叉积The Cross Product

机械臂手里拿着一块砖，肩关节要出多大的力来撑住它？这取决于砖有多重，也取决于砖离关节有多远、往哪个方向压。同样 10 牛的力，顺着手臂的方向推，关节纹丝不动；垂直于手臂推，关节被拧得最狠。"一个力对关节有多大的拧劲"，就是叉积算出来的：\(\boldsymbol\tau = \mathbf r \times \mathbf F\)。

同一个运算在机器人里反复出现：

• 力矩：\(\boldsymbol\tau = \mathbf r\times\mathbf F\)，一个力拧关节的力度。
• 旋转物体上一点的速度：转台、轮子、旋转的夹爪上，离轴心 \(\mathbf r\) 处的点速度为 \(\mathbf v = \boldsymbol\omega\times\mathbf r\)。
• 曲面的法向量：点云里取相邻三点，用叉积得到它们所在平面的法向，这就是抓取时的接近方向。
• 左转还是右转：二维路径规划里，判断拐角、求凸包、判断障碍物在直线哪一侧，用的都是叉积的符号。

更往后，叉积的矩阵形式 \(\skew{\boldsymbol\omega}\) 是旋转的"导数"，5.5 和 5.6 节的李代数、指数映射都建立在它上面。

A robot arm holds a brick. How hard must the shoulder joint work to keep it up? It depends on the weight, but also on how far the brick is from the joint and in which direction it pushes. A force of 10 N pushing along the arm leaves the joint unmoved; pushing across the arm twists it the hardest. "How much a force twists a joint" is computed by a cross product: \(\boldsymbol\tau = \mathbf r \times \mathbf F\).

The same operation keeps coming back in robotics:

• Torque: \(\boldsymbol\tau = \mathbf r\times\mathbf F\), how hard a force twists a joint.
• Velocity of a point on a spinning body: on a turntable, a wheel or a rotating gripper, the point at \(\mathbf r\) from the axis moves with \(\mathbf v = \boldsymbol\omega\times\mathbf r\).
• Surface normals: take three neighbouring points of a point cloud, and a cross product gives the normal of the plane through them: the approach direction for a grasp.
• Left turn or right turn: in 2D path planning, detecting corners, building convex hulls and deciding which side of a line an obstacle is on all use the sign of a cross product.

Further on, the matrix form \(\skew{\boldsymbol\omega}\) of the cross product is the "derivative" of rotation: the Lie algebra and the exponential map of Lessons 5.5 and 5.6 are built on it.

在平面里，两个向量围出一个平行四边形，面积是一个数（1.5 节的行列式）。到了三维，这个平行四边形可以朝任何方向"躺着"：只说面积还不够，还得说它躺在哪个平面里。叉积把两件事打包成一个向量：

• 长度是平行四边形的面积，\(\lVert\mathbf a\times\mathbf b\rVert = \lVert\mathbf a\rVert\lVert\mathbf b\rVert\sin\theta\)；
• 方向垂直于这个平面，也就是同时垂直于 \(\mathbf a\) 和 \(\mathbf b\)。垂直方向有两个（朝上或朝下），由右手定则选：四指从 \(\mathbf a\) 弯向 \(\mathbf b\)，拇指指向 \(\mathbf a\times\mathbf b\)。

如果 \(\mathbf a\)、\(\mathbf b\) 平行，平行四边形压成一条线，面积为零，叉积就是零向量。下面固定 a，让 b 在水平面里转一圈，看 a × b 怎么变。

In the plane, two vectors span a parallelogram, and its area is one number (the determinant of Lesson 1.5). In 3D that parallelogram can lie in any orientation: the area alone is not enough, you also need to say which plane it lies in. The cross product packs both facts into one vector:

• its length is the parallelogram's area, \(\lVert\mathbf a\times\mathbf b\rVert = \lVert\mathbf a\rVert\lVert\mathbf b\rVert\sin\theta\);
• its direction is perpendicular to that plane, i.e. to both \(\mathbf a\) and \(\mathbf b\). There are two perpendicular directions (up or down), and the right-hand rule picks one: curl your fingers from \(\mathbf a\) toward \(\mathbf b\), and your thumb points along \(\mathbf a\times\mathbf b\).

