(1)二维空间中的旋转矩阵
易得
{ x ′ = ∣ O P ′ ∣ c o s ( θ + β ) = ∣ O P ∣ ( c o s θ ⋅ c o s β − s i n θ ⋅ s i n β ) = x c o s β − y s i n β y ′ = ∣ O P ′ ∣ s i n ( θ + β ) = ∣ O P ∣ ( s i n θ ⋅ c o s β + c o s θ ⋅ s i n β ) = y c o s β + x s i n β \left\{ \begin{matrix} x'=|OP'|cos(\theta+\beta)=|OP|(cos\theta\cdot cos\beta-sin\theta\cdot sin\beta)=xcos\beta-ysin\beta \\ y'=|OP'|sin(\theta+\beta)=|OP|(sin\theta\cdot cos\beta+cos\theta\cdot sin\beta)=ycos\beta+xsin\beta \end{matrix} \right. {x′=∣OP′∣cos(θ+β)=∣OP∣(cosθ⋅cosβ−sinθ⋅sinβ)=xcosβ−ysinβy′=∣OP′∣sin(θ+β)=∣OP∣(sinθ⋅cosβ+cosθ⋅sinβ)=ycosβ+xsinβ
写为矩阵形式即
[ x ′ y ′ ] = [ c o s β − s i n β s i n β c o s β ] [ x y ] \left[\begin{array}{l} x' \\ y' \end{array}\right]= \left[\begin{array}{l} cos\beta & -sin\beta \\ sin\beta & cos\beta \end{array}\right] \left[\begin{array}{l} x \\ y \end{array}\right] [x′y′]=[cosβsinβ−sinβcosβ][xy]
(2)三维空间中的旋转矩阵
以绕 z z z 轴旋转为例:
将向量投影到 X-Y 平面,类似的有
{ x ′ = x c o s β − y s i n β y ′ = y c o s β + x s i n β z ′ = z (3-1-1) \left\{ \begin{matrix} x'=xcos\beta-ysin\beta \\ y'=ycos\beta+xsin\beta \\ z'=z \end{matrix} \right. \tag{3-1-1} ⎩ ⎨ ⎧x′=xcosβ−ysinβy′=ycosβ+xsinβz′=z(3-1-1)
写成矩阵形式
[ x ′ y ′ z ′ ] = [ c o s β − s i n β 0 s i n β c o s β 0 0 0 1 ] [ x y z ] (3-1-2) \left[\begin{array}{l} x' \\ y' \\ z' \end{array}\right]= \left[\begin{array}{l} cos\beta & -sin\beta & 0\\ sin\beta & cos\beta & 0\\ 0 & 0 & 1 \end{array}\right] \left[\begin{array}{l} x \\ y \\ z \end{array}\right] \tag{3-1-2} x′y′z′ = cosβsinβ0−sinβcosβ0001 xyz (3-1-2)
记为
a ′ = R a (3-1-3) \boldsymbol{a^{\prime}}=\boldsymbol{Ra} \tag{3-1-3} a′=Ra(3-1-3)
将式中 β \beta β 改为 − β -\beta −β,来表示向量 OP’ 顺时针旋转 − β -\beta −β 角度后,回到 OP 的过程,此时
[ x y z ] = [ c o s β s i n β 0 − s i n β c o s β 0 0 0 1 ] [ x ′ y ′ z ′ ] (3-1-4) \left[\begin{array}{l} x \\ y \\ z \end{array}\right]= \left[\begin{array}{l} cos\beta & sin\beta & 0\\ -sin\beta & cos\beta & 0\\ 0 & 0 & 1 \end{array}\right] \left[\begin{array}{l} x' \\ y' \\ z' \end{array}\right] \tag{3-1-4} xyz = cosβ−sinβ0sinβcosβ0001 x′y′z′ (3-1-4)
