MathsCBSE Class 12 · Three Dimensional Geometry

Equation of a Line in Space: Vector and Cartesian Forms (Class 12)

Updated 15 October 2026 · Fact-checked

A line through the point with position vector a, parallel to b, has vector equation r = a + λb. If a = (x₁, y₁, z₁) and b has direction ratios a, b, c, the Cartesian form is (x − x₁)/a = (y − y₁)/b = (z − z₁)/c. The angle between lines is that between their directions.

Understand Equation of a Line in Space

Direction cosines and ratios. If a line makes angles $α, β, γ$ with the positive x-, y- and z-axes, its direction cosines are $l = cos α$ , $m = cos β$ , $n = cos γ$ , with $l^{2} + m^{2} + n^{2} = 1$ . Any numbers $a, b, c$ proportional to them are direction ratios. From direction ratios, $l = \pm \frac{a}{a ^{2} + b ^{2} + c ^{2}}$ (and similarly for $m$ , $n$ ). For the line through $P (x_{1}, y_{1}, z_{1})$ and $Q (x_{2}, y_{2}, z_{2})$ , the direction ratios are $x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1}$ .

Vector form. A line through the point with position vector $a$ and parallel to $b$ is
$r = a + λ b$ , $λ \in R$ .
Each value of $λ$ gives one point of the line. Through two points: $r = a + λ (b - a)$ .

Cartesian form. With $a = (x_{1}, y_{1}, z_{1})$ and direction ratios $a, b, c$ :
$\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c}$ $(= λ)$ .
Setting each fraction equal to $λ$ gives the general point $(x_{1} + aλ, y_{1} + bλ, z_{1} + c λ)$ , the most useful tool in this chapter.

Reading a line correctly. Before reading the point and direction ratios, every numerator must be of the form $x - x_{1}$ with coefficient 1. For $\frac{2 x - 1}{4} = \frac{y + 2}{3} = \frac{3 - z}{2}$ , rewrite as $\frac{x - \frac{1}{2}}{2} = \frac{y + 2}{3} = \frac{z - 3}{- 2}$ : the point is $(\frac{1}{2}, - 2, 3)$ and the direction ratios are $2, 3, - 2$ .

Angle between two lines. With direction vectors $b_{1}$ and $b_{2}$ (or direction ratios $a_{1}, b_{1}, c_{1}$ and $a_{2}, b_{2}, c_{2}$ ),
$cos θ = \frac{b _{1} \cdot b _{2}}{∣ b _{1} ∣∣ b _{2} ∣}$ ,
taking the acute angle. Lines are perpendicular if $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$ and parallel if $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ .

A point at a given distance. Along $r = a + λ b$ , the distance from the point $a$ is $∣ λ ∣∣ b ∣$ , so points at distance $d$ have $λ = \pm \frac{d}{∣ b ∣}$ .

Planes are not in the 2026-27 syllabus; the chapter is about lines only (NCERT Sections 11.2 to 11.4).

Equation of a Line in Space: formulas to remember

Vector equation: $r = a + λ b$; Through $a$ , parallel to $b$ .
Cartesian equation: $\frac{x - x _{1}}{a} = \frac{y - y _{1}}{b} = \frac{z - z _{1}}{c}$; Coefficients of x, y, z must be 1 before reading.
Through two points: $\frac{x - x _{1}}{x _{2} - x _{1}} = \frac{y - y _{1}}{y _{2} - y _{1}} = \frac{z - z _{1}}{z _{2} - z _{1}}$; Vector form: $r = a + λ (b - a)$ .
Direction cosines: $l = \frac{a}{a ^{2} + b ^{2} + c ^{2}}, m = \frac{b}{\cdot}, n = \frac{c}{\cdot}, l^{2} + m^{2} + n^{2} = 1$; Up to an overall sign.
Angle between lines: $cos θ = \frac{∣ a _{1} a _{2} + b _{1} b _{2} + c _{1} c _{2} ∣}{a _{1}^{2} + b _{1}^{2} + c _{1}^{2} a _{2}^{2} + b _{2}^{2} + c _{2}^{2}}$; Acute angle between the lines.
Perpendicular and parallel: $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0; \frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$; Use direction ratios only, not points.

