Shortest Distance Between Two Skew Lines in 3D

We have two lines in 3D space given in vector form:

$$\vec r_1 = \vec a_1 + \lambda \vec b_1, \qquad \vec r_2 = \vec a_2 + \mu \vec b_2$$

• $\vec a_1, \vec a_2$ are fixed position vectors (points on each line).
• $\vec b_1, \vec b_2$ are direction vectors of the lines.
• $\lambda, \mu$ are real parameters.

We want the shortest distance between these two lines when they are skew (not parallel, and they do not intersect).

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Geometric Idea

In 3D, two lines can be:

1. Intersecting – they cross at some point.
2. Parallel – same direction, possibly different positions.
3. Skew – not parallel and do not intersect; they are “offset” in 3D.

For skew lines, the shortest segment connecting them:

• Is perpendicular to both lines.
• Lies along the direction of the cross product of their direction vectors.

Let

$$\vec n = \vec b_1 \times \vec b_2$$

Then $\vec n$ is perpendicular to both $\vec b_1$ and $\vec b_2$. The shortest segment between the lines must be parallel to $\vec n$.

We also have the vector between two reference points (one on each line):

$$\vec d = \vec a_2 - \vec a_1$$

This $\vec d$ connects a point on line 1 to a point on line 2.

The component of $\vec d$ along $\vec n$ gives the shortest distance between the lines, because that’s the part of $\vec d$ that is perpendicular to both lines.

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Deriving the Formula

1. Compute the normal vector to both lines:

$$\vec n = \vec b_1 \times \vec b_2$$

• If $\vec n = \vec 0$ then $\vec b_1$ and $\vec b_2$ are parallel, and the formula below does not apply (handle the parallel case separately).

2. The signed scalar projection of $\vec d = \vec a_2 - \vec a_1$ on $\vec n$ is:

$$\text{proj length} = \frac{\vec d \cdot \vec n}{\lVert \vec n \rVert}$$

3. The shortest distance is the absolute value of this projection:

$$\text{Distance} = \left|\text{proj length}\right| = \frac{\left|\vec d \cdot \vec n\right|}{\lVert \vec n \rVert}$$

Substituting $\vec d = \vec a_2 - \vec a_1$ and $\vec n = \vec b_1 \times \vec b_2$:

$$\boxed{\text{Distance} = \dfrac{\left|\,(\vec a_2 - \vec a_1) \cdot (\vec b_1 \times \vec b_2)\,\right|}{\left\|\vec b_1 \times \vec b_2\right\|}}$$

This is exactly the formula in the tip.

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Why Students Mix Up Parallel and Skew

• Parallel lines in 3D share the same direction ($\vec b_1$ is a scalar multiple of $\vec b_2$), so their cross product is zero: $\vec b_1 \times \vec b_2 = \vec 0$.
• Skew lines have $\vec b_1 \times \vec b_2 \neq \vec 0$ and do not intersect.

The formula uses $\vec b_1 \times \vec b_2$ in both numerator and denominator. If the lines are parallel, the denominator becomes zero, and the formula breaks down — this is a signal that you must treat the parallel case separately (distance from a point to a line).

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Visual Interpretation (What the Visualization Shows)

The visualization animates the following:

• Two skew lines in 3D projected onto the 2D canvas.
• A moving point on each line (controlled by parameters $\lambda$ and $\mu$).
• The shortest connecting segment between the lines, which is perpendicular to both:
  • This segment aligns with $\vec n = \vec b_1 \times \vec b_2$.
  • Its length equals the formula:

$$\text{Distance} = \frac{|(\vec a_2 - \vec a_1) \cdot (\vec b_1 \times \vec b_2)|}{\lVert \vec b_1 \times \vec b_2 \rVert}$$

By adjusting direction vectors and positions, you can see:

• When $\vec b_1$ and $\vec b_2$ get closer to parallel, the cross product shrinks and the geometry degenerates.
• How the shortest segment "slides" along the lines but always stays perpendicular to both.

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Step‑by‑Step Use of the Formula

1. Identify $\vec a_1, \vec a_2, \vec b_1, \vec b_2$ from the line equations.
2. Compute:
   • $\vec d = \vec a_2 - \vec a_1$
   • $\vec n = \vec b_1 \times \vec b_2$
3. Check $\vec n \neq \vec 0$ (if zero, lines are parallel).
4. Compute the dot product $\vec d \cdot \vec n$.
5. Compute $\lVert \vec n \rVert$.
6. Plug into:

$$\text{Distance} = \frac{|\vec d \cdot \vec n|}{\lVert \vec n \rVert}$$

This is the minimal distance between the two skew lines.