MathsCBSE Class 12 · Three Dimensional Geometry

Shortest Distance Between Two Lines: Skew, Parallel and Intersecting (Class 12)

Updated 15 October 2026 · Fact-checked

For lines r = a₁ + λb₁ and r = a₂ + μb₂ that are not parallel, the shortest distance is |(a₂ − a₁) · (b₁ × b₂)| / |b₁ × b₂|. If it is zero, the lines intersect; otherwise they are skew. For parallel lines with direction b, it is |(a₂ − a₁) × b| / |b|.

Understand Shortest Distance Between Two Lines

Two lines in space can be parallel, intersecting, or skew (neither parallel nor meeting, like a road and an overpass going in different directions). The shortest distance between two lines is the length of the segment perpendicular to both.

Skew (non-parallel) lines. For $r = a_{1} + λ b_{1}$ and $r = a_{2} + μ b_{2}$ with $b_{1} \times b_{2} \neq = 0$ , the common perpendicular has direction $b_{1} \times b_{2}$ . The shortest distance is the projection of $a_{2} - a_{1}$ (any segment joining the lines) on that direction:
$d = \frac{( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} )}{∣ b _{1} \times b _{2} ∣}$ .
In Cartesian form, with points $(x_{1}, y_{1}, z_{1})$ , $(x_{2}, y_{2}, z_{2})$ and direction ratios $(a_{1}, b_{1}, c_{1})$ , $(a_{2}, b_{2}, c_{2})$ , the numerator is the determinant
$x_{2} - x_{1} a_{1} a_{2} y_{2} - y_{1} b_{1} b_{2} z_{2} - z_{1} c_{1} c_{2}$ .

Intersecting lines. If $d = 0$ , the (non-parallel) lines meet. To find the point, equate the general points $a_{1} + λ b_{1} = a_{2} + μ b_{2}$ , solve two of the component equations for $λ$ and $μ$ , and check the third. If the third fails, the lines do not meet (and $d$ would not have been 0).

Parallel lines. If $b_{1} ∥ b_{2}$ (take the common direction $b$ ), the cross product is zero and the formula above breaks down. Use
$d = \frac{∣ ( a _{2} - a _{1} ) \times b ∣}{∣ b ∣}$ ,
which is the distance from any point of one line to the other line. If this is 0, the lines coincide.

Order of work. Always check the directions first: parallel or not? This decides the formula. Then compute $a_{2} - a_{1}$ , the cross product and its magnitude, and the dot product. Keep the answer as a surd fraction like $\frac{17}{35}$ (rationalising is optional).

Shortest distance is one of the most-asked 5-mark 3D questions: it appeared in the 2023, 2024, 2025 and 2026 board papers and in the sample papers (NCERT Section 11.5).

Shortest Distance Between Two Lines: formulas to remember

Skew lines (vector form): $d = \frac{( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} )}{∣ b _{1} \times b _{2} ∣}$; Valid when $b_{1}$ and $b_{2}$ are not parallel.
Skew lines (Cartesian numerator): $x_{2} - x_{1} a_{1} a_{2} y_{2} - y_{1} b_{1} b_{2} z_{2} - z_{1} c_{1} c_{2}$; Divide its absolute value by $∣ b_{1} \times b_{2} ∣$ .
Parallel lines: $d = \frac{∣ ( a _{2} - a _{1} ) \times b ∣}{∣ b ∣}$; $b$ is the common direction.
Intersection test: $(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = 0$; For non-parallel lines, they then intersect.
Cross product magnitude: $∣ b_{1} \times b_{2} ∣ = (b_{1} c_{2} - c_{1} b_{2})^{2} + (c_{1} a_{2} - a_{1} c_{2})^{2} + (a_{1} b_{2} - b_{1} a_{2})^{2}$; Compute the components carefully.
Meeting point: $a_{1} + λ b_{1} = a_{2} + μ b_{2}$; Solve two components, check the third.

How to solve Shortest Distance Between Two Lines questions

The 5-mark routine for shortest-distance and intersection questions.

1Write each line in vector form; read $a_{1}, b_{1}, a_{2}, b_{2}$ (clean the Cartesian forms first).
2Check whether $b_{1}$ and $b_{2}$ are parallel (proportional components).
3If not parallel: find $a_{2} - a_{1}$ , $b_{1} \times b_{2}$ and $∣ b_{1} \times b_{2} ∣$ ; compute $d = \frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}$ .
4If parallel: compute $d = \frac{∣ ( a _{2} - a _{1} ) \times b ∣}{∣ b ∣}$ .
5If $d = 0$ for non-parallel lines, find the intersection by equating general points.
6Conclude: 'the lines are skew', 'intersect at ...' or 'are parallel at distance ...'.