可以看出
a = R T a ′ (3-1-5) \boldsymbol{a}=\boldsymbol{R^Ta'} \tag{3-1-5} a=RTa′(3-1-5)
由于 R R R 是正交矩阵,也即
a = R − 1 a ′ (3-1-6) \boldsymbol{a}=\boldsymbol{R^{-1}a'} \tag{3-1-6} a=R−1a′(3-1-6)
也就是说, R − 1 \boldsymbol{R^{-1}} R−1 表示向量经反方向旋转回到原向量的过程。
当然,也可以单纯从数学角度推导,将式(3-1-3) 两端分别左乘 R − 1 \boldsymbol{R^{-1}} R−1,同样可以得到式 (3-1-5)。
(1)只需证明 R R T = I \boldsymbol{RR^T=I} RRT=I 即证明 R T = R − 1 \boldsymbol{R^T = R^{-1}} RT=R−1 即可。
(2)分块矩阵转置:先对整体进行转置,再对每一个分块进行转置,例如向量 A = [ A 1 , A 2 , A 3 , . . . , A N ] \boldsymbol{A=[A_1, A_2, A_3,...,A_N]} A=[A1,A2,A3,...,AN],则其转置为
A T = [ A 1 T A 2 T A 3 T . . . A N T ] (3-2-1) \boldsymbol{A}^T=\left[\begin{array}{l} \boldsymbol{A_1}^T\\ \boldsymbol{A_2}^T \\ \boldsymbol{A_3}^T \\ ...\\ \boldsymbol{A_N}^T \end{array}\right] \tag{3-2-1} AT= A1TA2TA3T...ANT (3-2-1)
(3) ( A B ) T = B T A T \boldsymbol{(AB)^T=B^TA^T} (AB)T=BTAT
(4)下面证明旋转矩阵 R \boldsymbol{R} R 是正交矩阵:
不妨设变换前后坐标系基底分别为 [ e 1 , e 2 , e 3 ] [\boldsymbol{e_1}, \boldsymbol{e_2}, \boldsymbol{e_3}] [e1,e2,e3] 和 [ e 1 ′ , e 2 ′ , e 3 ′ ] [\boldsymbol{e_1'}, \boldsymbol{e_2'}, \boldsymbol{e_3'}] [e1′,e2′,e3′],由于向量本身并未改变,则有
[ e 1 , e 2 , e 3 ] [ a 1 a 2 a 3 ] = [ e 1 ′ , e 2 ′ , e 3 ′ ] [ a 1 ′ a 2 ′ a 3 ′ ] (3-2-2) \left[\boldsymbol{e}_{1}, \boldsymbol{e}_{2}, \boldsymbol{e}_{3}\right]\left[\begin{array}{l} a_{1} \\ a_{2} \\ a_{3} \end{array}\right]=\left[\boldsymbol{e}_{1}^{\prime}, \boldsymbol{e}_{2}^{\prime}, \boldsymbol{e}_{3}^{\prime}\right]\left[\begin{array}{c} a_{1}^{\prime} \\ a_{2}^{\prime} \\ a_{3}^{\prime} \end{array}\right] \tag{3-2-2} [e1,e2,e3] a1a2a3 =[e1′,e2′,e3′] a1′a2′a3′ (3-2-2)
将上式两端分别左乘 [ e 1 T e 2 T e 3 T ] \left[\begin{array}{l} \boldsymbol{e_{1}^T} \\ \boldsymbol{e}_{2}^T \\ \boldsymbol{e}_{3}^T \end{array}\right] e1Te2Te3T 得到,
