Shortest distance between two lines
The shortest distance between skew lines is the length of their common perpendicular: d = |(b⃗₁ × b⃗₂) · (a⃗₂ − a⃗₁)| / |b⃗₁ × b⃗₂|.
What shortest distance means
Join a point of one line to a point of the other. Of all such segments, one is the shortest. It is perpendicular to both lines, so it is called the common perpendicular. Its length is the shortest distance (SD). For intersecting lines the SD is 0. For skew lines it is positive. For parallel lines it is the constant gap between them.
How to draw the figure
- Draw two lines l₁ and l₂ that do not meet, one slanting across the other.
- Mark the known points S (position vector a⃗₁) on l₁ and T (position vector a⃗₂) on l₂.
- Draw the segment PQ perpendicular to both lines, with P on l₁ and Q on l₂; mark the two right angles.
- Label the directions b⃗₁ along l₁ and b⃗₂ along l₂, and write 'PQ = d' on the common perpendicular.
- Join S to T with a dashed line: d is the projection of ST⃗ on PQ.
Derivation of the formula for skew lines
Let l₁: r⃗ = a⃗₁ + λb⃗₁ and l₂: r⃗ = a⃗₂ + μb⃗₂ be skew. S (position vector a⃗₁) is on l₁ and T (position vector a⃗₂) is on l₂.
- 1The common perpendicular PQ is perpendicular to both b⃗₁ and b⃗₂, so it is along b⃗₁ × b⃗₂.
- 2Unit vector along PQ: n̂ = (b⃗₁ × b⃗₂)/|b⃗₁ × b⃗₂|.
- 3ST⃗ = a⃗₂ − a⃗₁.
- 4PQ is the projection of ST⃗ on n̂, so d = |ST⃗ · n̂|.
- 5★So, d = |(b⃗₁ × b⃗₂) · (a⃗₂ − a⃗₁)| / |b⃗₁ × b⃗₂|.
★ key step
Example 1: skew lines in vector form
Find the shortest distance between r⃗ = (î − k̂) + λ(î + ĵ) and r⃗ = (3î + ĵ + k̂) + μ(ĵ + k̂).
- a⃗₁ = î − k̂, b⃗₁ = î + ĵ; a⃗₂ = 3î + ĵ + k̂, b⃗₂ = ĵ + k̂
- a⃗₂ − a⃗₁ = 2î + ĵ + 2k̂
- b⃗₁ × b⃗₂ = (1 × 1 − 0 × 1)î − (1 × 1 − 0 × 0)ĵ + (1 × 1 − 1 × 0)k̂ = î − ĵ + k̂
- |b⃗₁ × b⃗₂| = √(1 + 1 + 1) = √3
- (b⃗₁ × b⃗₂) · (a⃗₂ − a⃗₁) = 2 − 1 + 2 = 3
- d = |3|/√3 = √3
d = √3 units.
Example 2: skew lines in Cartesian form
Find the shortest distance between (x − 1)/2 = y/1 = (z + 1)/(−1) and x/1 = (y − 2)/1 = (z − 1)/2.
- (x₁, y₁, z₁) = (1, 0, −1) with DRs 2, 1, −1; (x₂, y₂, z₂) = (0, 2, 1) with DRs 1, 1, 2
- First row: x₂ − x₁, y₂ − y₁, z₂ − z₁ = −1, 2, 2
- Determinant = −1(1 × 2 − (−1) × 1) − 2(2 × 2 − (−1) × 1) + 2(2 × 1 − 1 × 1) = −3 − 10 + 2 = −11
- b₁c₂ − b₂c₁ = 2 + 1 = 3, c₁a₂ − c₂a₁ = −1 − 4 = −5, a₁b₂ − a₂b₁ = 2 − 1 = 1
- Denominator = √(9 + 25 + 1) = √35
- d = |−11|/√35 = 11/√35
d = 11/√35 units (about 1.86 units, rounded to 2 decimal places).
Example 3: parallel lines
Find the distance between the parallel lines r⃗ = (î + k̂) + λ(2î − ĵ + 2k̂) and r⃗ = (2î + ĵ − k̂) + μ(2î − ĵ + 2k̂).
- b⃗ = 2î − ĵ + 2k̂, |b⃗| = 3
- a⃗₂ − a⃗₁ = î + ĵ − 2k̂
- b⃗ × (a⃗₂ − a⃗₁) = ((−1)(−2) − 2 × 1)î − (2 × (−2) − 2 × 1)ĵ + (2 × 1 − (−1) × 1)k̂ = 0î + 6ĵ + 3k̂
- |b⃗ × (a⃗₂ − a⃗₁)| = √(36 + 9) = √45 = 3√5
- d = 3√5/3 = √5
d = √5 units.
Pipes and electric cables that cross at different heights in a factory or a metro tunnel run along skew lines. Engineers use the shortest distance formula to check the safe gap between them.
2 marks: “Find the distance between the parallel lines r⃗ = λ(î + 2ĵ + 2k̂) and r⃗ = (î + ĵ) + μ(î + 2ĵ + 2k̂).” b⃗ = î + 2ĵ + 2k̂, |b⃗| = 3, a⃗₂ − a⃗₁ = î + ĵ. b⃗ × (a⃗₂ − a⃗₁) = −2î + 2ĵ − k̂, with modulus 3. d = 3/3 = 1 unit.
3 marks: “Find the shortest distance between r⃗ = (ĵ + 2k̂) + λ(î + ĵ + k̂) and r⃗ = (2î + 3ĵ + k̂) + μ(î − ĵ).” b⃗₁ × b⃗₂ = î + ĵ − 2k̂, modulus √6. a⃗₂ − a⃗₁ = 2î + 2ĵ − k̂. Dot product = 2 + 2 + 2 = 6. d = 6/√6 = √6 units.
5 marks: “Find the shortest distance between the lines (x − 2)/1 = (y + 1)/2 = z/(−2) and x/2 = (y − 1)/1 = (z − 3)/2.” b⃗₁ × b⃗₂ = 6î − 6ĵ − 3k̂, modulus 9. a⃗₂ − a⃗₁ = −2î + 2ĵ + 3k̂. Dot product = −12 − 12 − 9 = −33. d = 33/9 = 11/3 units.
Cross the directions, dot the gap, divide by the cross.