How to solve Equation of a Line in Space questions

Use this routine for writing, reading and comparing lines.

1Identify a point on the line and a direction vector (from a given parallel vector or from two points).
2Write the vector form $r = a + λ b$ and the Cartesian form; they must match.
3When reading a given Cartesian line, first make each numerator $x - x_{1}$ , $y - y_{1}$ , $z - z_{1}$ with coefficient 1, adjusting denominators.
4For a general point, set each fraction equal to $λ$ and write $(x_{1} + aλ, y_{1} + bλ, z_{1} + c λ)$ .
5For angles, use only the direction ratios: dot product over the product of magnitudes.
6For conditions (perpendicular, parallel, point on line), write the equation and solve for the unknown.

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.

Points on a line at a given distance

Suppose a laser in a lab is fixed at $(0, 0, 1)$ (in metres) and its beam points along $2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ .

Write the equation of the beam and find the point on the beam 14 m from the laser.

Try it first, then show the solutionSolution

$r = \hat{k} + λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$ , $λ \geq 0$ (the beam goes forward only).
$∣2 \hat{i} + 3 \hat{j} + 6 \hat{k} ∣ = 4 + 9 + 36 = 7$ , so distance $= 7 λ$ .
$7 λ = 14 \Rightarrow λ = 2$ : point $(4, 6, 13)$ .

Answer: $r = \hat{k} + λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$ ; the point is $(4, 6, 13)$ .

Memory hook: Distance along a line is λ times the length of the direction vector.

Perpendicular lines

Imagine a plumber joining two straight pipes whose directions are $\hat{i} + 2 \hat{j} + 2 \hat{k}$ and $2 \hat{i} - 2 \hat{j} + \hat{k}$ .

Is the joint a right angle?

Try it first, then show the solutionSolution

Dot product of the directions: $(1) (2) + (2) (- 2) + (2) (1) = 2 - 4 + 2 = 0$ .
A zero dot product means the directions are perpendicular.

Answer: Yes, the pipes meet at $90°$ .

Memory hook: Perpendicular lines: direction ratios multiply and add to zero.

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 plus λ direction
Every line is 'a point plus λ times a direction': $r = a + λ b$ .
Use it when: Writing any line.
Clean numerators first
Make each numerator exactly $x - x_{1}$ (coefficient 1, x first) before reading points or direction ratios.
Use it when: Lines like $\frac{3 - z}{2}$ or $\frac{2 x - 1}{4}$ .
Angles ignore points
Only the denominators (direction ratios) decide angles, parallelism and perpendicularity.
Use it when: Angle and condition questions.
General point is your best friend
Put the fractions equal to λ to get a moving point; most 3D problems start here.
Use it when: Distances, feet of perpendiculars, intersections.

Equation of a Line in Space: solved examples

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

Example 1: Line through a point parallel to a vector · 2 marks · Very short answerBoard level

Find the vector and Cartesian equations of the line through $(2, - 1, 3)$ parallel to $3 \hat{i} + 2 \hat{j} - \hat{k}$ .

Try it first, then show the solutionSolution

Vector: $r = (2 \hat{i} - \hat{j} + 3 \hat{k}) + λ (3 \hat{i} + 2 \hat{j} - \hat{k})$ . (1 mark)
Cartesian: $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{- 1}$ . (1 mark)

Answer: $r = (2 \hat{i} - \hat{j} + 3 \hat{k}) + λ (3 \hat{i} + 2 \hat{j} - \hat{k})$ ; $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{- 1}$

Example 2: Reading a line in non-standard form · 2 marks · Very short answerBoard level

For the line $\frac{2 x - 1}{4} = \frac{y + 2}{3} = \frac{3 - z}{2}$ , find a point on it, its direction ratios and its direction cosines.

