Line of shortest distance between two skew lines
Beyond NCERT In the JEE Main syllabus; the Class 12 book has no section for it.
Quick answer
The line of shortest distance between two skew lines is their common perpendicular, the one line that meets both at right angles. Take P = a + λb and Q = a + μb, solve PQ · b = 0 and PQ · b = 0 for λ and μ, and P and Q are its feet, with d = |PQ|. Its equation is r = p + t(b × b), through P and parallel to b × b.
What is the line of shortest distance?
Line of shortest distance
For skew lines, the line of the shortest distance will be perpendicular to both the lines.
NCERT Mathematics Part II (Class 12), Section 11.5, page 386
- So the line of shortest distance is the common perpendicular of the two skew lines: the unique line that meets both at right angles. Its feet P and Q are the closest points.
The 2026–27 NCERT book finds the shortest distance between skew lines but not where it is. JEE Main 2027 lists skew lines, the shortest distance between them “and its equation” (Unit 11), so this page finds the segment itself. Apart from the NCERT lines cited, it is beyond the 2026–27 NCERT text.
In plain words: once the feet P and Q are known, the Class 11 distance formula gives the length PQ, and that is the shortest distance.
Finding the feet
Write a general point of each line: P = a + λb and Q = a + μb. The segment PQ must be perpendicular to both lines:
Feet of the common perpendicular
PQ · b = 0 and PQ · b = 0
two linear equations in λ and μ; then P, Q and d = |PQ|
The answer can be checked two ways: PQ must be parallel to b × b, and |PQ| must equal |(b × b) · (a − a)| / |b × b|.
The equation of the line of shortest distance
The line of shortest distance passes through P (or Q) with direction PQ, which is parallel to b × b:
Line of shortest distance
r = p + t(b × b), where p is the position vector of P
In Cartesian form, use the coordinates of P and the components of b × b (or of PQ) as direction ratios.
Solved examples
Example 1: An axis and a slanting line
Find the line of shortest distance between the x-axis and the line through (0, 2, 3) parallel to ĵ + k̂.
P = (λ, 0, 0) and Q = (0, 2 + μ, 3 + μ), so PQ = −λî + (2 + μ)ĵ + (3 + μ)k̂. PQ · î = 0 gives λ = 0; PQ · (ĵ + k̂) = 0 gives 5 + 2μ = 0, μ = −5/2. So P = O and Q = (0, −1/2, 1/2).
Answer: d = |PQ| = 1/√2, and the line of shortest distance is r = t(−ĵ + k̂), that is, x = 0 and y + z = 0.
Example 2: Feet with fractions (the lines of NCERT Exercise 11.2, Q14, page 390)
Find the feet of the common perpendicular to r = î + 2ĵ + 3k̂ + λ(î − 3ĵ + 2k̂) and r = 4î + 5ĵ + 6k̂ + μ(2î + 3ĵ + k̂), and the line of shortest distance.
With w = a − a = 3î + 3ĵ + 3k̂, PQ = w + μb − λb = (3 + 2μ − λ)î + (3 + 3μ + 3λ)ĵ + (3 + μ − 2λ)k̂. Then PQ · b = −14λ − 5μ and PQ · b = 18 + 5λ + 14μ, so the two conditions are 14λ + 5μ = 0 and −5λ − 14μ = 18. So λ = 10/19 and μ = −28/19, and P = (29/19, 8/19, 77/19), Q = (20/19, 11/19, 86/19).
Answer: PQ = (3/19)(−3î + ĵ + 3k̂), so d = 3/√19, which the formula |(b × b) · (a − a)| / |b × b| also gives; the line is (x − 29/19)/(−3) = (y − 8/19)/1 = (z − 77/19)/3.
Example 3: The given points are the feet (the lines of NCERT Exercise 11.2, Q13, page 390)
Find the feet of the common perpendicular to (x + 1)/7 = (y + 1)/(−6) = (z + 1)/1 and (x − 3)/1 = (y − 5)/(−2) = (z − 7)/1.
From the points A(−1, −1, −1) and B(3, 5, 7), AB = 4î + 6ĵ + 8k̂. Its dot products with 7î − 6ĵ + k̂ and with î − 2ĵ + k̂ are 28 − 36 + 8 = 0 and 4 − 12 + 8 = 0. So λ = μ = 0: A and B are themselves the feet.
Answer: P = (−1, −1, −1), Q = (3, 5, 7), d = √116 = 2√29, and the line of shortest distance is (x + 1)/2 = (y + 1)/3 = (z + 1)/4.
Common mistakes to avoid
Common mistake
Using one parameter for both lines
P and Q move independently, so they need two parameters, λ and μ. Writing both with λ forces them to move together and usually gives no solution.