[ a 1 a 2 a 3 ] = [ e 1 T e 1 ′ e 1 T e 2 ′ e 1 T e 3 ′ e 2 T e 1 ′ e 2 T e 2 ′ e 2 T e 3 ′ e 3 T e 1 ′ e 3 T e 2 ′ e 3 T e 3 ′ ] [ a 1 ′ a 2 ′ a 3 ′ ] (3-2-3) \left[\begin{array}{l} a_{1} \\ a_{2} \\ a_{3} \end{array}\right]=\left[\begin{array}{lll} \boldsymbol{e}_{1}^T \boldsymbol{e}_{1}^{\prime} & \boldsymbol{e}_{1}^T \boldsymbol{e}_{2}^{\prime} & \boldsymbol{e}_{1}^T \boldsymbol{e}_{3}^{\prime} \\ \boldsymbol{e}_{2}^T \boldsymbol{e}_{1}^{\prime} & \boldsymbol{e}_{2}^T \boldsymbol{e}_{2}^{\prime} & \boldsymbol{e}_{2}^T \boldsymbol{e}_{3}^{\prime} \\ \boldsymbol{e}_{3}^T \boldsymbol{e}_{1}^{\prime} & \boldsymbol{e}_{3}^T \boldsymbol{e}_{2}^{\prime} & \boldsymbol{e}_{3}^T \boldsymbol{e}_{3}^{\prime} \end{array}\right]\left[\begin{array}{c} a_{1}^{\prime} \\ a_{2}^{\prime} \\ a_{3}^{\prime} \end{array}\right] \tag{3-2-3} a1a2a3 = e1Te1′e2Te1′e3Te1′e1Te2′e2Te2′e3Te2′e1Te3′e2Te3′e3Te3′ a1′a2′a3′ (3-2-3)
则
R = [ e 1 T e 1 ′ e 1 T e 2 ′ e 1 T e 3 ′ e 2 T e 1 ′ e 2 T e 2 ′ e 2 T e 3 ′ e 3 T e 1 ′ e 3 T e 2 ′ e 3 T e 3 ′ ] (3-2-4) \boldsymbol{R}=\left[\begin{array}{lll} \boldsymbol{e}_{1}^T \boldsymbol{e}_{1}^{\prime} & \boldsymbol{e}_{1}^T \boldsymbol{e}_{2}^{\prime} & \boldsymbol{e}_{1}^T \boldsymbol{e}_{3}^{\prime} \\ \boldsymbol{e}_{2}^T \boldsymbol{e}_{1}^{\prime} & \boldsymbol{e}_{2}^T \boldsymbol{e}_{2}^{\prime} & \boldsymbol{e}_{2}^T \boldsymbol{e}_{3}^{\prime} \\ \boldsymbol{e}_{3}^T \boldsymbol{e}_{1}^{\prime} & \boldsymbol{e}_{3}^T \boldsymbol{e}_{2}^{\prime} & \boldsymbol{e}_{3}^T \boldsymbol{e}_{3}^{\prime} \end{array}\right] \tag{3-2-4} R= e1Te1′e2Te1′e3Te1′e1Te2′e2Te2′e3Te2′e1Te3′e2Te3′e3Te3′ (3-2-4)
这是向量 a ′ \boldsymbol{a'} a′ 经旋转得到 a \boldsymbol{a} a 的旋转矩阵。
再将式 (3-8) 两端分别左乘 [ e 1 ′ T e 2 ′ T e 3 ′ T ] \left[\begin{array}{l} \boldsymbol{e_{1}'^T} \\ \boldsymbol{e}_{2}'^T \\ \boldsymbol{e}_{3}'^T \end{array}\right] e1′Te2′Te3′T 得到,
