MathsISC Class 12 · Three-dimensional Geometry

Equation of a Line in Space: Forms, Intersection and Shortest Distance

Updated 15 October 2026 · Fact-checked

A line in space is fixed by a point $\vec{a}$ on it and a direction $\vec{b}$. Its vector equation is $\vec{r} = \vec{a} + \lambda\vec{b}$; with $\vec{a} = (x_1, y_1, z_1)$ and $\vec{b} = (a, b, c)$, the Cartesian form is $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, where a, b, c are direction ratios.

Understand Equation of a Line in Space

Point plus direction. Start at a point $A (a)$ and move any multiple of a direction vector $b$ : every point of the line is $r = a + λ b$ , $λ \in R$ . Writing $r = (x, y, z)$ and equating components, $x = x_{1} + λa$ , $y = y_{1} + λb$ , $z = z_{1} + λ c$ ; eliminating $λ$ gives the Cartesian (symmetric) form
$\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c} .$
The denominators are the direction ratios. A zero denominator is allowed: $\frac{x - 1}{0}$ just means $x = 1$ all along the line.

Through two points. Through $A (a)$ and $B (b)$ : $r = a + λ (b - a)$ , or $\frac{x - x _{1}}{x _{2} - x _{1}} = \frac{y - y _{1}}{y _{2} - y _{1}} = \frac{z - z _{1}}{z _{2} - z _{1}}$ . Three points $P, Q, R$ are collinear when $P Q$ and $QR$ are proportional. Use these displacement vectors, not the position vectors; that was the error the ISC 2025 examiners reported.

Standard form check. Before reading direction ratios, every numerator must be "variable minus number" with coefficient 1. For example, $\frac{3 - x}{2} = \frac{2 y + 1}{4}$ becomes $\frac{x - 3}{- 2} = \frac{y + \frac{1}{2}}{2}$ , so the direction ratios are $- 2, 2, \dots$ , not $2, 4$ .

Axes. The x-axis is $\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$ , that is $y = 0, z = 0$ ; a line parallel to the x-axis through $(x_{1}, y_{1}, z_{1})$ is $y = y_{1}$ , $z = z_{1}$ .

Angle between two lines. With directions $b_{1}$ and $b_{2}$ , $cos θ = \frac{∣ b _{1} \cdot b _{2} ∣}{∣ b _{1} ∣∣ b _{2} ∣} = \frac{∣ a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2} ∣}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}$ . Perpendicular: $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$ . Parallel: $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ . A line parallel to a given one shares its direction ratios, as in the ISC 2025 honeybee question.

How two lines can sit. In space two lines are either parallel, intersecting, or skew (neither parallel nor meeting, so they lie in no common plane). Parallel and intersecting lines are coplanar.

Do they intersect? Write the general point of each line with its own parameter, equate two coordinates to find $λ$ and $μ$ , and test the third. If it agrees, the lines meet at that point; if not, and they are not parallel, they are skew.

Shortest distance between skew lines. The shortest segment between $r = a_{1} + λ b_{1}$ and $r = a_{2} + μ b_{2}$ is perpendicular to both lines, so it lies along $b_{1} \times b_{2}$ . Its length is the projection of $a_{2} - a_{1}$ on that direction:
$d = \frac{( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} )}{∣ b _{1} \times b _{2} ∣} .$
In Cartesian form the numerator is the determinant with rows $(x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1})$ , $(a_{1}, b_{1}, c_{1})$ , $(a_{2}, b_{2}, c_{2})$ . The lines intersect (are coplanar) exactly when this numerator is zero. For parallel lines with direction $b$ , $d = \frac{∣ b \times ( a _{2} - a _{1} ) ∣}{∣ b ∣}$ .

Distance of a point from a line. For $P (p)$ and the line $a + λ b$ , the general point $Q = a + λ b$ is the foot of the perpendicular when $P Q \cdot b = 0$ ; solve for $λ$ , then $P Q$ is the distance. Directly, $d = \frac{∣ ( p - a ) \times b ∣}{∣ b ∣}$ .

Equation of a Line in Space: formulas to remember

Vector form: $r = a + λ b$; Through two points: $b$ replaced by $b - a$ .
Cartesian form: $\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c}$; Coefficients of x, y, z must be 1 first.
Angle between lines: $cos θ = \frac{∣ b _{1} \cdot b _{2} ∣}{∣ b _{1} ∣∣ b _{2} ∣}$; Perpendicular: $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$ .
Skew lines SD: $d = \frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}$; Zero numerator: the lines intersect.
Parallel lines: $d = \frac{∣ b \times ( a _{2} - a _{1} ) ∣}{∣ b ∣}$; Same direction $b$ .
Point to line: $d = \frac{∣ ( p - a ) \times b ∣}{∣ b ∣}$; Or find the foot by $P Q \cdot b = 0$ .

