Shortest distance
\[d=\frac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}\]
Three Dimensional Geometry
Shortest Distance Between Two Lines
Understand
Lines in space can be intersecting, parallel, or skew (neither). The shortest distance between two lines is the length of their common perpendicular.
\[\vec r=\vec a_1+\lambda\vec b_1,\ \vec r=\vec a_2+\mu\vec b_2:\qquad d=\left|\frac{(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)}{|\vec b_1\times\vec b_2|}\right|\]
If \(d=0\), the lines intersect. For parallel lines \(\vec r=\vec a_1+\lambda\vec b\), \(\vec r=\vec a_2+\mu\vec b\): \(d=\frac{|\vec b\times(\vec a_2-\vec a_1)|}{|\vec b|}\).
Distance of a point from a line / foot of the perpendicular: take the general point \(Q\) of the line, impose \(\overrightarrow{PQ}\cdot\vec b=0\), solve for \(\lambda\).
Key Concepts
- Skew: not parallel and not intersecting.
- SD formula uses b₁ × b₂.
- d = 0 ⇔ the lines intersect (if not parallel).
- Parallel lines: |b × (a₂ − a₁)| / |b|.
Formula Bank
Key Points
Intersecting lines
Two non-parallel lines intersect exactly when their shortest distance is 0 (they are coplanar).
Solved Examples
Shortest distance
\(\vec r=(\hat i+\hat k)+\lambda(\hat i+\hat j)\) and \(\vec r=(2\hat j-\hat k)+\mu(\hat j+\hat k)\).
\(\vec b_1\times\vec b_2=\hat i-\hat j+\hat k\), \(\vec a_2-\vec a_1=-\hat i+2\hat j-2\hat k\); \(d=\frac{|-1-2-2|}{\sqrt3}=\frac{5}{\sqrt3}\).
Practice & Topic Test
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Compute b₁ × b₂, dot with a₂ − a₁, divide by |b₁ × b₂|; zero means the lines meet.
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