If \(\mathbf a\) and \(\mathbf b\) are parallel, the parallelogram collapses to a line, its area is zero, and the cross product is the zero vector. Below, a is fixed and b sweeps once around the horizontal plane. Watch a × b.

把 \(\mathbf a\)、\(\mathbf b\) 都放进 xy 平面（\(z=0\)）。它们围出的平行四边形躺在这个平面里，叉积必然沿着 z 轴，只剩一个分量要算：

Put both \(\mathbf a\) and \(\mathbf b\) in the xy-plane (\(z=0\)). The parallelogram they span lies in this plane, so the cross product must point along the z-axis, and only one component is left to compute:

这正是 1.5 节的行列式：面积的正负号里藏着方向信息。结果为正，说明 \(\mathbf b\) 在 \(\mathbf a\) 的左边（从 \(\mathbf a\) 逆时针转不到 180° 就能到 \(\mathbf b\)）；为负，\(\mathbf b\) 在右边；为零，两者共线。拖动下图的两个箭头尖。

This is exactly the determinant of Lesson 1.5: the sign of the area carries orientation. A positive value means \(\mathbf b\) is on the left of \(\mathbf a\) (you reach \(\mathbf b\) by turning counter-clockwise from \(\mathbf a\) by less than 180°); a negative value means b is on the right; zero means they are collinear. Drag the two arrow tips below.

一台机器人沿着路径点走。在每个拐角处，设进来的方向是 \(\mathbf u = p_i - p_{i-1}\)，出去的方向是 \(\mathbf v = p_{i+1}-p_i\)，则 \(\mathbf u\times\mathbf v>0\) 是左转，\(<0\) 是右转。不用算角度，不用 \(\arctan\)，一次乘减就够了。

"搭路径"模式下拖动路径点，实时看每个拐角；"抢答"模式下看一眼就判断。

A robot follows a path through waypoints. At each corner let the incoming direction be \(\mathbf u = p_i - p_{i-1}\) and the outgoing one \(\mathbf v = p_{i+1}-p_i\). Then \(\mathbf u\times\mathbf v>0\) is a left turn and \(<0\) a right turn. No angles and no \(\arctan\): one multiply-and-subtract is enough.

In "Build" mode drag the waypoints and watch every corner live; in "Quiz" mode make the call at a glance.

固定 \(\mathbf a\)，看 \(\mathbf b\mapsto\mathbf a\times\mathbf b\)。这个映射对 \(\mathbf b\) 是线性的：\(\mathbf a\times(s\mathbf b+t\mathbf c) = s\,\mathbf a\times\mathbf b + t\,\mathbf a\times\mathbf c\)。1.3 节说过，线性映射都是矩阵。它的矩阵由 \(\mathbf a\) 的三个分量排成，叫做 \(\mathbf a\) 的叉积矩阵（hat 矩阵）：

Fix \(\mathbf a\) and look at \(\mathbf b\mapsto\mathbf a\times\mathbf b\). This map is linear in \(\mathbf b\): \(\mathbf a\times(s\mathbf b+t\mathbf c) = s\,\mathbf a\times\mathbf b + t\,\mathbf a\times\mathbf c\). By Lesson 1.3, every linear map is a matrix. Its matrix is built from the three components of \(\mathbf a\) and is called the cross-product matrix (or hat matrix) of \(\mathbf a\):

用下面的图验证：改 \(\mathbf a\) 和 \(\mathbf b\)（左右拖动数字，或直接输入），看矩阵乘出来的结果是不是琥珀色的 \(\mathbf a\times\mathbf b\)。淡琥珀色的圆盘是单位球经过 \(\skew{\mathbf a}\) 后的像：球被压扁成一个垂直于 \(\mathbf a\) 的圆盘，半径 \(\lVert\mathbf a\rVert\)。矩阵的秩是 2，沿 \(\mathbf a\) 的方向被完全抹掉。

Check it in the figure below: change \(\mathbf a\) and \(\mathbf b\) (scrub the numbers sideways or type them) and see that the matrix product is the amber \(\mathbf a\times\mathbf b\). The pale amber disc is the image of the unit sphere under \(\skew{\mathbf a}\): the sphere is flattened into a disc perpendicular to \(\mathbf a\), of radius \(\lVert\mathbf a\rVert\). The matrix has rank 2, and the direction of \(\mathbf a\) is wiped out completely.

