3d 空间中 3 个点之间的角度 [英] Angle between 3 points in 3d space

问题描述

我有 3 个包含 X、Y、Z 坐标的点:

I have 3 points containing X, Y, Z coordinates:

var A = {x: 100, y: 100, z: 80},
    B = {x: 100, y: 175, z: 80},
    C = {x: 100, y: 100, z: 120};

坐标是来自 3d CSS 变换的像素.如何获得向量 BA 和 BC 之间的角度?一个数学公式就行了，JavaScript 代码会更好.谢谢.

The coordinates are pixels from a 3d CSS transform. How can I get the angle between vectors BA and BC? A math formula will do, JavaScript code will be better. Thank you.

推荐答案

在伪代码中，向量 BA(称为 v1)是:

In pseudo-code, the vector BA (call it v1) is:

v1 = {A.x - B.x, A.y - B.y, A.z - B.z}

同样，向量 BC(称为 v2)是:

Similarly the vector BC (call it v2) is:

v2 = {C.x - B.x, C.y - B.y, C.z - B.z}

v1 和 v2 的点积是它们之间夹角余弦的函数(按它们的大小乘积缩放).所以首先规范化v1和v2:

The dot product of v1 and v2 is a function of the cosine of the angle between them (it's scaled by the product of their magnitudes). So first normalize v1 and v2:

v1mag = sqrt(v1.x * v1.x + v1.y * v1.y + v1.z * v1.z)
v1norm = {v1.x / v1mag, v1.y / v1mag, v1.z / v1mag}

v2mag = sqrt(v2.x * v2.x + v2.y * v2.y + v2.z * v2.z)
v2norm = {v2.x / v2mag, v2.y / v2mag, v2.z / v2mag}

然后计算点积:

res = v1norm.x * v2norm.x + v1norm.y * v2norm.y + v1norm.z * v2norm.z

最后，恢复角度:

angle = acos(res)

这篇关于3d 空间中 3 个点之间的角度的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！