[ e 1 ′ T e 1 e 1 ′ T e 2 e 1 ′ T e 3 e 2 ′ T e 1 e 2 ′ T e 2 e 2 ′ T e 3 e 3 ′ T e 1 e 3 ′ T e 2 e 3 ′ T e 3 ] [ a 1 a 2 a 3 ] = [ a 1 ′ a 2 ′ a 3 ′ ] (3-2-5) \left[\begin{array}{lll} \boldsymbol{e}_{1}'^T \boldsymbol{e}_{1} & \boldsymbol{e}_{1}'^T \boldsymbol{e}_{2} & \boldsymbol{e}_{1}'^T \boldsymbol{e}_{3}\\ \boldsymbol{e}_{2}'^T \boldsymbol{e}_{1} & \boldsymbol{e}_{2}'^T \boldsymbol{e}_{2}& \boldsymbol{e}_{2}'^T \boldsymbol{e}_{3} \\ \boldsymbol{e}_{3}'^T \boldsymbol{e}_{1} & \boldsymbol{e}_{3}'^T \boldsymbol{e}_{2} & \boldsymbol{e}_{3}'^T \boldsymbol{e}_{3} \end{array}\right]\left[\begin{array}{c} a_{1} \\ a_{2}\\ a_{3} \end{array}\right]= \left[\begin{array}{l} a_{1}' \\ a_{2}' \\ a_{3}' \end{array}\right] \tag{3-2-5} e1′Te1e2′Te1e3′Te1e1′Te2e2′Te2e3′Te2e1′Te3e2′Te3e3′Te3 a1a2a3 = a1′a2′a3′ (3-2-5)
显然,这是向量 a \boldsymbol{a} a 经过反方向旋转得到 a ′ \boldsymbol{a'} a′ 的过程,那么有
R − 1 = [ e 1 ′ T e 1 e 1 ′ T e 2 e 1 ′ T e 3 e 2 ′ T e 1 e 2 ′ T e 2 e 2 ′ T e 3 e 3 ′ T e 1 e 3 ′ T e 2 e 3 ′ T e 3 ] (3-2-6) \boldsymbol{R^{-1}}=\left[\begin{array}{lll} \boldsymbol{e}_{1}'^T \boldsymbol{e}_{1} & \boldsymbol{e}_{1}'^T \boldsymbol{e}_{2} & \boldsymbol{e}_{1}'^T \boldsymbol{e}_{3}\\ \boldsymbol{e}_{2}'^T \boldsymbol{e}_{1} & \boldsymbol{e}_{2}'^T \boldsymbol{e}_{2}& \boldsymbol{e}_{2}'^T \boldsymbol{e}_{3} \\ \boldsymbol{e}_{3}'^T \boldsymbol{e}_{1} & \boldsymbol{e}_{3}'^T \boldsymbol{e}_{2} & \boldsymbol{e}_{3}'^T \boldsymbol{e}_{3} \end{array}\right] \tag{3-2-6} R−1= e1′Te1e2′Te1e3′Te1e1′Te2e2′Te2e3′Te2e1′Te3e2′Te3e3′Te3 (3-2-6)
根据分块矩阵转置规则,易得
R T = R − 1 (3-2-7) \boldsymbol{R^T = R^{-1}} \tag{3-2-7} RT=R−1(3-2-7)
证毕。
(1)初等行变换法求矩阵的逆:将矩阵 A \boldsymbol{A} A 写成增广矩阵的形式即 ( A ∣ I ) \boldsymbol{(A | I)} (A∣I) ,经初等行变换,变为 ( I ∣ A − 1 ) \boldsymbol{(I | A^{-1})} (I∣A−1),则求出矩阵 A \boldsymbol{A} A 的逆。
(2)已知三维空间变换矩阵为:
T = [ R 3 × 3 t 3 × 1 0 T 1 ] (3-3-1) \boldsymbol{T}=\left[\begin{array}{ll} \boldsymbol{R}_{3\times3} & \boldsymbol{t}_{3\times1} \\ \mathbf{0}^{\mathrm{T}} & 1 \end{array}\right] \tag{3-3-1} T=[R3×30Tt3×11](3-3-1)
写成增广矩阵
