91Ó°ÊÓ

In the direction of the negative\(y\)-axis, vector\({{\bf{e}}_1}\)rotates at\(30^\circ \). The point where the vector ends is shown below:

\(\begin{array}{c}\left( {\cos \varphi ,\sin \varphi ,0} \right) \equiv \left( {\cos \left( { - 30^\circ } \right),\sin \left( { - 30^\circ } \right)} \right)\\ \equiv \left( {\sqrt 3 /2, - 1/2,0} \right)\end{array}\)

And in the direction of the positive\(x\)-axis, vector\({{\bf{e}}_2}\)rotates at\(60^\circ \)\(\left( {90^\circ - 30^\circ = 60^\circ } \right)\). The point where the vector ends is shown below:

\(\begin{array}{c}\left( {\cos \varphi ,\sin \varphi ,0} \right) \equiv \left( {\cos 60^\circ ,\sin 60^\circ ,0} \right)\\ \equiv \left( {1/2,\sqrt 3 /2,0} \right)\end{array}\)

Vector\({{\bf{e}}_3}\)on the\(z\)-axis does not move by the rotation.

Construct the\(3 \times 3\)matrix for the rotation, as shown below:

\(A = \left[ {\begin{array}{*{20}{c}}{\sqrt 3 /2}&{1/2}&0\\{ - 1/2}&{\sqrt 3 /2}&0\\0&0&1\end{array}} \right]\)

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

Obtain the\(4 \times 4\)matrix that rotates the points in\({\mathbb{R}^3}\)as shown below:

\(\left[ {\begin{array}{*{20}{c}}A&0\\{{0^T}}&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\sqrt 3 /2}&{1/2}&0&0\\{ - 1/2}&{\sqrt 3 /2}&0&0\\0&0&1&0\\0&0&0&1\end{array}} \right]\)

Thus, the \(4 \times 4\) matrix that rotates the points in \({\mathbb{R}^3}\) is \(\left[ {\begin{array}{*{20}{c}}{\sqrt 3 /2}&{1/2}&0&0\\{ - 1/2}&{\sqrt 3 /2}&0&0\\0&0&1&0\\0&0&0&1\end{array}} \right]\).

The translation by vector\({\bf{p}} = \left( {5, - 2,1} \right)\).

Compare\(\left[ {\begin{array}{*{20}{c}}{\sqrt 3 /2}&{1/2}&0&5\\{ - 1/2}&{\sqrt 3 /2}&0&{ - 2}\\0&0&1&1\\0&0&0&1\end{array}} \right]\)with the matrix \(\left[ {\begin{array}{*{20}{c}}A&{\bf{p}}\\{{0^T}}&1\end{array}} \right]\).

So,\(A = \left[ {\begin{array}{*{20}{c}}{\sqrt 3 /2}&{1/2}&0\\{ - 1/2}&{\sqrt 3 /2}&0\\0&0&1\end{array}} \right]\)and\({\bf{p}} = \left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\1\end{array}} \right]\).

Thetranslation matrix is represented as

\(\left[ {\begin{array}{*{20}{c}}1&0&0&5\\0&1&0&{ - 2}\\0&0&1&1\\0&0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{p}}\\{{0^T}}&1\end{array}} \right]\).

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

The completetransformation is represented inhomogeneous coordinates in the following matrix form:

\(\left[ {\begin{array}{*{20}{c}}1&0&0&5\\0&1&0&{ - 2}\\0&0&1&1\\0&0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\sqrt 3 /2}&{1/2}&0&0\\{ - 1/2}&{\sqrt 3 /2}&0&0\\0&0&1&0\\0&0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\sqrt 3 /2}&{1/2}&0&5\\{ - 1/2}&{\sqrt 3 /2}&0&{ - 2}\\0&0&1&1\\0&0&0&1\end{array}} \right]\)

Therefore, the required matrix is \(\left[ {\begin{array}{*{20}{c}}{\sqrt 3 /2}&{1/2}&0&5\\{ - 1/2}&{\sqrt 3 /2}&0&{ - 2}\\0&0&1&1\\0&0&0&1\end{array}} \right]\).