3D 空间中两条线之间的最短距离 | 12 年级数学
在 3D 空间中，两条线可以在某个点相互交叉、相互平行，或者它们既不交叉也不相互平行，也称为斜线。
- 在相交线的情况下，它们之间的最短距离为 0。
- 对于平行线，连接两条平行线的线的长度或垂直于两条平行线的线的长度具有最短距离。
- 在斜线的情况下，最短距离是垂直于两条给定线的线。
Note: The alphabets written in bold represent vector. ‘x’ denotes cross product(vector product).
两条平行线之间的最短距离
将向量形式的 2 行视为：
v1 = a1 + c * b
v2 = a2 + d * b
这里，c 和 d 是常数。
b =向量v1和v2的平行向量
a1, a2分别是v1和v2上某个点的位置向量
Shortest distance = |b x (a2 – a1)| / |b|
例子
示例1：对于 3D 空间中的以下行。
v1 = i – 2j + i – j + k
v2 = i – 3j + k + i – j + k
找到这些线之间的最短距离？
解决方案：
v1: i – 2j + i – j + k
v2: i – 3j + k + i – j + k
b = i – j + k
a1 = i -2j
a2 = i – 3j + k
a2 – a1 = -j + k
|b| = √3 = 1.73
|b x (a2 – a1)| = √2 = 1.41
Shortest distance = |b x (a2 – a1)|/|b| = 1.41/1.73 = 0.815
示例 2：对于 3D 空间中的以下行。
v1 = i – j – k + 2i – 3j + 4k
v2 = 2i – 3j + k + 6i – 9j + 12k
找到这些线之间的最短距离？
解决方案：
The vector can be written in form as:
v1 = i – j – k + 2i – 3j + 4k
v2 = 2i – 3j + k + 3 * (2i – 3j + 4k)
b = 2i – 3j + 4k
|b| = √(2)2 + (-3)2+ (4)2 = 5.385
a1 = i – j – k
a2 = 2i – 3j + k
a2 – a1 = i – 2j + 2k
b x (a2 – a1) = 2i – k
|b x (a2 – a1)| = √(2)2 + (1)2 = 2.236
Now applying the shortest distance formula for parallel lines = |b x (a2 – a1)|/|b| = 2.236/5.385 = 0.415
示例 3：给定笛卡尔格式的两行：
V1: (x – 2)/2 = (y – 1)/3 = (z)/4
V2：(x – 3)/4 = (y – 2)/6 = (z – 5)/8
找出这些线之间的最短距离。
解决方案：
The displacement vector of V1 is 2i + 3j + 4k, for V2 is 4i + 6j + 8k
The displacement vector V2 is a multiple of V1 as,
4i + 6j + 8k = 2 * (2i + 3j + 4k)
So the two given lines are parallel to each other.
a1 = 2i + j + 0k
a2 = 3i + 2j + 5k
a2 – a1 = i + j +5k
b = 2i + 3j + 4k
|b| = √(2)2 + (3)2 + (4)2 = 5.385
b x (a2 – a1) = 11i – 6j – k
|b x (a2 – a1)| = 12.569
shortest distance = |b x (a2 – a1)|/|b| = 12.569/5.385 = 2.334
斜线之间的最短距离
将向量形式的 2 行视为：
v1 = a1 + c * b1
v2 = a2 + d * b2
这里，c 和 d 是常数。
The shortest distance 2 skew lines = |(b1 x b2)(a2 – a1)|/|(b1 x b2)|
Note: If two lines are intersecting then’s the shortest distance considering the two lines skew will automatically come out to be zero.
例子
示例 1：给定向量形式的两条线：
V1: i - j + 2i + j + k
V2: i + j + 3i – j – k
找出这些线之间的最短距离。
解决方案：
The given lines are skew lines.
b1 = 2i + j + k
b2 = 3i – j – k
a2 = i + j
a1 = i – j
a2 – a1 = 2j
(b1 x b2) = 5j – 5k
Shortest distance = |(b1 x b2)(a2 – a1)|/|(b1 x b2)| = 10/7.07 = 1.41
示例 2：给定向量形式的两条线：
V1：2i – j + 5 * (3i – j + 2k)
v2: i – j + 2k + 2* (i + 3j + 4k)
找出这些线之间的最短距离。
解决方案：
The given lines are skew lines.
Shortest distance = |(b1 x b2)(a2 – a1)|/|(b1 x b2)|
b1 = 3i – j + 2k
b2 = i + 3j + 4k
a1 = 2i – j
a2 = i – j + 2k
a2 – a1 = -i + 2k
(b1 x b2) = -10i – 10j + 10k
|b1 x b2| = 17.320
|(b1 x b2)(a2 – a1)| = 40
Shortest distance = 40/17.320 = 2.309
示例 3：给定 2 条笛卡尔形式的线，找出它们之间的最短距离。
V1: (x – 1)/2 = (y – 1)/3 = (z)/4
V2：(x)/1 = (y – 2)/2 = (z – 1)/3
解决方案：
a1 = i + j
a2 = -2j + k
b1 = 2i + 3j + 4k
b2 = i + 2j + 3k
a2 – a1 = -3i – j + k
(b1 x b2) = i – 2j + k
|b1 x b2| = 2.44
Shortest distance =|(i – 2j + k)( -3i – j + k)|/2.44 = 0
Since, shortest distance is zero it means these two lines are intersecting lines.