每个分量是"另外两个下标"的交叉相乘再相减，下标按 1→2→3→1 循环。更好记的是一个"形式上的行列式"，第一行放的不是数而是基向量：

Each component is a cross-multiplication of the "other two indices", with the indices running cyclically 1→2→3→1. Easier to remember is a "formal determinant" whose first row holds basis vectors instead of numbers:

其中 \(\hl{theta}{\theta}\in[0,\pi]\) 是 \(\mathbf a\)、\(\mathbf b\) 的夹角。理由：以 \(\lVert\mathbf a\rVert\) 为底，平行四边形的高是 \(\lVert\mathbf b\rVert\sin\theta\)，面积 = 底 × 高。把它和点积（1.7 节）放在一起看：

Here \(\hl{theta}{\theta}\in[0,\pi]\) is the angle between \(\mathbf a\) and \(\mathbf b\). Reason: with base \(\lVert\mathbf a\rVert\), the parallelogram's height is \(\lVert\mathbf b\rVert\sin\theta\), and area = base × height. Put it next to the dot product (Lesson 1.7):

其中 \(V_{\mathbf a,\mathbf b,\mathbf c}\) 是 \(\mathbf a,\mathbf b,\mathbf c\) 张成的平行六面体的体积，三重积是带符号的体积。\(\mathbf b\times\mathbf c\) 的长度是底面积，方向垂直于底面；再和 \(\mathbf a\) 点积，就是 \(\mathbf a\) 在垂直方向的分量（高）乘底面积，即体积。正负号反映 \((\mathbf a,\mathbf b,\mathbf c)\) 是不是右手系，这就是 1.5 节说的三维行列式。例：\(\mathbf a=(1,0,0)\)，\(\mathbf b=(1,2,0)\)，\(\mathbf c=(1,1,3)\)，则 \(\mathbf b\times\mathbf c=(6,-3,-1)\)，\(\mathbf a\cdot(\mathbf b\times\mathbf c)=6\)；矩阵是上三角的，行列式 \(1\cdot2\cdot3=6\)，一致。

Here \(V_{\mathbf a,\mathbf b,\mathbf c}\) is the volume of the parallelepiped spanned by \(\mathbf a,\mathbf b,\mathbf c\): the triple product is a signed volume. The length of \(\mathbf b\times\mathbf c\) is the base area and its direction is perpendicular to the base; dotting with \(\mathbf a\) picks the component of \(\mathbf a\) along that perpendicular (the height) times the base area: the volume. The sign says whether \((\mathbf a,\mathbf b,\mathbf c)\) is right-handed, which is the 3D determinant of Lesson 1.5. Example: \(\mathbf a=(1,0,0)\), \(\mathbf b=(1,2,0)\), \(\mathbf c=(1,1,3)\) give \(\mathbf b\times\mathbf c=(6,-3,-1)\) and \(\mathbf a\cdot(\mathbf b\times\mathbf c)=6\); the matrix is upper triangular with determinant \(1\cdot2\cdot3=6\), the same.

第二条（斜对称）读矩阵就能看出：对角线是 0，关于对角线对称位置的两个数互为相反数。第三条是 \(\mathbf a\times\mathbf a=\mathbf 0\)。因为 \(\skew{\mathbf a}\) 把 \(\mathbf a\) 变成零，它不可逆，秩为 2，零空间是 \(\mathbf a\) 所在的直线（1.6 节）。第四条由 BAC-CAB 得到：\(\skew{\mathbf a}^2\mathbf b=\mathbf a\times(\mathbf a\times\mathbf b)=\mathbf a(\mathbf a\cdot\mathbf b)-\mathbf b\lVert\mathbf a\rVert^2\)。第五条再乘一次 \(\skew{\mathbf a}\)，利用 \(\skew{\mathbf a}\mathbf a=\mathbf 0\)：\(\skew{\mathbf a}^3=\skew{\mathbf a}\mathbf a\mathbf a\T-\lVert\mathbf a\rVert^2\skew{\mathbf a}=-\lVert\mathbf a\rVert^2\skew{\mathbf a}\)。

