向量、坐标与距离Vectors, Coordinates & Distance
怎样用一组数表达位置、位移与两点之间的距离？How can a list of numbers describe position, displacement and distance?
前置知识：Prerequisites: B01：函数与平面坐标。B01: functions and Cartesian coordinates.
复习前置：Review prerequisites: B01 · 函数、坐标与参数Functions, Coordinates & Parameters
学完后你能：After this lesson you can: 计算并相加位移，利用勾股定理计算两点间的距离。Calculate and add displacements, and use Pythagoras to find distance.
动手观察Explore
拖动点 A 或 B，也可用滑块调整坐标。预测 B 与 A 重合时距离是多少。Drag A or B, or use the sliders to adjust coordinates. Predict the distance when B coincides with A.
从观察到解释From observation to explanation
点与箭头Points and arrows
点 A = (1, 2) 表示位置。若 B = (4, 6)，从 A 到 B 的位移是 (4−1, 6−2) = (3, 4)。位移箭头可以平移，只要方向和长度保持不变。A = (1, 2) describes a position. If B = (4, 6), the displacement from A to B is (4−1, 6−2) = (3, 4). A displacement arrow can be moved without changing its length or direction.
把位移接起来Add displacements
从原点先走 (1, 2)，再走 (3, 1)，终点是 (4, 3)，所以 (1, 2) + (3, 1) = (4, 3)。数乘 2(1, 2) = (2, 4) 将每个分量都乘以 2。Starting at the origin, move (1, 2) and then (3, 1): the endpoint is (4, 3), so (1, 2) + (3, 1) = (4, 3). Scaling 2(1, 2) = (2, 4) multiplies each component by 2.
勾股定理给距离Distance from Pythagoras
水平位移 3、竖直位移 4 构成直角三角形，距离是 √(3² + 4²) = 5。一个向量的长度是它表示的点到原点的距离。Horizontal displacement 3 and vertical displacement 4 form a right triangle, giving distance √(3² + 4²) = 5. A vector’s length is the distance from its represented point to the origin.
公式与符号Formula and symbols
a、b 是两个点的位置向量；下标 1、2 表示横纵坐标；b−a 是从 A 到 B 的位移，d(a,b) 是两点的距离。a and b are position vectors; subscripts 1 and 2 are horizontal and vertical coordinates; b−a is the displacement from A to B, and d(a,b) is their distance.
留意这个区别Watch for this
点的位置不等于两点间的位移。距离依赖坐标单位：把米换成厘米而不统一尺度，会改变直接计算出的数值。A point’s position is not the displacement between two points. Distance depends on coordinate units: changing meters to centimeters without a consistent scale changes the numerical result.
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