QUESION3 Global Rotation Choose one $\cdot 5$ points Write the global rotation matrix of $45^{\circ}$ about the worldy-axis, ${ }^{G} R_{y}(45)$ \[ \left(\begin{array}{ccc} \frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2} \\ 0 & 1 & 0 \\ -\frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2} \end{array}\right) \] $\left(\begin{array}{ccc}\frac{\sqrt{2}}{2} & 0 & -\frac{\sqrt{2}}{2} \\ 0 & 1 & 0 \\ \frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2}\end{array}\right)$ \[ \left(\begin{array}{ccc} \frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2} \\ 0 & 1 & 0 \\ \frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2} \end{array}\right) \]

Step 1 ：The problem is asking for the global rotation matrix of 45 degrees about the world y-axis.

Step 2 ：The rotation matrix for a rotation about the y-axis by an angle \(\theta\) is given by: \[\begin{bmatrix} cos(\theta) & 0 & sin(\theta) \\ 0 & 1 & 0 \\ -sin(\theta) & 0 & cos(\theta) \end{bmatrix}\]

Step 3 ：In this case, \(\theta\) is 45 degrees. We need to convert this to radians before calculating the cosine and sine.

Step 4 ：\(\theta = 0.7853981633974483\)

Step 5 ：Substituting \(\theta\) into the rotation matrix, we get: \[\begin{bmatrix} 0.70710678 & 0 & 0.70710678 \\ 0 & 1 & 0 \\ -0.70710678 & 0 & 0.70710678 \end{bmatrix}\]

Step 6 ：Final Answer: The global rotation matrix of $45^\circ$ about the world y-axis, ${ }^{G} R_{y}(45)$ is \[\boxed{\begin{bmatrix} 0.70710678 & 0 & 0.70710678 \\ 0 & 1 & 0 \\ -0.70710678 & 0 & 0.70710678 \end{bmatrix}}\]