Common mistake
Making PQ perpendicular to only one line
Both conditions are needed. With only PQ · b = 0, Q could be any point of the second line, and PQ is just a perpendicular from Q to the first line.
Common mistake
Forgetting to check
Check that PQ is parallel to b × b and that |PQ| equals the formula distance; an arithmetic slip shows up at once (Example 2).
Quick revision summary
- The line of shortest distance between skew lines is their common perpendicular; its feet P and Q are the closest points.
- With P = a + λb and Q = a + μb, solve PQ · b = 0 and PQ · b = 0 for λ and μ.
- Then d = |PQ|, and the line is r = p + t(b × b) through P.
Formula sheet
- PQ · b = 0 and PQ · b = 0
- d = |PQ| = |(b × b) · (a − a)| / |b × b|
- r = p + t(b × b)
Questions students ask
How do you find the foot of the perpendicular from a point to a line in 3D?
Beyond the 2026–27 NCERT text, the foot of the perpendicular from a point P to the line r = a + λb is its point Q for which PQ · b = 0, one linear equation in λ; then |PQ| is the distance of P from the line, also |b × (a − p)|/|b| with p the position vector of P, and with P at O it is the line’s distance from the origin. For P(−1, 3, 2) and the line through (−2, −1, 1) with direction ratios 1, −2, −2, the general point gives PQ · b = 9λ + 9 = 0, so λ = −1, the foot is Q(−3, 1, 3) and the distance is |−2î − 2ĵ + k̂| = 3. The image of P in the line is (−5, −1, 4), since Q is the mid-point of P and its image.
Why is the shortest distance between skew lines measured along their common perpendicular?
The shortest distance between two skew lines is measured along their common perpendicular because the shortest segment PQ, with P on the first line and Q on the second, must be perpendicular to both, as NCERT states (Section 11.5, page 386): otherwise moving Q, or P, a little along its line in the right direction would shorten it. In calculus terms, for a fixed P the square |PQ| is a quadratic in μ with derivative 2PQ · b, so it is least where PQ · b = 0, and for a fixed Q it is least where PQ · b = 0. The closest pair must satisfy both, and these are exactly the two conditions for the feet on this page.
What happens to the line of shortest distance when the two lines intersect?
Beyond the NCERT text, when two non-parallel lines intersect, the two conditions PQ · b = 0 and PQ · b = 0 give P = Q, the common point, so the shortest distance is 0 and there is no segment between two feet. The line through that point along b × b still meets both lines at right angles, and it is the only such line. For the line through (5, −2, 1) with direction ratios 2, −3, 1 and the line through (−1, 2, 5) with direction ratios 1, 1, −3, the conditions give λ = −2 and μ = 2, both feet are the common point (1, 4, −1), and the perpendicular through it has direction ratios 8, 7, 5, from b × b.
Practice questions
Try each question first, then open the answer to check your working.
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Q1. Find the feet of the common perpendicular to r = î + λ(î + ĵ) and r = 2î + μ(ĵ + k̂) (the skew lines of practice Q3 on the shortest-distance page).
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PQ = (1 − λ)î + (μ − λ)ĵ + μk̂. The conditions are (1 − λ) + (μ − λ) = 0 and (μ − λ) + μ = 0, so λ = 2/3 and μ = 1/3. P = (5/3, 2/3, 0) and Q = (2, 1/3, 1/3), with PQ = (î − ĵ + k̂)/3 and d = 1/√3.
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Q2. Find the line of shortest distance between the z-axis and the line through (2, 0, 1) parallel to î + ĵ.
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P = (0, 0, λ) and Q = (2 + μ, μ, 1). PQ · k̂ = 0 gives λ = 1, and PQ · (î + ĵ) = 2 + 2μ = 0 gives μ = −1. So P = (0, 0, 1), Q = (1, −1, 1) and d = √2, and the line is r = k̂ + t(î − ĵ), that is, z = 1 and x + y = 0.
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Q3. Why is the direction of the line of shortest distance always parallel to b × b?
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It is perpendicular to both b and b, and every vector perpendicular to two non-parallel vectors is a multiple of their cross product.
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Q4. For the lines of NCERT Exercise 11.2, Q12 (page 390), r = î + 2ĵ + k̂ + λ(î − ĵ + k̂) and r = 2î − ĵ − k̂ + μ(2î + ĵ + 2k̂), find the feet of the common perpendicular.
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The conditions give λ = 11/6 and μ = 7/6, so P = (17/6, 1/6, 17/6) and Q = (13/3, 1/6, 4/3). Then PQ = (3/2)î − (3/2)k̂, of length 3/√2, as the formula gives.