对单位向量 \(\hat{\mathbf u}\)，\(K=\skew{\hat{\mathbf u}}\) 满足 \(K^3=-K\)。于是 \(K\) 的高次幂只在 \(K\) 和 \(K^2\) 之间来回，指数级数 \(e^{\theta K}\) 能合并成 \(I+\sin\theta\,K+(1-\cos\theta)K^2\)：这就是 5.6 节的罗德里格斯公式。

The second one (skew-symmetry) can be read off the matrix: the diagonal is 0 and entries mirrored across the diagonal are negatives of each other. The third is \(\mathbf a\times\mathbf a=\mathbf 0\). Because \(\skew{\mathbf a}\) turns \(\mathbf a\) into zero, it is not invertible: its rank is 2 and its null space is the line through \(\mathbf a\) (Lesson 1.6). The fourth follows from BAC-CAB: \(\skew{\mathbf a}^2\mathbf b=\mathbf a\times(\mathbf a\times\mathbf b)=\mathbf a(\mathbf a\cdot\mathbf b)-\mathbf b\lVert\mathbf a\rVert^2\). The fifth multiplies by \(\skew{\mathbf a}\) once more and uses \(\skew{\mathbf a}\mathbf a=\mathbf 0\): \(\skew{\mathbf a}^3=\skew{\mathbf a}\mathbf a\mathbf a\T-\lVert\mathbf a\rVert^2\skew{\mathbf a}=-\lVert\mathbf a\rVert^2\skew{\mathbf a}\).

For a unit vector \(\hat{\mathbf u}\), \(K=\skew{\hat{\mathbf u}}\) satisfies \(K^3=-K\). Higher powers of \(K\) then just alternate between \(K\) and \(K^2\), and the exponential series \(e^{\theta K}\) collapses into \(I+\sin\theta\,K+(1-\cos\theta)K^2\): Rodrigues' formula of Lesson 5.6.

关节在原点，连杆伸到作用点 \(\mathbf r\)，那里有一个力 \(\mathbf F\)。关节感受到的拧劲是 \(\boldsymbol\tau=\mathbf r\times\mathbf F\)，它沿着关节轴。在平面里它退化成一个数 \(\tau=r_xF_y-r_yF_x\)：正号表示逆时针拧。大小还可以写成 \(\tau=\lVert\mathbf F\rVert\cdot d\)，\(d\) 是关节到力的作用线的垂直距离（力臂）。只有垂直于连杆的那部分力才拧关节。

拖动连杆末端（青色点）和力的箭头尖（紫色）。切换到"转动"模式，可以看同一个叉积的另一种用法。

The joint sits at the origin, the link reaches out to the point of application \(\mathbf r\), and a force \(\mathbf F\) acts there. The twist the joint feels is \(\boldsymbol\tau=\mathbf r\times\mathbf F\), directed along the joint axis. In a plane it collapses to one number, \(\tau=r_xF_y-r_yF_x\), positive for a counter-clockwise twist. Its size can also be written \(\tau=\lVert\mathbf F\rVert\cdot d\), where \(d\) is the perpendicular distance from the joint to the line of action of the force (the lever arm). Only the part of the force across the link twists the joint.

Drag the end of the link (cyan dot) and the tip of the force arrow (purple). Switch to "Spinning" mode for another use of the same cross product.

点云里取相邻三点 \(p_1,p_2,p_3\)，两条边的叉积垂直于它们所在的平面：

Take three neighbouring points \(p_1,p_2,p_3\) of a point cloud; the cross product of two edges is perpendicular to the plane through them:

未归一化的 \(\mathbf N\) 顺手给出三角形的面积 \(A_\triangle\)（平行四边形的一半）。下面给三个点加一点测量噪声，看噪声怎么影响法向量。深色的 琥珀色箭头 是无噪声的法向量，青色小点是 36 次加噪后的单位法向量的尖端。

The unnormalised \(\mathbf N\) gives the triangle's area \(A_\triangle\) for free (half the parallelogram). Below, a little measurement noise is added to the three points to see how it affects the normal. The amber arrow is the noise-free normal; the small cyan dots are the tips of the unit normals from 36 noisy trials.