点积The Dot Product

机器人每时每刻都在回答一类很小的问题：

• 前方检测到一个障碍物，它在我的前面还是后面？
• 那个人在不在我的相机视野里？
• 我现在的速度，有多少是真正朝着目标的？
• 机械臂推着一个箱子，我的力里有多少在做功，有多少只是白白顶着墙？
• 这句指令和这张图像不像？（CLIP 这类模型正是这样把语言和图像配对的，见 Part V。）

这些问题看上去毫不相干，答案却都来自同一个运算：点积 (dot product)。它还有一个隐藏身份：1.3 节里矩阵乘向量的"行视角"，每一个分量就是一次点积。这一节把它的两种写法、两种读法，以及机器人最常用的几个场景，一次讲清楚。

A robot is constantly answering a family of tiny questions:

• An obstacle was detected: is it in front of me or behind me?
• Is that person inside my camera's field of view?
• How much of my current velocity actually points toward the goal?
• The arm pushes a box: how much of my force does work, and how much just presses against the wall?
• Does this instruction match this image? (Models like CLIP pair language with images exactly this way; see Part V.)

These questions look unrelated, yet every answer comes from one operation: the dot product. It also has a hidden second identity: the "row view" of a matrix–vector product from Lesson 1.3 computes each entry as a dot product. This lesson covers its two formulas, its two readings, and the places robots use it most.

取两支箭头 \(\mathbf a\) 和 \(\mathbf b\)，夹角为 \(\theta\)。把 \(\mathbf b\) 垂直地"投"到 \(\mathbf a\) 所在的直线上，得到一段影子，它的带符号长度是 \(|\mathbf b|\cos\theta\)。再乘上 \(\mathbf a\) 自己的长度，就是点积：

Take two arrows \(\mathbf a\) and \(\mathbf b\) with angle \(\theta\) between them. Drop \(\mathbf b\) perpendicularly onto the line through \(\mathbf a\) to get a shadow whose signed length is \(|\mathbf b|\cos\theta\). Multiply by the length of \(\mathbf a\) itself and you have the dot product:

影子和 \(\mathbf a\) 同向，结果为正；影子反向，结果为负；\(\mathbf b\) 垂直于 \(\mathbf a\) 时，影子缩成一个点，结果为 0。

最具体的例子是推车。小车只能沿导轨走，你推它的力 \(\mathbf F\) 却常常是斜的。真正推动小车的只有沿导轨的那一份，垂直于导轨的那一份只是顶着导轨。物理里"力做的功"正是 \(W = \mathbf F\cdot\mathbf s\)，\(\mathbf s\) 是小车的位移。拖动力的箭头，看看什么时候做正功、零功、负功。

A shadow pointing the same way as \(\mathbf a\) gives a positive result, an opposing shadow gives a negative one, and when \(\mathbf b\) is perpendicular to \(\mathbf a\) the shadow shrinks to a point and the result is 0.

The most concrete example is pushing a cart. The cart can only roll along its rail, but your force \(\mathbf F\) is often slanted. Only the part of the force along the rail moves the cart; the part across the rail just presses on it. In physics the work done by a force is exactly \(W = \mathbf F\cdot\mathbf s\), where \(\mathbf s\) is the cart's displacement. Drag the force arrow and see when the work is positive, zero, or negative.

实验室右边有两个公式。代数公式 \(a_1b_1 + a_2b_2\) 要用到坐标，换一套坐标轴，每个分量都变了。几何公式 \(|\mathbf a||\mathbf b|\cos\theta\) 只用到长度和夹角，而把整幅图转个角度，长度和夹角都不会变。两者要想永远相等，分量的乘积就必须"此消彼长"，使总和保持不变。

下面把 \(\mathbf a\) 和 \(\mathbf b\) 当作一个刚性的整体一起转动，用三根柱子看这件事。

The lab shows two formulas. The algebraic formula \(a_1b_1 + a_2b_2\) uses coordinates: pick different axes and every component changes. The geometric formula \(|\mathbf a||\mathbf b|\cos\theta\) uses only lengths and the angle, and rotating the whole picture changes neither. For the two to agree forever, the component products must trade off against each other so that their sum stays put.

