繞軸旋轉矩陣

Last updated 2 years ago

Was this helpful?

線代 ⟩ 矩陣 ⟩ 旋轉矩陣 ⟩ 繞軸旋轉

若以單位向量 $\mathbf{u} = [u_1 \ u_2 \ u_3]^T$ 為軸，將 $\mathbf{P}$ 點旋轉 $\theta$ 角，得到 $\mathbf{P'}$ 點，則： $(1) \ \ \mathbf{P'} = ({\color{orange} \cos\theta}) \ \mathbf{P} \ + \ ({\color{orange}1-\cos\theta}) \ (\mathbf{P\cdot u}) \mathbf{u} + \ ({\color{orange}\sin\theta}) \ (\mathbf{u\times P})$ $(2) \ \ \mathbf{P'} = [ \ ({\color{orange}\cos\theta}) \ \mathbf{I} \ + \ ({\color{orange}1-\cos\theta}) \ \mathbf{u} \mathbf{u}^T + \ ({\color{orange}\sin\theta}) \ \mathbf{M} \ ] \ \mathbf{P}$
其中： $\mathbf{I} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ (單位矩陣) $\mathbf{uu^T} = \begin{bmatrix} u_{1}^2 & u_{1} u_2 & u_{1} u_3 \\ u_{2} u_1 & u_{2}^2 & u_{2} u_3 \\ u_{3} u_1 & u_{3} u_2 & u_{3}^2 \end{bmatrix}$ ( 注意：① 行向量都平行 $\mathbf{u}$ ② $\mathbf{M^T} = \mathbf{M}$ ) $\mathbf{M} = \begin{bmatrix} 0 & -u_3 & u_2 \\ u_3 & 0 & -u_1 \\ -u_2 & u_1 & 0 \end{bmatrix}$ ( 注意：① 行向量都垂直 $\mathbf{u}$ ② $\mathbf{M^T} = -\mathbf{M}$ )

Last updated 2 years ago

Was this helpful?

線代 ⟩ 矩陣 ⟩ 旋轉矩陣 ⟩ 繞軸旋轉

若以單位向量 $\mathbf{u} = [u_1 \ u_2 \ u_3]^T$ 為軸，將 $\mathbf{P}$ 點旋轉 $\theta$ 角，得到 $\mathbf{P'}$ 點，則： $(1) \ \ \mathbf{P'} = ({\color{orange} \cos\theta}) \ \mathbf{P} \ + \ ({\color{orange}1-\cos\theta}) \ (\mathbf{P\cdot u}) \mathbf{u} + \ ({\color{orange}\sin\theta}) \ (\mathbf{u\times P})$ $(2) \ \ \mathbf{P'} = [ \ ({\color{orange}\cos\theta}) \ \mathbf{I} \ + \ ({\color{orange}1-\cos\theta}) \ \mathbf{u} \mathbf{u}^T + \ ({\color{orange}\sin\theta}) \ \mathbf{M} \ ] \ \mathbf{P}$
其中： $\mathbf{I} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ (單位矩陣) $\mathbf{uu^T} = \begin{bmatrix} u_{1}^2 & u_{1} u_2 & u_{1} u_3 \\ u_{2} u_1 & u_{2}^2 & u_{2} u_3 \\ u_{3} u_1 & u_{3} u_2 & u_{3}^2 \end{bmatrix}$ ( 注意：① 行向量都平行 $\mathbf{u}$ ② $\mathbf{M^T} = \mathbf{M}$ ) $\mathbf{M} = \begin{bmatrix} 0 & -u_3 & u_2 \\ u_3 & 0 & -u_1 \\ -u_2 & u_1 & 0 \end{bmatrix}$ ( 注意：① 行向量都垂直 $\mathbf{u}$ ② $\mathbf{M^T} = -\mathbf{M}$ )