Try it first, then show the solutionSolution

Rewrite: $\frac{x - \frac{1}{2}}{2} = \frac{y + 2}{3} = \frac{z - 3}{- 2}$ . (1 mark)
Point $(\frac{1}{2}, - 2, 3)$ ; direction ratios $2, 3, - 2$ ; direction cosines $\frac{2}{17}, \frac{3}{17}, \frac{- 2}{17}$ . (1 mark)

Answer: Point $(\frac{1}{2}, - 2, 3)$ ; d.r. $2, 3, - 2$ ; d.c. $\frac{1}{17} (2, 3, - 2)$

Example 3: Line through two points and a point on it · 3 marks · Short answerHarder board level

Find the equation of the line through $A (1, 2, - 1)$ and $B (3, - 2, 4)$ , and the point on it whose z-coordinate is 9.

Try it first, then show the solutionSolution

Direction ratios: $3 - 1, - 2 - 2, 4 + 1 = 2, - 4, 5$ . Line: $\frac{x - 1}{2} = \frac{y - 2}{- 4} = \frac{z + 1}{5}$ . (1 mark)
General point: $(1 + 2 λ, 2 - 4 λ, - 1 + 5 λ)$ . (1 mark)
$- 1 + 5 λ = 9 \Rightarrow λ = 2$ : point $(5, - 6, 9)$ . (1 mark)

Answer: $\frac{x - 1}{2} = \frac{y - 2}{- 4} = \frac{z + 1}{5}$ ; point $(5, - 6, 9)$

Example 4: Angle between two lines · 3 marks · Short answerHarder board level

Find the angle between the lines $\frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{- 2}$ and $\frac{x + 2}{4} = \frac{y - 4}{- 1} = \frac{z + 1}{1}$ .

Try it first, then show the solutionSolution

Direction ratios $(2, 1, - 2)$ and $(4, - 1, 1)$ . (½ mark)
Dot product $= 8 - 1 - 2 = 5$ ; magnitudes $3$ and $18 = 32$ . (1½ marks)
$cos θ = \frac{5}{9 2} = \frac{5 2}{18}$ , so $θ = cos^{- 1} \frac{5 2}{18}$ . (1 mark)

Answer: $θ = cos^{- 1} (\frac{5 2}{18})$

Example 5: A parameter for perpendicular lines · 3 marks · Short answerHOTS / competency

Find $k$ so that the lines $\frac{x - 3}{k} = \frac{y + 1}{2} = \frac{z}{- 1}$ and $\frac{x + 2}{2} = \frac{y}{k} = \frac{z - 1}{4}$ are perpendicular. Show also that they can never be parallel.

Try it first, then show the solutionSolution

Perpendicular: $2 k + 2 k - 4 = 0 \Rightarrow k = 1$ . (1 mark)
Parallel would need $\frac{k}{2} = \frac{2}{k} = \frac{- 1}{4}$ . (1 mark)
$\frac{k}{2} = - \frac{1}{4}$ gives $k = - \frac{1}{2}$ , but then $\frac{2}{k} = - 4 \neq = - \frac{1}{4}$ . So no k makes them parallel. (1 mark)

Answer: $k = 1$ ; never parallel.

Example 6: Case study: a drone's straight flight · 4 marks · Case-basedHOTS / competency

Suppose a drone starts at $P (1, 0, 2)$ and flies straight in the direction $2 \hat{i} + \hat{j} - 2 \hat{k}$ (units in hundreds of metres).
(i) Write the vector equation of its path. (1 mark)
(ii) Find the points on the line at distance 6 units from P. (2 marks)
(iii) Does the point $(7, 3, - 4)$ lie on the path? (1 mark)