[ R 3 × 3 t 3 × 1 E 3 × 3 0 0 T 1 0 1 ] (3-3-2) \left[\begin{array}{cc|cc} \boldsymbol{R}_{3\times3} & \boldsymbol{t}_{3\times1} & \boldsymbol{E}_{3\times3} & 0 \\ \mathbf{0}^{\mathrm{T}} & 1 & 0 & 1\\ \end{array}\right] \tag{3-3-2} [R3×30Tt3×11E3×3001](3-3-2)
经过初等行变换,将前两列变为单位矩阵时,后两列即为 T − 1 \boldsymbol{T^{-1}} T−1。
首先,将第一行左乘 R 3 × 3 − 1 \boldsymbol{R}_{3\times3}^{-1} R3×3−1,上式变为
[ E 3 × 3 R 3 × 3 − 1 t 3 × 1 R 3 × 3 − 1 0 3 × 1 0 T 1 0 1 ] (3-3-3) \left[\begin{array}{cc|cc} \boldsymbol{E}_{3\times3} & \boldsymbol{R}_{3\times3}^{-1}\boldsymbol{t}_{3\times1} & \boldsymbol{R}_{3\times3}^{-1} & \boldsymbol{0}_{3\times1} \\ \mathbf{0}^{\mathrm{T}} & 1 & 0 & 1\\ \end{array}\right] \tag{3-3-3} [E3×30TR3×3−1t3×11R3×3−1003×11](3-3-3)
再将第二行乘 − R 3 × 3 − 1 t 3 × 1 -\boldsymbol{R}_{3\times3}^{-1}\boldsymbol{t}_{3\times1} −R3×3−1t3×1 ,加到第一行上,得
[ E 3 × 3 0 3 × 1 R 3 × 3 − 1 − R 3 × 3 − 1 t 3 × 1 0 T 1 0 1 ] (3-3-4) \left[\begin{array}{cc|cc} \boldsymbol{E}_{3\times3} & \boldsymbol{0}_{3\times1} & \boldsymbol{R}_{3\times3}^{-1} & -\boldsymbol{R}_{3\times3}^{-1}\boldsymbol{t}_{3\times1}\\ \mathbf{0}^{\mathrm{T}} & 1 & 0 & 1\\ \end{array}\right] \tag{3-3-4} [E3×30T03×11R3×3−10−R3×3−1t3×11](3-3-4)
又因为 R \boldsymbol{R} R 为正交矩阵,有 R − 1 = R T \boldsymbol{R^{-1}=R^T} R−1=RT。
则变换矩阵的逆矩阵为
T − 1 = [ R T − R T t 0 T 1 ] (3-3-5) \boldsymbol{T}^{-1}=\left[\begin{array}{cc} \boldsymbol{R}^{\mathrm{T}} & -\boldsymbol{R}^{\mathrm{T}} \boldsymbol{t} \\ \mathbf{0}^{\mathrm{T}} & 1 \end{array}\right] \tag{3-3-5} T−1=[RT0T−RTt1](3-3-5)
(1)注意绕轴旋转的方式,(类似直线绕轴旋转形成圆锥的过程),也就是说,该直线是母线,所绕的轴是中心线。
(2)向量点乘有交换律 a ⃗ ⋅ b ⃗ = b ⃗ ⋅ a ⃗ \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a} a⋅b=b⋅a ;没有结合律,即 ( a ⃗ ⋅ b ⃗ ) ⋅ c ⃗ ≠ a ⃗ ⋅ ( b ⃗ ⋅ c ⃗ ) (\vec{a} \cdot \vec{b})\cdot\vec{c} \not=\vec{a} \cdot (\vec{b} \cdot \vec{c}) (a⋅b)⋅c=a⋅(b⋅c)。矩阵乘法有结合律即 ( A B ) C = A ( B C ) \boldsymbol{(AB)C=A(BC)} (AB)C=A(BC)。
(3)向量投影定理
如图,有
∣ a ⃗ ∣ c o s θ = a ⃗ ⋅ b ⃗ ∣ b ⃗ ∣ = a ⃗ ⋅ b 0 ⃗ (3-4-1) |\vec{a}|cos\theta=\frac {\vec{a}\cdot\vec{b}}{|\vec{b}|}=\vec{a}\cdot\vec{b_0} \tag{3-4-1} ∣a∣cosθ=∣b∣a⋅b=a⋅b0(3-4-1)