Shortest Distance Between Two Lines in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Clearance between skew lines

Suppose a straight road runs along the x-axis at ground level, and a straight flyover passes above it along the line through $(0, 0, 6)$ in the direction $\hat{i} + \hat{j}$ (in metres).

Find the shortest distance between the road and the flyover (the vertical clearance).

Try it first, then show the solutionSolution

$a_{1} = 0$ , $b_{1} = \hat{i}$ ; $a_{2} = 6 \hat{k}$ , $b_{2} = \hat{i} + \hat{j}$ .
$b_{1} \times b_{2} = \hat{k}$ , magnitude 1.
$d = ∣6 \hat{k} \cdot \hat{k} ∣ = 6$ m.

Answer: 6 m

Memory hook: The common perpendicular of skew lines is the shortest gap, like the clearance under a flyover.

Do two paths meet?

Imagine two drones moving along $r = t (\hat{i} + \hat{j} + \hat{k})$ and $r = 4 \hat{i} + s (- \hat{i} + \hat{j} + \hat{k})$ (in metres).

Do their paths cross? If so, where?

Try it first, then show the solutionSolution

$b_{1} \times b_{2} = (1, 1, 1) \times (- 1, 1, 1) = (0, - 2, 2)$ ; $a_{2} - a_{1} = (4, 0, 0)$ .
Dot product $= 0$ , so the shortest distance is 0 and the paths intersect.
$t = 4 - s$ , $t = s$ , $t = s$ give $t = s = 2$ : the point $(2, 2, 2)$ .

Answer: Yes; the paths cross at $(2, 2, 2)$ (a collision is possible if both drones arrive there together).

Memory hook: Zero shortest distance means the paths meet.

Tricks to remember Shortest Distance Between Two Lines

Short memory aids. Say each one aloud once, then use it on a question the same day.

Join, cross, dot, divide
Join the points ( $a_{2} - a_{1}$ ), cross the directions, dot the two, divide by the cross's length.
Use it when: Skew-line formula.
Parallel? Cross the join with b
For parallel lines, the cross product of directions is zero; use $\frac{∣ ( a _{2} - a _{1} ) \times b ∣}{∣ b ∣}$ instead.
Use it when: Choosing the right formula.
Zero means meet
Non-parallel lines with shortest distance 0 intersect; find the point by equating general points.
Use it when: Intersection questions.
Two equations, then check the third
Solve for λ and μ from two components and verify with the third.
Use it when: Avoiding false intersection points.

Shortest Distance Between Two Lines: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Shortest distance between skew lines · 3 marks · Short answerBoard level

Find the shortest distance between $r = (\hat{i} + 2 \hat{k}) + λ (\hat{i} + 2 \hat{j} + \hat{k})$ and $r = (3 \hat{i} + \hat{j}) + μ (2 \hat{i} - \hat{j} + \hat{k})$ .

Try it first, then show the solutionSolution

$a_{2} - a_{1} = 2 \hat{i} + \hat{j} - 2 \hat{k}$ . (½ mark)
$b_{1} \times b_{2} = \hat{i} (2 + 1) - \hat{j} (1 - 2) + \hat{k} (- 1 - 4) = 3 \hat{i} + \hat{j} - 5 \hat{k}$ , magnitude $35$ . (1 mark)
$(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = 6 + 1 + 10 = 17$ . (1 mark)
$d = \frac{17}{35}$ units. (½ mark)

Answer: $\frac{17}{35}$

Example 2: Distance between parallel lines · 3 marks · Short answerBoard level

Find the distance between the parallel lines $r = (\hat{i} + \hat{j}) + λ (2 \hat{i} - \hat{j} + 2 \hat{k})$ and $r = (2 \hat{i} + \hat{j} - \hat{k}) + μ (2 \hat{i} - \hat{j} + 2 \hat{k})$ .

