我希望这是正确的位置来问这个问题，这是的同这个，但EX pressed纯数学，而不是图形（至少我希望我正确的翻译问题，以数学）。

I hope this is the proper location to ask this question which is the same as this one, but expressed as pure math instead of graphically (at least I hope I translated the problem to math correctly).

考虑：

问题：什么是Y1和向上之间的角度的值

Question: what is the value of the angle between Y1 and Up?

正如数学家会同意，这是一个非常基本的问题，但我一直在抓我的头，至少有两个星期没有能够直观地了解项目Ÿ到P ...也许现在太老了寻求解决办法学校练习。

As mathematicians will agree, this is a very basic question, but I've been scratching my head for at least two weeks without being able to visualize how to project Y onto P... maybe now too old for finding solutions to school exercises.

我要寻找的三角函数解决方案，而不是使用矩阵解决方案。谢谢你。

I'm looking for the trigonometric solution, not a solution using a matrix. Thanks.

修改：我发现我需要确定角度的符号，相对于旋转轴这不得不看。我张贴的最后code对我的链接的问题（见以上链接）。感谢那些谁帮助。我AP preciate你的时间。

Edit: I found that I needed to determine the sign of the angle, relative to a rotation axis which had to be Look. I posted the final code on my linked question (see link above). Thanks to those who helped. I appreciate your time.

推荐答案

我只是在做这在纸面上。我希望这是正确的。

I'm just doing this on paper. I hope it's right.

让我们假设时望向归一，也就是，长度为1比方说，平面P包含原点，L为正常。 Y是（0，1，0）

Let's assume Up and Look are normalized, that is, length 1. Let's say that plane P contains the origin, and L is its normal. Y is (0, 1, 0)

要投射Ÿ到P，求其距离P ...

To project Y onto P, find its distance to P...

d = Y dot L = ly

...然后缩放正常通过-d得到Y1（即，Y对P的投影）

...and then scale the normal by -d to get the Y1 (that is, the projection of Y on P)

Y1 = (lx * ly, ly * ly, lz * ly)

现在归Y1，即，（1 /长度）缩放。如果它的长度为0，那么你的运气了。

Now normalize Y1, that is, scale it by (1 / length). If its length was 0 then you're out of luck.

Y1和向上的点积=的角度的余弦值。因此，

The dot product of Y1 and Up = the cosine of the angle. So

angle = acos(Y1 dot Up)