其中 b 0 ⃗ \vec{b_0} b0为与向量 b ⃗ \vec{b} b 同向的单位向量。
式(3-19)表示向量 a ⃗ \vec{a} a 在向量 b ⃗ \vec{b} b 上的投影的模长为 a ⃗ ⋅ b 0 ⃗ \vec{a}\cdot\vec{b_0} a⋅b0,那么,其在 b ⃗ \vec{b} b 上的投影向量为 ( a ⃗ ⋅ b 0 ⃗ ) ⋅ b 0 ⃗ (\vec{a}\cdot\vec{b_0})\cdot\vec{b_0} (a⋅b0)⋅b0。
(4)下面进行罗德里格斯公式的推导:
如上图所示,向量 v ⃗ \vec{v} v 绕 u ⃗ \vec{u} u 旋转 θ \theta θ 得到 v ′ ⃗ \vec{v'} v′,其中 u ⃗ \vec{u} u 是单位向量 。
① 向量 v ⃗ \vec{v} v分解: v ⃗ = v ⃗ ∥ + v ⃗ ⊥ \vec{v}=\vec{v}_{\parallel}+\vec{v}_{\perp} v=v∥+v⊥;
② 根据向量投影定理, v ⃗ ∥ = ( v ⃗ ⋅ u ⃗ ) ⋅ u ⃗ \vec{v}_{\parallel}=(\vec{v}\cdot\vec{u})\cdot\vec{u} v∥=(v⋅u)⋅u
③ 综合 ① ② ,得
v ⃗ ⊥ = v ⃗ − v ⃗ ∥ = v ⃗ − ( v ⃗ ⋅ u ⃗ ) ⋅ u ⃗ (3-4-2) \vec{v}_{\perp}=\vec{v}-\vec{v}_{\parallel}=\vec{v}-(\vec{v}\cdot\vec{u})\cdot\vec{u} \tag{3-4-2} v⊥=v−v∥=v−(v⋅u)⋅u(3-4-2)
④ 借助辅助向量 w ⃗ \vec{w} w,且 w ⃗ = u ⃗ × v ⃗ ⊥ \vec{w}=\vec{u}\times\vec{v}_{\perp} w=u×v⊥,即向量 w ⃗ \vec{w} w、 u ⃗ \vec{u} u、 v ⃗ ⊥ \vec{v}_{\perp} v⊥ 两两垂直。俯视图如下:
则有
w ⃗ = u ⃗ × v ⃗ ⊥ = u ⃗ × ( v ⃗ − v ⃗ ∥ ) = u ⃗ × v ⃗ − u ⃗ × v ⃗ ∥ = u ⃗ × v ⃗ (3-4-3) \begin{aligned} \vec{w}=\vec{u}\times\vec{v}_{\perp}=&\vec{u}\times(\vec{v}-\vec{v}_{\parallel})\\ =& \vec{u}\times\vec{v}-\vec{u}\times\vec{v}_{\parallel}\\ =& \vec{u}\times\vec{v} \end{aligned} \tag{3-4-3} w=u×v⊥===u×(v−v∥)u×v−u×v∥u×v(3-4-3)
⑤ 将 v ′ ⃗ ⊥ \vec{v'}_{\perp} v′⊥ 投影到 v ⃗ ⊥ \vec{v}_{\perp} v⊥ 和 $\vec{w} $ 上,则有
v ′ ⃗ ⊥ = v ⃗ v ′ + v ⃗ w ′ (3-4-4) \vec{v'}_{\perp}=\vec{v}_v'+\vec{v}_w' \tag{3-4-4} v′⊥=vv′+vw′(3-4-4)
⑥ 由上图
∣ v ′ ⃗ ⊥ ∣ = ∣ w ⃗ ∣ = ∣ v ⃗ ⊥ ∣ (3-4-5) |\vec{v'}_{\perp}|=|\vec{w}|=|\vec{v}_{\perp}| \tag{3-4-5} ∣v′⊥∣=∣w∣=∣v⊥∣(3-4-5)
且
∣ v ⃗ w ′ ∣ = ∣ v ′ ⃗ ⊥ ∣ s i n θ = ∣ w ⃗ ∣ s i n θ |\vec{v}_w'|=|\vec{v'}_{\perp}|sin\theta=|\vec{w}|sin\theta ∣vw′∣=∣v′⊥∣sinθ=∣w∣sinθ
结合式(2-23),有
v ⃗ w ′ = w ⃗ s i n θ (3-4-6) \vec{v}_w'=\vec{w}sin\theta \tag{3-4-6} vw′=wsinθ(3-4-6)
同理
∣ v ⃗ v ′ ∣ = ∣ v ⃗ ⊥ ∣ c o s θ |\vec{v}_v'|=|\vec{v}_{\perp}|cos\theta ∣vv′∣=∣v⊥∣cosθ