Try it first, then show the solutionSolution

Common direction $b = 2 \hat{i} - \hat{j} + 2 \hat{k}$ , $∣ b ∣ = 3$ ; $a_{2} - a_{1} = \hat{i} - \hat{k}$ . (1 mark)
$(\hat{i} - \hat{k}) \times b = \hat{i} (0 - 1) - \hat{j} (2 + 2) + \hat{k} (- 1 - 0) = - \hat{i} - 4 \hat{j} - \hat{k}$ , magnitude $18 = 32$ . (1 mark)
$d = \frac{3 2}{3} = 2$ units. (1 mark)

Answer: $2$

Example 3: Cartesian lines that are skew · 5 marks · Long answerHarder board level

Find the shortest distance between the lines $\frac{x}{1} = \frac{y - 1}{1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y}{- 1} = \frac{z - 1}{1}$ .

Try it first, then show the solutionSolution

$a_{1} = (0, 1, - 1)$ , $b_{1} = (1, 1, 2)$ ; $a_{2} = (2, 0, 1)$ , $b_{2} = (1, - 1, 1)$ . Directions are not proportional, so the lines are not parallel. (1 mark)
$a_{2} - a_{1} = (2, - 1, 2)$ . (½ mark)
$b_{1} \times b_{2} = \hat{i} (1 + 2) - \hat{j} (1 - 2) + \hat{k} (- 1 - 1) = 3 \hat{i} + \hat{j} - 2 \hat{k}$ , magnitude $14$ . (1½ marks)
Dot product $= 6 - 1 - 4 = 1$ . (1 mark)
$d = \frac{1}{14}$ units; non-zero, so the lines are skew. (1 mark)

Answer: $\frac{1}{14}$ ; the lines are skew.

Example 4: Lines that intersect · 5 marks · Long answerHarder board level

Show that the lines $r = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} + \hat{j} + \hat{k})$ and $r = (4 \hat{j} + 4 \hat{k}) + μ (2 \hat{i} - \hat{j})$ intersect, and find the point of intersection.

Try it first, then show the solutionSolution

$b_{1} \times b_{2} = \hat{i} (0 + 1) - \hat{j} (0 - 2) + \hat{k} (- 1 - 2) = \hat{i} + 2 \hat{j} - 3 \hat{k}$ . (1 mark)
$a_{2} - a_{1} = - \hat{i} + 2 \hat{j} + \hat{k}$ ; dot product $= - 1 + 4 - 3 = 0$ . (1 mark)
The lines are not parallel and the shortest distance is 0, so they intersect. (½ mark)
Equate: $1 + λ = 2 μ$ , $2 + λ = 4 - μ$ , $3 + λ = 4$ . The third gives $λ = 1$ , then $μ = 1$ ; the second: $3 = 3$ ✓. (1½ marks)
Point: $(1 + 1, 2 + 1, 3 + 1) = (2, 3, 4)$ . (1 mark)

Answer: They intersect at $(2, 3, 4)$ .

Example 5: Finding k so that lines meet · 5 marks · Long answerHOTS / competency

Find $k$ so that the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, and find their point of intersection.

Try it first, then show the solutionSolution

$a_{1} = (1, - 1, 1)$ , $b_{1} = (2, 3, 4)$ ; $a_{2} = (3, k, 0)$ , $b_{2} = (1, 2, 1)$ ; $b_{1} \times b_{2} = (3 - 8, 4 - 2, 4 - 3) = (- 5, 2, 1)$ . (1½ marks)
$a_{2} - a_{1} = (2, k + 1, - 1)$ ; dot product $= - 10 + 2 k + 2 - 1 = 2 k - 9$ . (1 mark)
Intersect when $2 k - 9 = 0$ : $k = \frac{9}{2}$ . (½ mark)
Equate general points: $1 + 2 λ = 3 + μ$ , $- 1 + 3 λ = \frac{9}{2} + 2 μ$ , $1 + 4 λ = μ$ . From the first and third: $1 + 2 λ = 4 + 4 λ \Rightarrow λ = - \frac{3}{2}$ , $μ = - 5$ ; second: $- \frac{11}{2} = - \frac{11}{2}$ ✓. (1½ marks)
Point: $(1 - 3, - 1 - \frac{9}{2}, 1 - 6) = (- 2, - \frac{11}{2}, - 5)$ . (½ mark)

Answer: $k = \frac{9}{2}$ ; point $(- 2, - \frac{11}{2}, - 5)$

Example 6: Case study: two flight paths · 4 marks · Case-basedHOTS / competency

Suppose two small aircraft fly along straight paths $r = \hat{k} + λ (\hat{i} + \hat{j})$ and $r = (\hat{i} + 3 \hat{k}) + μ (\hat{j} + \hat{k})$ (in km). Air-traffic rules need paths to stay at least 1 km apart.
(i) Show that the paths are not parallel. (1 mark)
(ii) Find $b_{1} \times b_{2}$ . (1 mark)
(iii) Find the shortest distance between the paths. (1 mark)
(iv) Do the paths satisfy the rule? (1 mark)

