1/2}&0\\0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&{ - 6}\\0&1&{ - 8}\\0&0&1\end{array}} \right]\), as shown below:

\[\]

Thus, the translation matrix is \(\left[ {\begin{array}{*{20}{c}}{1/2}&{ - \sqrt 3 /2}&{3 + 4\sqrt 3 }\\{\sqrt 3 /2}&{1/2}&{4 - 3\sqrt 3 }\\0&0&1\end{array}} \right]\).

Write the translation matrix

Recall that the transformation onhomogeneous coordinates for graphics has the matrix of the form\(\left[ {\begin{array}{*{20}{c}}A&{\bf{p}}\\{{0^T}}&1\end{array}} \right]\).

Compare\(\left[ {\begin{array}{*{20}{c}}{1/2}&{ - \sqrt 3 /2}&{3 + 4\sqrt 3 }\\{\sqrt 3 /2}&{1/2}&{4 - 3\sqrt 3 }\\0&0&1\end{array}} \right]\)with \(\left[ {\begin{array}{*{20}{c}}A&{\bf{p}}\\{{0^T}}&1\end{array}} \right]\).

This gives\(A = \left[ {\begin{array}{*{20}{c}}{1/2}&{ - \sqrt 3 /2}\\{\sqrt 3 /2}&{1/2}\end{array}} \right]\)and\({\bf{p}} = \left[ {\begin{array}{*{20}{c}}{3 + 4\sqrt 3 }\\{4 - 3\sqrt 3 }\end{array}} \right]\).

Thetranslation matrix is represented as

\(\left[ {\begin{array}{*{20}{c}}1&0&{3 + 4\sqrt 3 }\\0&1&{4 - 3\sqrt 3 }\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{p}}\\{{0^T}}&1\end{array}} \right]\).

The matrix can also be represented as\(\left[ {\begin{array}{*{20}{c}}I&{\bf{p}}\\{{0^T}}&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&0\\{{0^T}}&1\end{array}} \right]\).

Thus, the transformation is equivalent to rotation about the origin.

And \({\bf{p}} = \left[ {\begin{array}{*{20}{c}}{3 + 4\sqrt 3 }\\{4 - 3\sqrt 3 }\end{array}} \right]\).