得
v ⃗ v ′ = v ⃗ ⊥ c o s θ (3-4-7) \vec{v}_v'=\vec{v}_{\perp}cos\theta \tag{3-4-7} vv′=v⊥cosθ(3-4-7)
⑦ 综合式(3-22)、(3-24)、(3-25),得
v ′ ⃗ ⊥ = v ⃗ v ′ + v ⃗ w ′ = v ⃗ ⊥ c o s θ + w ⃗ s i n θ (3-4-8) \vec{v'}_{\perp}=\vec{v}_v'+\vec{v}_w'=\vec{v}_{\perp}cos\theta+\vec{w}sin\theta \tag{3-4-8} v′⊥=vv′+vw′=v⊥cosθ+wsinθ(3-4-8)
⑧ 综上,
v ′ ⃗ = v ⃗ ∥ + v ′ ⃗ ⊥ = ( v ⃗ ⋅ u ⃗ ) ⋅ u ⃗ + v ⃗ ⊥ c o s θ + w ⃗ s i n θ = ( v ⃗ ⋅ u ⃗ ) ⋅ u ⃗ + ( v ⃗ − ( v ⃗ ⋅ u ⃗ ) ⋅ u ⃗ ) c o s θ + u ⃗ × v ⃗ s i n θ = ( v ⃗ ⋅ u ⃗ ) ⋅ u ⃗ + s i n θ u ⃗ × v ⃗ + c o s θ v ⃗ − c o s θ ( v ⃗ ⋅ u ⃗ ) ⋅ u ⃗ = c o s θ v ⃗ + ( 1 − c o s θ ) ( v ⃗ ⋅ u ⃗ ) ⋅ u ⃗ + s i n θ u ⃗ × v ⃗ (3-4-9) \begin{aligned} \vec{v'}=&\vec{v}_{\parallel}+\vec{v'}_{\perp}\\ =& (\vec{v}\cdot\vec{u})\cdot\vec{u}+\vec{v}_{\perp}cos\theta+\vec{w}sin\theta\\ =& (\vec{v}\cdot\vec{u})\cdot\vec{u}+(\vec{v}-(\vec{v}\cdot\vec{u})\cdot\vec{u})cos\theta+\vec{u}\times\vec{v}sin\theta\\ =& (\vec{v}\cdot\vec{u})\cdot\vec{u}+sin\theta\vec{u}\times\vec{v}+cos\theta\vec{v}-cos\theta(\vec{v}\cdot\vec{u})\cdot\vec{u}\\ =& cos\theta\vec{v}+(1-cos\theta)(\vec{v}\cdot\vec{u})\cdot\vec{u}+sin\theta\vec{u}\times\vec{v} \end{aligned} \tag{3-4-9} v′=====v∥+v′⊥(v⋅u)⋅u+v⊥cosθ+wsinθ(v⋅u)⋅u+(v−(v⋅u)⋅u)cosθ+u×vsinθ(v⋅u)⋅u+sinθu×v+cosθv−cosθ(v⋅u)⋅ucosθv+(1−cosθ)(v⋅u)⋅u+sinθu×v(3-4-9)
⑨ 设矩阵
u = [ u x u y u z ] , v = [ v x v y v z ] \boldsymbol{u}=\left[\begin{array}{c} u_{x} \\ u_{y}\\ u_{z} \end{array}\right], \boldsymbol{v}=\left[\begin{array}{c} v_{x} \\ v_{y}\\ v_{z} \end{array}\right] u= uxuyuz ,v= vxvyvz
1)将向量点乘转换为矩阵乘法,(此处应用向量交换律和矩阵结合律)
( v ⃗ ⋅ u ⃗ ) ⋅ u ⃗ = u ⃗ ⋅ ( v ⃗ ⋅ u ⃗ ) = u ( u T v ) = u u T v (3-4-10) (\vec{v}\cdot\vec{u})\cdot\vec{u}=\vec{u}\cdot(\vec{v}\cdot\vec{u})=\boldsymbol{u}(\boldsymbol{u}^T\boldsymbol{v})=\boldsymbol{u}\boldsymbol{u}^T\boldsymbol{v} \tag{3-4-10} (v⋅u)⋅u=u⋅(v⋅u)=u(uTv)=uuTv(3-4-10)
2)写成反对称矩阵形式