Try it first, then show the solutionSolution

(i) Directions $(1, 1, 0)$ and $(0, 1, 1)$ are not proportional. (1 mark)
(ii) $b_{1} \times b_{2} = \hat{i} (1 - 0) - \hat{j} (1 - 0) + \hat{k} (1 - 0) = \hat{i} - \hat{j} + \hat{k}$ , magnitude $3$ . (1 mark)
(iii) $a_{2} - a_{1} = \hat{i} + 2 \hat{k}$ ; dot product $= 1 + 0 + 2 = 3$ ; $d = \frac{3}{3} = 3 \approx 1.73$ km. (1 mark)
(iv) $3 > 1$ , so yes, the paths are always at least 1.73 km apart. (1 mark)

Answer: (ii) $\hat{i} - \hat{j} + \hat{k}$ (iii) $3$ km (iv) yes.

Where students lose marks in Shortest Distance Between Two Lines

The slips examiners see most often, and what to write instead.

Mistake: Using the skew formula for parallel lines
For parallel lines $b_{1} \times b_{2} = 0$ , so the formula divides by zero.
Wrong working
Parallel lines: $d = \frac{0}{0}$ , 'distance 0'.
Right working
$d = \frac{∣ ( a _{2} - a _{1} ) \times b ∣}{∣ b ∣} = 2$ .
Do this: Check for parallel directions first.
Usually costs: The whole question.
Mistake: Cross-product sign errors
The middle component of the cross product carries a minus sign.
Wrong working
$b_{1} \times b_{2} = 3 \hat{i} - \hat{j} - 5 \hat{k}$ .
Right working
$= 3 \hat{i} + \hat{j} - 5 \hat{k}$ (since $- (1 - 2) = + 1$ ).
Do this: Check that $b_{1} \times b_{2}$ is perpendicular to both $b_{1}$ and $b_{2}$ .
Usually costs: The numerator, and the answer.
Mistake: Reading points from unclean Cartesian forms
$\frac{y}{- 1}$ means the point has $y = 0$ and the ratio is $- 1$ .
Wrong working
$\frac{x - 2}{1} = \frac{y}{- 1} = \frac{z - 1}{1}$ : point $(2, - 1, 1)$ .
Right working
Point $(2, 0, 1)$ , direction $(1, - 1, 1)$ .
Do this: Write each numerator as $x - x_{1}$ first.
Usually costs: Most marks.
Mistake: Not checking the third equation for intersection
Two equations always give some λ and μ; only the third confirms the meeting.
Wrong working
Solved two equations, declared the point.
Right working
Substitute into the third equation; if it fails, the lines do not meet.
Do this: Always write 'check in the third equation'.
Usually costs: 1 mark.

Exam tips

Shortest-distance questions were 5-mark questions in the 2023, 2024 and 2026 board papers (3-mark questions in 2025); lay out $a_{2} - a_{1}$ , $b_{1} \times b_{2}$ , its magnitude and the dot product on separate lines.
State whether the lines are skew, parallel or intersecting in your conclusion.
Keep the absolute value in the formula; a negative dot product still gives a positive distance.
If your distance is 0, find the intersection point; examiners often ask for it next.
Leave answers like $\frac{17}{35}$ ; rationalising is not required.

