QUESION2 Global Rotation Choose one $\cdot 5$ points Write the global rotation matrix of $60^{\circ}$ about the world $z$-aris, $G_{R_{3}}(60)$ \[ \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{1}{2} & \frac{\sqrt{8}}{2} \\ 0 & -\frac{\sqrt{3}}{2} & \frac{1}{2} \end{array}\right) \] \[ \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{\sqrt{B}}{2} & -\frac{1}{2} \\ 0 & \frac{1}{2} & \frac{\sqrt{8}}{2} \end{array}\right) \] \[ \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \frac{1}{2} & -\frac{\sqrt{B}}{2} \\ 0 & \frac{\sqrt{B}}{2} & \frac{1}{2} \end{array}\right) \]
Solution
Step 1 ：The problem is asking for the global rotation matrix of 60 degrees about the world z-axis. This is a standard rotation matrix in 3D space.
Step 2 ：The rotation matrix for a rotation of θ degrees about the z-axis is given by: \[\begin{bmatrix} cos(\theta) & -sin(\theta) & 0 \\ sin(\theta) & cos(\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix}\]
Step 3 ：We need to substitute θ = 60 degrees into this matrix. However, we need to convert the degrees to radians first, because the trigonometric functions use radians. The conversion is given by radians = degrees * π / 180.
Step 4 ：Substituting θ = 60 degrees into the rotation matrix, we get: \[\begin{bmatrix} 0.5 & -0.8660254 & 0 \\ 0.8660254 & 0.5 & 0 \\ 0 & 0 & 1 \end{bmatrix}\]
Step 5 ：Final Answer: The global rotation matrix of $60^{\circ}$ about the world $z$-axis, $G_{R_{3}}(60)$ is \[\boxed{\begin{bmatrix} 0.5 & -0.8660254 & 0 \\ 0.8660254 & 0.5 & 0 \\ 0 & 0 & 1 \end{bmatrix}}\]