u ⃗ × v ⃗ = ∣ i j k u x u y u z v x v y v z ∣ = [ u y v z − u z v y u z v x − u x v z u x v y − u y v x ] = [ 0 − u z u y u z 0 − u x − u y u x 0 ] [ v x v y v z ] = u ∧ v (3-4-11) \vec{u}\times\vec{v}=\left| \begin{array}{cccc} \boldsymbol{i} & \boldsymbol{j} & \boldsymbol{k} \\ u_{x} & u_{y} & u_{z}\\ v_{x} & v_{y} & v_{z} \end{array} \right|=\left[\begin{array}{c} u_{y}v_{z}-u_{z}v_{y} \\ u_{z}v_{x}-u_{x}v_{z}\\ u_{x}v_{y}-u_{y}v_{x} \end{array}\right]= \left[\begin{array}{c} 0 & -u_{z} & u_{y} \\ u_{z} & 0 & -u_{x}\\ -u_{y} & u_{x} & 0 \end{array}\right]\left[\begin{array}{c} v_{x}\\ v_{y}\\ v_{z} \end{array}\right]=\boldsymbol{u}^{\wedge}\boldsymbol{v} \tag{3-4-11} u×v= iuxvxjuyvykuzvz = uyvz−uzvyuzvx−uxvzuxvy−uyvx = 0uz−uy−uz0uxuy−ux0 vxvyvz =u∧v(3-4-11)
⑩ 因为 $ \boldsymbol{v’}=\boldsymbol{R}\boldsymbol{v}$
综上所述,式(3-27)可化为
R = c o s θ I + ( 1 − c o s θ ) u u T + s i n θ u ∧ (3-4-12) \boldsymbol{R}=cos\theta\boldsymbol{I}+(1-cos\theta)\boldsymbol{u}\boldsymbol{u}^T+sin\theta\boldsymbol{u}^{\wedge} \tag{3-4-12} R=cosθI+(1−cosθ)uuT+sinθu∧(3-4-12)
证毕。
(5)旋转矩阵到旋转向量的转换
将式(3-4-12)两边取迹,得
t r ( R ) = c o s θ t r ( I ) + ( 1 − c o s θ ) t r ( u u T ) + s i n θ t r ( u ∧ ) = 3 c o s θ + ( 1 − c o s θ ) + 0 = 1 + 2 c o s θ (3-4-13) \begin{aligned} tr(\boldsymbol{R})=& cos\theta tr(\boldsymbol{I})+(1-cos\theta)tr(\boldsymbol{u}\boldsymbol{u}^T)+sin\theta tr(\boldsymbol{u}^{\wedge})\\ =&3cos\theta+(1-cos\theta)+0\\ =&1+2cos\theta \end{aligned} \tag{3-4-13} tr(R)===cosθtr(I)+(1−cosθ)tr(uuT)+sinθtr(u∧)3cosθ+(1−cosθ)+01+2cosθ(3-4-13)
其中, t r ( u u T ) = u x 2 + u y 2 + u z 2 = 1 tr(\boldsymbol{u}\boldsymbol{u}^T)=u_x^2+u_y^2+u_z^2=1 tr(uuT)=ux2+uy2+uz2=1(向量 u \boldsymbol{u} u是单位向量,模长为1)。
那么,对于转角 θ \theta θ 有
θ = a r c c o s ( t r ( R ) − 1 2 ) (3-4-14) \theta=arccos(\frac {tr(\boldsymbol{R})-1} {2}) \tag{3-4-14} θ=arccos(2tr(R)−1)(3-4-14)
对于转轴 u \boldsymbol{u} u 有,
R u = u (3-4-15) \boldsymbol{Ru=u} \tag{3-4-15} Ru=u(3-4-15)
即转轴 u \boldsymbol{u} u 是旋转矩阵 R \boldsymbol{R} R 特征值为 1 对应的特征向量。