How to solve Equation of a Line in Space questions

Most line questions follow one of three routines.

1Equation: find one point and one direction (from two points, or from a parallel line), then write both the vector and Cartesian forms.
2Relative position: compare directions first. Proportional means parallel (or the same line). Otherwise test for intersection with general points.
3Intersection: put $(x_{1} + a_{1} λ, \dots) = (x_{2} + a_{2} μ, \dots)$ ; solve two equations, check the third.
4Shortest distance: compute $b_{1} \times b_{2}$ , then $(a_{2} - a_{1}) \cdot (b_{1} \times b_{2})$ , divide by the modulus of the cross product and take the absolute value.
5Foot of a perpendicular: general point Q, vector PQ, set $P Q \cdot b = 0$ , solve for the parameter, find Q and PQ.
6Check: substitute the point you found into each line.

Equation of a Line in Space 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.

Where a line meets a coordinate plane

Suppose a laser beam in a hall passes through the points (1, 2, 5) and (3, 3, 3), in metres, with the floor being the plane z = 0.

Find where the beam hits the floor.

Try it first, then show the solutionSolution

Direction: $(2, 1, - 2)$ ; line: $(1 + 2 λ, 2 + λ, 5 - 2 λ)$ .
Floor: $5 - 2 λ = 0$ , so $λ = \frac{5}{2}$ .
Point: $(1 + 5, 2 + 2.5, 0) = (6, 4.5, 0)$ .

Answer: At (6, 4.5, 0).

Memory hook: To meet a plane, set its coordinate condition in the general point.

Angle between two lines

Imagine a tent pole whose top is at (0, 0, 4) with two guy ropes pegged at (3, 0, 0) and (0, 4, 0) (metres).

Find the angle between the two ropes.

Try it first, then show the solutionSolution

Directions: $(3, 0, - 4)$ and $(0, 4, - 4)$ .
$cos θ = \frac{0 + 0 + 16}{5 \times 4 2} = \frac{4}{5 2} = \frac{2 2}{5} \approx 0.566$ .

Answer: $θ = cos^{- 1} \frac{2 2}{5} \approx 55.6°$ .

Memory hook: Angle between lines = dot of directions over product of lengths.

Tricks to remember Equation of a Line in Space

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

Point + λ direction
Every line equation is "a point you know, plus a free multiple of the way it points".
Use it when: Writing any line.
Clean before you read
Make each numerator "x − number", "y − number", "z − number" with coefficient 1 before reading direction ratios.
Use it when: Lines given in awkward form.
P-I-S order
Parallel? (directions proportional) then Intersect? (general points agree) else Skew.
Use it when: Classifying two lines.
SD = shift · cross ÷ |cross|
Shift between the given points, dotted with the cross of the directions, over its length.
Use it when: Shortest distance.

Equation of a Line in Space: solved examples

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

Example 1: Vector to Cartesian · 1 mark · MCQBoard level

The Cartesian form of $r = (2 \hat{i} - \hat{j} + 4 \hat{k}) + λ (\hat{i} + 2 \hat{j} - \hat{k})$ is:
(a) $\frac{x - 1}{2} = \frac{y - 2}{- 1} = \frac{z + 1}{4}$
(b) $\frac{x - 2}{1} = \frac{y + 1}{2} = \frac{z - 4}{- 1}$
(c) $\frac{x + 2}{1} = \frac{y - 1}{2} = \frac{z + 4}{- 1}$
(d) $\frac{x - 2}{1} = \frac{y - 1}{2} = \frac{z - 4}{1}$

Try it first, then show the solutionSolution

Point (2, −1, 4), direction (1, 2, −1).

Answer: (b).

Example 2: Line through two points and collinearity · 2 marks · Short answerBoard level

Find the vector and Cartesian equations of the line through $A (1, - 1, 2)$ and $B (4, 1, 0)$ , and show that $C (7, 3, - 2)$ lies on it.

Try it first, then show the solutionSolution

$A B = (3, 2, - 2)$ : $r = (\hat{i} - \hat{j} + 2 \hat{k}) + λ (3 \hat{i} + 2 \hat{j} - 2 \hat{k})$ ; $\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 2}{- 2}$ . (1 mark)
$A C = (6, 4, - 4) = 2 A B$ , so C is on the line ( $λ = 2$ ). (1 mark)

Answer: As above; C corresponds to $λ = 2$ .

Example 3: Angle and a perpendicularity condition · 3 marks · Short answerHarder board level

Find the angle between the lines $\frac{x - 1}{2} = \frac{y + 2}{1} = \frac{z}{- 2}$ and $\frac{x}{4} = \frac{y - 3}{4} = \frac{z + 1}{2}$ . Find k if the line $\frac{x - 2}{k} = \frac{y}{3} = \frac{z + 1}{- 1}$ is perpendicular to the first line.

