1. **Problem statement:** (a) Given the transformation matrix from frame A to frame B: $$T_A^B = \begin{bmatrix} 1 & 0 & 0 & 2 \\ 0 & \cos(\theta) & -\sin(\theta) & 3 \\ 0 & \sin(\theta) & \cos(\theta) & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ Find the inverse transformation matrix $T_B^A$. (b) Given $\theta = 45^\circ$ and a point in frame B: $P^B = \begin{bmatrix}6 \\ 4 \\ 5 \\ \end{bmatrix}$, compute the coordinates of the point in frame A, $P^A$. 2. **Formula and rules:** The inverse of a homogeneous transformation matrix $T = \begin{bmatrix} R & t \\ 0 & 1 \end{bmatrix}$, where $R$ is a rotation matrix and $t$ is a translation vector, is given by: $$T^{-1} = \begin{bmatrix} R^T & -R^T t \\ 0 & 1 \end{bmatrix}$$ where $R^T$ is the transpose of $R$. 3. **Step-by-step solution for (a):** - Extract rotation matrix $R$ and translation vector $t$ from $T_A^B$: $$R = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos(\theta) & -\sin(\theta) \\ 0 & \sin(\theta) & \cos(\theta) \end{bmatrix}, \quad t = \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}$$ - Compute $R^T$ (transpose of $R$): $$R^T = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos(\theta) & \sin(\theta) \\ 0 & -\sin(\theta) & \cos(\theta) \end{bmatrix}$$ - Compute $-R^T t$: $$-R^T t = - \begin{bmatrix} 1 & 0 & 0 \\ 0 & \cos(\theta) & \sin(\theta) \\ 0 & -\sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} = - \begin{bmatrix} 2 \\ 3\cos(\theta) + 1\sin(\theta) \\ -3\sin(\theta) + 1\cos(\theta) \end{bmatrix} = \begin{bmatrix} -2 \\ -3\cos(\theta) - \sin(\theta) \\ 3\sin(\theta) - \cos(\theta) \end{bmatrix}$$ - Assemble $T_B^A$: $$T_B^A = \begin{bmatrix} R^T & -R^T t \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & -2 \\ 0 & \cos(\theta) & \sin(\theta) & -3\cos(\theta) - \sin(\theta) \\ 0 & -\sin(\theta) & \cos(\theta) & 3\sin(\theta) - \cos(\theta) \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ 4. **Step-by-step solution for (b):** - Substitute $\theta = 45^\circ$ (convert to radians if needed, but cosine and sine of 45° are well-known): $$\cos(45^\circ) = \frac{\sqrt{2}}{2}, \quad \sin(45^\circ) = \frac{\sqrt{2}}{2}$$ - Substitute into $T_B^A$ translation part: $$-3\cos(45^\circ) - \sin(45^\circ) = -3 \times \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -2\sqrt{2}$$ $$3\sin(45^\circ) - \cos(45^\circ) = 3 \times \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = \sqrt{2}$$ - So the matrix becomes: $$T_B^A = \begin{bmatrix} 1 & 0 & 0 & -2 \\ 0 & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & -2\sqrt{2} \\ 0 & -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & \sqrt{2} \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ - To find $P^A$, augment $P^B$ with 1 for homogeneous coordinates: $$P^B_h = \begin{bmatrix} 6 \\ 4 \\ 5 \\ 1 \end{bmatrix}$$ - Compute: $$P^A = T_B^A P^B_h = \begin{bmatrix} 1 & 0 & 0 & -2 \\ 0 & \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & -2\sqrt{2} \\ 0 & -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & \sqrt{2} \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 6 \\ 4 \\ 5 \\ 1 \end{bmatrix} = \begin{bmatrix} 6 - 2 \\ \frac{\sqrt{2}}{2} \times 4 + \frac{\sqrt{2}}{2} \times 5 - 2\sqrt{2} \\ -\frac{\sqrt{2}}{2} \times 4 + \frac{\sqrt{2}}{2} \times 5 + \sqrt{2} \\ 1 \end{bmatrix}$$ - Simplify each component: $$x = 4$$ $$y = \frac{\sqrt{2}}{2} (4 + 5) - 2\sqrt{2} = \frac{9\sqrt{2}}{2} - 2\sqrt{2} = \frac{9\sqrt{2} - 4\sqrt{2}}{2} = \frac{5\sqrt{2}}{2}$$ $$z = \frac{\sqrt{2}}{2} (-4 + 5) + \sqrt{2} = \frac{\sqrt{2}}{2} (1) + \sqrt{2} = \frac{\sqrt{2}}{2} + \sqrt{2} = \frac{3\sqrt{2}}{2}$$ 5. **Final answers:** - (a) $$T_B^A = \begin{bmatrix} 1 & 0 & 0 & -2 \\ 0 & \cos(\theta) & \sin(\theta) & -3\cos(\theta) - \sin(\theta) \\ 0 & -\sin(\theta) & \cos(\theta) & 3\sin(\theta) - \cos(\theta) \\ 0 & 0 & 0 & 1 \end{bmatrix}$$ - (b) $$P^A = \begin{bmatrix} 4 \\ \frac{5\sqrt{2}}{2} \\ \frac{3\sqrt{2}}{2} \end{bmatrix}$$