91Ó°ÊÓ

From the given figure, vector\({{\bf{e}}_1}\)on the\(x\)-axis is not moving by rotation.

In the direction of the positive\(z\)-axis, vector\({{\bf{e}}_2}\)rotates at\(60^\circ \). The point where the vector ends is shown below:

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

Also, in the direction of the negative\(y\)-axis, vector\({{\bf{e}}_3}\)rotates at\(60^\circ \). The point where the vector ends at\(\varphi = 150^\circ \)is shown below:

\(\begin{array}{c}\left( {0,\cos \varphi ,\sin \varphi } \right) \equiv \left( {0,\cos 150^\circ ,\sin 150^\circ } \right)\\ \equiv \left( {0, - \sqrt 3 /2,1/2} \right)\end{array}\)

Now, construct a\(3 \times 3\)matrix for the rotation, as shown below:

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

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:

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

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