Below, \(\mathbf a\) and \(\mathbf b\) are rotated together as one rigid pair. Watch the three bars.

1.3 节按列读矩阵乘向量：\(A\mathbf v\) 是 \(A\) 各列的线性组合。同一个乘积还有第二种读法，也是代码里真正在算的那一种：\(A\mathbf v\) 的第 \(i\) 个分量，是 \(A\) 的第 \(i\) 行与 \(\mathbf v\) 的点积。

可以把每一行想成一个"探测器"，它有自己的朝向 \(\mathbf r_i\)，读出的是 \(\mathbf v\) 在自己方向上有多少。下面左图画出两行 \(\mathbf r_1, \mathbf r_2\) 和 \(\mathbf v\)，右图用列视角拼出同一个 \(A\mathbf v\)。两边的结果必须一样。

Lesson 1.3 read a matrix–vector product by columns: \(A\mathbf v\) is a combination of the columns of \(A\). The same product has a second reading, the one code actually computes: the \(i\)-th entry of \(A\mathbf v\) is the dot product of row \(i\) of \(A\) with \(\mathbf v\).

Think of each row as a "detector" with its own direction \(\mathbf r_i\); it reads out how much of \(\mathbf v\) lies along that direction. On the left, the two rows \(\mathbf r_1, \mathbf r_2\) and \(\mathbf v\); on the right, the column view assembles the same \(A\mathbf v\). Both sides must agree.

• 对称：\(\mathbf a\cdot\mathbf b = \mathbf b\cdot\mathbf a\)。
• 线性（对每个位置分别线性）：\(\mathbf a\cdot(\mathbf b + \mathbf c) = \mathbf a\cdot\mathbf b + \mathbf a\cdot\mathbf c\)，\((s\mathbf a)\cdot\mathbf b = s\,(\mathbf a\cdot\mathbf b)\)。
• 与长度的关系：\(\mathbf a\cdot\mathbf a = |\mathbf a|^2\)，所以 \(|\mathbf a| = \sqrt{\mathbf a\cdot\mathbf a}\)。

• Symmetric: \(\mathbf a\cdot\mathbf b = \mathbf b\cdot\mathbf a\).
• Linear (in each slot separately): \(\mathbf a\cdot(\mathbf b + \mathbf c) = \mathbf a\cdot\mathbf b + \mathbf a\cdot\mathbf c\), \((s\mathbf a)\cdot\mathbf b = s\,(\mathbf a\cdot\mathbf b)\).
• Link to length: \(\mathbf a\cdot\mathbf a = |\mathbf a|^2\), so \(|\mathbf a| = \sqrt{\mathbf a\cdot\mathbf a}\).

"clamp"把括号里的数夹在 \([-1, 1]\) 之内（原因见"常见误区"）。例：\(\mathbf a = (3,1)\)，\(\mathbf b = (1,2)\)，则 \(\mathbf a\cdot\mathbf b = 5\)，\(|\mathbf a||\mathbf b| = \sqrt{10}\sqrt{5} = \sqrt{50}\)，\(\cos\theta = 5/\sqrt{50} = \tfrac{\sqrt2}{2}\)，\(\theta = 45°\)。

当 \(\mathbf u\) 是单位向量时，\(\mathbf u\cdot\mathbf b\) 就是 \(\mathbf b\) 在 \(\mathbf u\) 方向上的带符号长度。把这个长度再乘回 \(\mathbf u\)，得到完整的"投影向量" \((\mathbf u\cdot\mathbf b)\,\mathbf u\)，那是 1.9 节的主题。

"Clamp" squeezes the quantity in the parentheses into \([-1, 1]\) (see the pitfalls for why). Example: \(\mathbf a = (3,1)\), \(\mathbf b = (1,2)\) give \(\mathbf a\cdot\mathbf b = 5\), \(|\mathbf a||\mathbf b| = \sqrt{10}\sqrt{5} = \sqrt{50}\), \(\cos\theta = 5/\sqrt{50} = \tfrac{\sqrt2}{2}\), so \(\theta = 45°\).