Try it first, then show the solutionSolution

(i) $r = (\hat{i} + 2 \hat{k}) + λ (2 \hat{i} + \hat{j} - 2 \hat{k})$ . (1 mark)
(ii) $∣2 \hat{i} + \hat{j} - 2 \hat{k} ∣ = 3$ , so distance $= 3∣ λ ∣ = 6 \Rightarrow λ = \pm 2$ . (1 mark)
$λ = 2$ : $(5, 2, - 2)$ ; $λ = - 2$ : $(- 3, - 2, 6)$ . (1 mark)
(iii) $1 + 2 λ = 7 \Rightarrow λ = 3$ ; then $y = 3$ and $z = 2 - 6 = - 4$ . Yes, it lies on the path. (1 mark)

Answer: (i) $r = (\hat{i} + 2 \hat{k}) + λ (2 \hat{i} + \hat{j} - 2 \hat{k})$ (ii) $(5, 2, - 2)$ and $(- 3, - 2, 6)$ (iii) yes, with $λ = 3$ .

Where students lose marks in Equation of a Line in Space

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

Mistake: Reading direction ratios from an unclean numerator
Coefficients of x, y, z must be 1 and the variable must come first.
Wrong working
$\frac{3 - z}{2}$ : direction ratio for z is 2.
Right working
$\frac{3 - z}{2} = \frac{z - 3}{- 2}$ : direction ratio $- 2$ .
Do this: Rewrite every numerator before reading.
Usually costs: 1 to 2 marks and every later step.
Mistake: Using the point instead of the direction
Angles and parallelism depend only on directions.
Wrong working
Angle between lines through $(1, - 1, 3)$ and $(- 2, 4, - 1)$ computed from these points.
Right working
Use the denominators $(2, 1, - 2)$ and $(4, - 1, 1)$ .
Do this: Circle the denominators before computing angles.
Usually costs: The answer.
Mistake: Forgetting the negative λ for distance
There are two points at a given distance, one on each side.
Wrong working
Only $(5, 2, - 2)$ .
Right working
$(5, 2, - 2)$ and $(- 3, - 2, 6)$ .
Do this: Use $λ = \pm \frac{d}{∣ b ∣}$ unless the direction is restricted.
Usually costs: 1 mark.
Mistake: Mixing vector and Cartesian forms
Writing $r = \frac{x - 2}{3} \dots$ combines two notations.
Wrong working
$r = \frac{x - 2}{3} = \frac{y + 1}{2}$ .
Right working
Vector: $r = a + λ b$ ; Cartesian: $\frac{x - 2}{3} = \dots$ written separately.
Do this: Write each form on its own line.
Usually costs: Half a mark.

Exam tips

Equation-of-a-line and direction-cosine questions appeared in the 2023, 2024 and 2026 papers, often as 1-mark MCQs and as the first part of 5-mark questions.
Always write the general point of a line; it is the starting point for most longer 3D questions.
Give both vector and Cartesian forms when asked for 'the equation' without specifying.
Planes are not in the 2026-27 syllabus; skip any old question that needs a plane equation.
Leave non-standard angles as $cos^{- 1} (\dots)$ .