Practice questions on Shortest Distance Between Two Lines

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The shortest distance between two intersecting lines is:
(a) 1
(b) 0
(c) undefined
(d) the distance between their given points
Hint
They share a point.
Answer
(b) 0.
Q2 · 2 marks · Very short answerBoard level
Find the shortest distance between the x-axis and the line through $(0, 0, 3)$ parallel to the y-axis.
Hint
$\hat{i} \times \hat{j} = \hat{k}$ .
Answer
$a_{2} - a_{1} = 3 \hat{k}$ , $b_{1} \times b_{2} = \hat{k}$ : $d = 3$ .
Q3 · 3 marks · Short answerBoard level
Find the distance between the parallel lines $r = λ (\hat{i} + 2 \hat{j} + 2 \hat{k})$ and $r = (\hat{i} + \hat{k}) + μ (\hat{i} + 2 \hat{j} + 2 \hat{k})$ .
Hint
$∣ b ∣ = 3$ .
Answer
$(\hat{i} + \hat{k}) \times (\hat{i} + 2 \hat{j} + 2 \hat{k}) = - 2 \hat{i} - \hat{j} + 2 \hat{k}$ , magnitude 3; $d = \frac{3}{3} = 1$ .
Q4 · 3 marks · Short answerHarder board level
Find the shortest distance between $r = \hat{i} + λ (\hat{j} + \hat{k})$ and $r = \hat{j} + μ (\hat{i} + \hat{k})$ . What does your answer tell you?
Hint
Compute the dot product carefully.
Answer
$b_{1} \times b_{2} = \hat{i} + \hat{j} - \hat{k}$ ; $a_{2} - a_{1} = - \hat{i} + \hat{j}$ ; dot product $= - 1 + 1 + 0 = 0$ . So $d = 0$ : the lines intersect, at $(1, 1, 1)$ (λ = 1, μ = 1).
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x - 1}{2} = \frac{y - 2}{4} = \frac{z - 3}{6}$ are the same line.
Reason (R): Two parallel lines that have a common point coincide.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Is $(1, 2, 3)$ on the first line?
Answer
(a). The directions $(1, 2, 3)$ and $(2, 4, 6)$ are parallel, and $(1, 2, 3)$ lies on both lines, so by R they coincide.
Q6 · 5 marks · Long answerHarder board level
Find the shortest distance between $r = (2 \hat{i} - \hat{j}) + λ (\hat{i} + 2 \hat{j} - \hat{k})$ and $r = (\hat{j} + 3 \hat{k}) + μ (2 \hat{i} + \hat{k})$ .
Hint
$b_{1} \times b_{2} = 2 \hat{i} - 3 \hat{j} - 4 \hat{k}$ .
Answer
$a_{2} - a_{1} = - 2 \hat{i} + 2 \hat{j} + 3 \hat{k}$ ; $b_{1} \times b_{2} = 2 \hat{i} - 3 \hat{j} - 4 \hat{k}$ , magnitude $29$ ; dot product $= - 4 - 6 - 12 = - 22$ ; $d = \frac{22}{29}$ .
Q7 · 5 marks · Long answerHOTS / competency
Find $k$ so that the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\frac{x - 2}{1} = \frac{y - k}{1} = \frac{z - 4}{2}$ intersect, and find the point of intersection.
Hint
$b_{1} \times b_{2} = (1, 1, - 1)$ .
Answer
$a_{2} - a_{1} = (1, k - 2, 1)$ ; dot with $(1, 1, - 1)$ : $1 + k - 2 - 1 = k - 2 = 0 \Rightarrow k = 2$ . Equating: $1 + λ = 2 + μ$ , $2 + 2 λ = 2 + μ$ , $3 + 3 λ = 4 + 2 μ$ ; the first two give $λ = - 1$ , $μ = - 2$ , and the third checks ( $0 = 0$ ). Point: $(0, 0, 0)$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Shortest distance formula for skew lines?Show answer
$\frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}$ .
2.Formula for parallel lines?Show answer
$\frac{∣ ( a _{2} - a _{1} ) \times b ∣}{∣ b ∣}$ .
3.What are skew lines?Show answer
Lines that are neither parallel nor intersecting.
4.What does a shortest distance of 0 mean for non-parallel lines?Show answer
They intersect.
5.Direction of the common perpendicular?Show answer
$b_{1} \times b_{2}$ .
6.Shortest distance between the x-axis and the line through $(0, 0, 5)$ parallel to the x-axis?Show answer
5 units (they are parallel lines).

Shortest Distance Between Two Lines: frequently asked questions

What are skew lines?

Lines in space that are not parallel and do not intersect. They lie in different planes, like a road and a flyover crossing it at an angle.

How do I know which shortest-distance formula to use?

First compare the direction ratios. If they are proportional, the lines are parallel and you use |(a₂ − a₁) × b|/|b|. Otherwise use the skew-line formula with b₁ × b₂.

How do I find where two lines intersect?

Equate their general points, a₁ + λb₁ = a₂ + μb₂, solve two of the component equations for λ and μ, and check that the third equation is satisfied. Substitute λ back to get the point.

Which NCERT section covers shortest distance?

Section 11.5 (Shortest Distance between Two Lines), including 11.5.1 for skew lines and 11.5.2 for parallel lines, in Chapter 11 of NCERT Mathematics Part II for Class 12.

Is the Cartesian determinant formula needed?

It is the same calculation written as a determinant. You may use either the vector form or the determinant form; both earn full marks if worked correctly.