来自《耶鲁大学开放课程：基础物理 第2集 第2集》

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  and the length of the vector I'm going to call by A.
  然后我把矢量的模长记为 A
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  A is the length of A.
  A 表示矢量 A 的模长
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  If you drop the arrow,
  如果你把箭头去掉
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  it's the usual convention,
  就表示矢量 A 的模长
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  if you're talking about the length of the vector A.
  这是约定俗成的
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  First of all, it's clear from the Pythagoras' theorem
  首先 根据毕达哥拉斯定理
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  that a is the square root of (Ax)^2 + (Ay)^2.
  显然有 A 等于 (Ax)^2 + (Ay)^2 的平方根
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  The angle θ that it makes with the x-axis
  与 x 轴的夹角 θ
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  satisfies the condition tan is Ay over Ax..
  满足其正切等于 Ay 除以 Ax
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  What this means--Now here's the main point.
  这样的意义何在呢 下面我要讲出重点
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  If you give me a pair of numbers, Ax and Ay,
  如果给我一组数字 Ax 和 Ay
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  that's as good as giving me this arrow,
  就相当于给了我这个箭头示意图