Try it first, then show the solutionSolution

$b_{1} = (2, 1, - 2)$ , $b_{2} = (4, 4, 2)$ ; $b_{1} \cdot b_{2} = 8 + 4 - 4 = 8$ . (1 mark)
$cos θ = \frac{8}{3 \times 6} = \frac{4}{9}$ , $θ = cos^{- 1} \frac{4}{9}$ . (1 mark)
Perpendicular: $2 k + 3 + 2 = 0$ , so $k = - \frac{5}{2}$ . (1 mark)

Answer: $cos^{- 1} \frac{4}{9}$ ; $k = - \frac{5}{2}$ .

Example 4: Testing for intersection · 3 marks · Short answerHarder board level

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

Try it first, then show the solutionSolution

General points: $(1 + λ, λ, - 1 + 2 λ)$ and $(3 + μ, 1, 2 + μ)$ . (1 mark)
y: $λ = 1$ ; x: $2 = 3 + μ$ , so $μ = - 1$ . (1 mark)
z: $- 1 + 2 = 1$ and $2 - 1 = 1$ : equal, so the lines meet at $(2, 1, 1)$ . (1 mark)

Answer: They intersect at (2, 1, 1).

Example 5: Shortest distance between skew lines · 4 marks · Long answerHarder board level

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

Try it first, then show the solutionSolution

$b_{1} \times b_{2} = \hat{i} 12 \hat{j} - 1 1 \hat{k} 1 - 1 = (1 - 1) \hat{i} - (- 1 - 2) \hat{j} + (1 + 2) \hat{k} = 3 \hat{j} + 3 \hat{k}$ . (1 mark)
$∣ b_{1} \times b_{2} ∣ = 32$ ; $a_{2} - a_{1} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ . (1 mark)
$(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = 0 - 6 + 9 = 3$ . (1 mark)
$d = \frac{3}{3 2} = \frac{1}{2}$ ; non-zero, so the lines are skew. (1 mark)

Answer: $\frac{1}{2}$ units.

Example 6: Foot of the perpendicular from a point · 4 marks · Case-basedHOTS / competency

Suppose a straight water pipe runs along $\frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z - 2}{- 2}$ and a tap is at $P (2, 3, - 1)$ (metres). A short connecting pipe is laid from P perpendicular to the main pipe.
(i) Find the point Q where it joins the main pipe. (3 marks)
(ii) Find the length of the connecting pipe. (1 mark)

Try it first, then show the solutionSolution

General point $Q (1 + t, - 1 + 2 t, 2 - 2 t)$ ; $P Q = (t - 1, 2 t - 4, 3 - 2 t)$ .
$P Q \cdot (1, 2, - 2) = (t - 1) + 2 (2 t - 4) - 2 (3 - 2 t) = 9 t - 15 = 0$ , so $t = \frac{5}{3}$ .
$Q = (\frac{8}{3}, \frac{7}{3}, - \frac{4}{3})$ .
$P Q = (\frac{2}{3}, - \frac{2}{3}, - \frac{1}{3})$ , length $\frac{4 + 4 + 1}{9} = 1$ .

Answer: (i) $Q (\frac{8}{3}, \frac{7}{3}, - \frac{4}{3})$ ; (ii) 1 m.

Where students lose marks in Equation of a Line in Space

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

Mistake: Collinearity from position vectors
The ISC 2025 examiners reported candidates using the position vectors of P, Q, R instead of $P Q$ , $QR$ .
Do this: Subtract coordinates first; collinear when the displacement vectors are proportional.
Usually costs: 1 mark.
Mistake: Reading direction ratios from an unclean form
Numerators like $1 - y$ or $2 z + 1$ change the sign or scale of the direction ratio.
Wrong working
$\frac{1 - y}{4}$ read as direction ratio 4.
Right working
$\frac{y - 1}{- 4}$ : direction ratio −4.
Do this: Rewrite with coefficient +1 on each variable.
Usually costs: 1 to 2 marks.
Mistake: Missing the third-coordinate check
Finding λ and μ from two equations does not prove intersection.
Do this: Always substitute into the third equation and state the conclusion.
Usually costs: 1 mark.
Mistake: Forgetting the modulus in the SD formula
A negative dot product gives a negative "distance".
Do this: Write $d = ∣ \dots ∣$ and quote the positive value with units.
Usually costs: 1 mark.

Exam tips

Write both vector and Cartesian forms unless the question asks for one.
Lay out the cross product as a determinant; examiners award a mark for it.
State "the lines are skew" or "the lines intersect" explicitly after a shortest-distance calculation.
Use different parameters (λ and μ) for the two lines.

