Not to harp on the mathematical features of this, but cubing, AX*X*X you know, if you're starting to do AX star, X star, X, every time you want to cube some value in a program, it just feels like this is going to get a little messy looking, if nothing else.

不要总是说这个的数学特性，但是体积，你们懂的，如果你开始做，在一个程序中，每次你想算几个数值的体积，感觉它就变得,有一点凌乱的，如果没有其他的。

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So there is a harp, making its melodious harmonious sounds, and then you take an ax to the harp, bang bang bang, chop chop chop, or a hammer whatever.

假设这有一个竖琴,发出悦耳和谐的声音，然后你拿起一把斧子，梆梆邦，砍砍砍，或者拿一把锤子。

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Therefore, the vector A that you gave me, I have managed to write as i times Ax plus j times Ay.

这样,你给我的矢量 A,我已把它写成 i ? Ax + j ? Ay的形式

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You can ask yourself, "If you gave me a particular vector, what do I use for Ax and Ay?"

你可以问问自己,"如果给定一个矢量,该怎么确定 Ax 和 Ay"

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But then, I combine i times Ax with i times Bx, because that's the vector parallel to i, with the length Ax.

然后我把 i ? Ax 和 i ? Bx 加起来,因为这是平行于 i,模长为 Ax 的矢量

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That's clear. If I add them, I'll get a vector parallel to i with lengths Ax + Bx.

这很明显 如果把他们加起来,就得到一个平行于 i,模长为 Ax + Bx 的矢量

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But this is the same vector we are calling i times Ax plus j times Ay.

这和矢量 i ? Ax + j ? Ay 是一样的

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The point is the arrow A, somebody has chosen to write in terms of i prime and j prime as Ax prime and Ay prime.

解决问题的关键在于矢量 A,可以用这样的形式来描述,i' ? Ax' + j' ? Ay'

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In practice, most of the time we work with these two numbers, Ax and Ay.

在实际过程中,大多数时间我们就用 Ax 和 Ay 来计算

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If the Ax and Ay, some are positive and some are negative, this is the way by which we have learned we should combine multiples of i.

如果 Ax 和 Ay 有正有负,这就要用到我们所学过的方法,将所有 i 的倍数加起来

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But Ax prime and Ay prime will continue to be the coefficients.

但系数仍然是 Ax' 和 Ay'

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So whatever number it takes, Ax times i is this part.

因此无论系数是多少,Ax 乘以 i 表示这一部分

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If you give me a pair of numbers, Ax and Ay, that's as good as giving me this arrow, because I can find the length of the arrow by Pythagoras' theorem.

如果给我一组数字,Ax 和 Ay,就相当于给了我这个箭头示意图,因为我可以利用毕达哥拉斯定理定理求出模长

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That would come by saying, "I'm taking i times Ax + j times Ay + i times Bx + j times By, " and I'm trying to add all these guys.

那就这样做,"i ? Ax + j ? Ay + i ? Bx + j ? By",然后把它们都加起来

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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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Now, when you work with components, Ax and Ay, if I didn't mention it, they are the components of the vector, you can do all your bookkeeping in terms of Ax and Ay.

当你们在计算分量 Ax 和 Ay 的时候,即使我没有说明,你们也要记得它们是矢量的分量,你们可以都用 Ax 和 Ay 的形式来表示

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A very important result is that if two vectors are equal, if A = B, the only way it can happen is if separately Ax is equal to Bx and Ay is equal to By.

这里有一个相当重要的结论,如果两个矢量相等,例如 A = B,那么当且仅当,Ax = Bx 和 Ay = By 分别成立

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They won't invert the relation.

不是简单地交换 Ax 和 Ax'

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the Ax prime will drop off.

x 就被消去了

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I gave you a law of transformation of the components; namely, if the vector has components Ax and Ay in one reference frame and Ax prime and Ay prime in another reference frame, how are the two related?

我介绍过分量变换的法则,即如果矢量在一个坐标系的分量为 Ax 和 Ay,在另一坐标系中的分量为 Ax' 与 Ay',它们有着什么样的联系

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