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 $\frac{x - 1}{2} = \frac{3 - y}{1} = \frac{z + 4}{3}$ are:
(a) $2, 1, 3$
(b) $2, - 1, 3$
(c) $1, 3, - 4$
(d) $- 2, 1, 3$
Hint
Rewrite $\frac{3 - y}{1}$ as $\frac{y - 3}{- 1}$ .
Answer
(b) $2, - 1, 3$ .
Q2 · 2 marks · Very short answerBoard level
Find the direction cosines of a line that makes equal angles with the three coordinate axes.
Hint
$l = m = n$ and $l^{2} + m^{2} + n^{2} = 1$ .
Answer
$3 l^{2} = 1 \Rightarrow l = m = n = \pm \frac{1}{3}$ .
Q3 · 2 marks · Very short answerBoard level
Write the Cartesian equation of the line through $(1, 2, 3)$ and $(4, 5, 7)$ .
Hint
Direction ratios are the differences of coordinates.
Answer
$\frac{x - 1}{3} = \frac{y - 2}{3} = \frac{z - 3}{4}$ .
Q4 · 3 marks · Short answerHarder board level
Show that the lines $\frac{x - 1}{2} = \frac{y + 2}{- 4} = \frac{z - 3}{6}$ and $r = (\hat{i} + \hat{k}) + μ (- \hat{i} + 2 \hat{j} - 3 \hat{k})$ are parallel, and that they are not the same line.
Hint
Compare direction ratios; then test whether $(1, - 2, 3)$ lies on the second line.
Answer
$(2, - 4, 6) = - 2 (- 1, 2, - 3)$ , so the lines are parallel. On the second line, $x = 1 - μ = 1$ gives $μ = 0$ , the point $(1, 0, 1)$ , which is not $(1, - 2, 3)$ . So the lines are distinct parallel lines.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $\frac{1}{2}, \frac{1}{2}, \frac{1}{2}$ can be the direction cosines of a line.
Reason (R): Direction cosines satisfy $l^{2} + m^{2} + n^{2} = 1$ .
(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
Compute $\frac{1}{4} + \frac{1}{4} + \frac{1}{4}$ .
Answer
(d). The squares add to $\frac{3}{4} \neq = 1$ , so A is false; R is true.
Q6 · 2 marks · Very short answerHarder board level
Find the angle between two lines with direction ratios $1, 1, 0$ and $0, 1, 1$ .
Hint
Dot product 1; magnitudes $2$ each.
Answer
$cos θ = \frac{1}{2}$ , so $θ = \frac{π}{3}$ .
Q7 · 3 marks · Short answerHOTS / competency
Find the point where the line through $A (3, 4, 1)$ and $B (5, 1, 6)$ meets the xy-plane ( $z = 0$ ).
Hint
General point $(3 + 2 λ, 4 - 3 λ, 1 + 5 λ)$ .
Answer
$1 + 5 λ = 0 \Rightarrow λ = - \frac{1}{5}$ . Point: $(3 - \frac{2}{5}, 4 + \frac{3}{5}, 0) = (\frac{13}{5}, \frac{23}{5}, 0)$ .

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.Vector equation of a line through $a$ parallel to $b$ ?Show answer
$r = a + λ b$ .
2.Direction ratios of the line through $(1, 2, 3)$ and $(2, 0, 7)$ ?Show answer
$1, - 2, 4$ .
3.Sum of squares of direction cosines?Show answer
1.
4.Condition for perpendicular lines?Show answer
$a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$ .
5.General point of $\frac{x - 1}{2} = \frac{y}{3} = \frac{z + 1}{- 1}$ ?Show answer
$(1 + 2 λ, 3 λ, - 1 - λ)$ .
6.Direction cosines of the x-axis?Show answer
$1, 0, 0$ .

Equation of a Line in Space: frequently asked questions

What is the difference between direction cosines and direction ratios?

Direction cosines are the cosines of the angles a line makes with the axes, and their squares add up to 1. Direction ratios are any three numbers proportional to them; dividing by the square root of the sum of their squares gives the direction cosines.

How do I convert a vector equation into Cartesian form?

From r = a + λb, read the point (x₁, y₁, z₁) from a and the direction ratios from b, then write (x − x₁)/b₁ = (y − y₁)/b₂ = (z − z₁)/b₃.

Is the line through two points a separate topic for 2026-27?

NCERT dropped it as a separate section, but the syllabus includes direction ratios of a line joining two points, so you should be able to write such a line using the difference of coordinates as its direction.

Which NCERT sections cover this topic?

Sections 11.2 (Direction Cosines and Direction Ratios of a Line), 11.3 (Equation of a Line in Space) and 11.4 (Angle between Two Lines) of Chapter 11 in NCERT Mathematics Part II for Class 12.

Are planes in the 2026-27 syllabus?

No. Equations of planes, angles between planes and distances from planes were dropped. The chapter covers lines only.