Practice questions on Equation of a Line in Space

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 direction ratios of the line $\frac{x - 3}{2} = \frac{1 - y}{4} = \frac{2 z + 1}{6}$ are:
(a) 2, 4, 6
(b) 2, −4, 3
(c) 2, −4, 6
(d) 2, 4, 3
Hint
Rewrite as $\frac{y - 1}{- 4}$ and $\frac{z + \frac{1}{2}}{3}$ .
Answer
(b).
Q2 · 1 mark · Very short answerBoard level
Write the Cartesian equation of the line through $(2, - 1, 3)$ parallel to $\frac{x - 1}{3} = \frac{y + 2}{- 1} = \frac{z}{2}$ .
Hint
Same direction ratios.
Answer
$\frac{x - 2}{3} = \frac{y + 1}{- 1} = \frac{z - 3}{2}$ .
Q3 · 3 marks · Short answerHarder board level
Find k so that the lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{1}$ and $\frac{x - 2}{1} = \frac{y - k}{1} = \frac{z - 1}{- 1}$ intersect, and find the point of intersection.
Hint
Use the x and z equations to get both parameters.
Answer
$s = - \frac{1}{2}$ , $t = - \frac{3}{2}$ , $k = \frac{5}{2}$ ; point $(\frac{1}{2}, 1, \frac{5}{2})$ .
Q4 · 2 marks · Short answerHarder board level
Find the distance between the parallel lines $r = (\hat{i} + \hat{j}) + λ (2 \hat{i} - \hat{j} + \hat{k})$ and $r = (2 \hat{i} + \hat{j} + \hat{k}) + μ (4 \hat{i} - 2 \hat{j} + 2 \hat{k})$ .
Hint
Use $b = (2, - 1, 1)$ .
Answer
$b \times (1, 0, 1) = (- 1, - 1, 1)$ ; $d = \frac{3}{6} = \frac{1}{2}$ .
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 skew.
Reason (R): Two lines that do not intersect are skew.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) Both false.
Hint
Is (1, 2, 3) on the first line?
Answer
(d): the directions are proportional and (1, 2, 3) lies on both, so they are the same line; and parallel lines do not meet but are not skew.
Q6 · 2 marks · Short answerHOTS / competency
Find the distance of $P (1, 2, 3)$ from the line through the origin with direction ratios 2, 1, 2.
Hint
$\frac{∣ p \times b ∣}{∣ b ∣}$ .
Answer
$(1, 2, 3) \times (2, 1, 2) = (1, 4, - 3)$ ; $d = \frac{26}{3}$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose two drones fly along $L_{1} : r = 10 \hat{k} + t (\hat{i} + \hat{j})$ and $L_{2} : r = (10 \hat{i} + 12 \hat{k}) + s (- \hat{i} + \hat{j})$ (metres).
(i) Show that the paths are not parallel. (1 mark)
(ii) Find the shortest distance between the paths. (2 marks)
(iii) The safety rule needs at least 3 m separation. Is it met? (1 mark)
Hint
$b_{1} \times b_{2} = (0, 0, 2)$ .
Answer
(i) (1, 1, 0) and (−1, 1, 0) are not proportional. (ii) $\frac{∣ ( 10 , 0 , 2 ) \cdot ( 0 , 0 , 2 ) ∣}{2} = 2$ m. (iii) No: the paths come within 2 m.

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.Vector equation of a line through $a$ parallel to $b$ ?Show answer
$r = a + λ b$ .
2.Equations of the y-axis?Show answer
$x = 0$ , $z = 0$ .
3.Condition for two lines to be perpendicular?Show answer
$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$ .
4.What are skew lines?Show answer
Lines that are neither parallel nor intersecting; they lie in no common plane.
5.If $(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = 0$ for non-parallel lines, what follows?Show answer
The lines intersect (they are coplanar).

Equation of a Line in Space: frequently asked questions

How do I convert a Cartesian line to vector form?

Read the point from the numerators and the direction from the denominators (after cleaning the form), then write \vec{r} = \vec{a} + \lambda\vec{b}.

How do I know whether two lines are skew?

First check the directions: if they are not proportional, the lines are not parallel. Then compute (\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2): zero means they intersect, non-zero means they are skew.

Can a direction ratio be zero?

Yes. A zero in the denominator means that coordinate stays constant along the line; for example, a line with direction ratios 1, 0, 0 is parallel to the x-axis.

Is the distance of a point from a line in the ISC syllabus?

Yes. The 2027 syllabus lists the distance of a point from a line and the shortest distance between two lines alongside the equations of lines.

How many marks do lines carry in ISC?

Three-dimensional geometry carries 6 marks. In 2025 a 2-mark item asked for a line parallel to a given line and a 1-mark item tested collinearity.