When \(\mathbf u\) is a unit vector, \(\mathbf u\cdot\mathbf b\) is the signed length of \(\mathbf b\) along \(\mathbf u\). Multiply it back onto \(\mathbf u\) and you get the full "projection vector" \((\mathbf u\cdot\mathbf b)\,\mathbf u\), the subject of Lesson 1.9.

机器人的朝向是单位向量 \(\hat{\mathbf h}\)，相机的视野半角是 \(\alpha\)。物体相对机器人的位移是 \(\mathbf d = \mathbf p_{\text{obj}} - \mathbf p_{\text{robot}}\)。物体可见，当且仅当 \(\mathbf d\) 与 \(\hat{\mathbf h}\) 的夹角 \(\varphi\le\alpha\)。因为 \(\cos\) 在 \([0,\pi]\) 上递减，这等价于

The robot's heading is a unit vector \(\hat{\mathbf h}\) and the camera's field of view has half-angle \(\alpha\). The object's displacement from the robot is \(\mathbf d = \mathbf p_{\text{obj}} - \mathbf p_{\text{robot}}\). The object is visible exactly when the angle \(\varphi\) between \(\mathbf d\) and \(\hat{\mathbf h}\) satisfies \(\varphi\le\alpha\). Since \(\cos\) is decreasing on \([0,\pi]\), this is equivalent to

右边的 \(\cos\alpha\) 只算一次，每个物体只需要一次点积和一次开方，不需要 \(\operatorname{atan2}\) 或 \(\arccos\)。拖动机器人的朝向（红色箭头尖），再调节视野半角，看看右侧表格里每个物体的"点积 ≥ 阈值"是怎么变的。

\(\cos\alpha\) is computed once, and each object needs one dot product and one square root, with no \(\operatorname{atan2}\) or \(\arccos\). Drag the robot's heading (the red arrow tip), adjust the half-angle, and watch the "dot ≥ threshold" comparison for each object in the table on the right.

机器人在原点，目标在 \(\mathbf G\)，指向目标的单位向量是 \(\hat{\mathbf g} = \mathbf G/|\mathbf G|\)。速度 \(\mathbf v\) 中朝向目标的那一份，就是逼近速度 \(s = \mathbf v\cdot\hat{\mathbf g}\)：每秒大约离目标近多少米。假设速度大小不能超过 \(v_{\max}\)，什么方向最好？Cauchy–Schwarz 给出答案：\(s\le|\mathbf v||\hat{\mathbf g}|\le v_{\max}\)，当且仅当 \(\mathbf v = v_{\max}\hat{\mathbf g}\) 时取等号。

很多局部规划器的做法与此一脉相承：生成一批候选速度（图中速度圆上的小点），给每个候选打分，选分数最高的。点积 \(\mathbf v\cdot\hat{\mathbf g}\) 就是最基本的"朝向目标"得分。

The robot sits at the origin, the goal is at \(\mathbf G\), and the unit direction to the goal is \(\hat{\mathbf g} = \mathbf G/|\mathbf G|\). The part of the velocity \(\mathbf v\) that points at the goal is the closing speed \(s = \mathbf v\cdot\hat{\mathbf g}\): roughly how many meters per second you get closer. If the speed cannot exceed \(v_{\max}\), which direction is best? Cauchy–Schwarz answers: \(s\le|\mathbf v||\hat{\mathbf g}|\le v_{\max}\), with equality exactly when \(\mathbf v = v_{\max}\hat{\mathbf g}\).

Many local planners follow the same idea: generate candidate velocities (the dots on the speed circle in the figure), score each one, and pick the best. The dot product \(\mathbf v\cdot\hat{\mathbf g}\) is the most basic "toward the goal" score.