Equation of a line in space: vector and Cartesian forms

What is the vector equation of a line?

In plain words: a point and a direction fix a line in space, just as the Class 11 point-slope form fixes a line in the plane. Let A have position vector a, and let the line through A be parallel to b. A point P with position vector r is on the line exactly when AP is parallel to b, that is, AP = λb for some real λ. Since AP = r − a (page 382, Fig 11.3):

NCERT slip (page 382): equation (1) prints the parameter as “»” (r = a + » b). It is λ, as in the line above it (λb = r − a) and in the Summary (page 392). In an exam, write λ.

If b = aî + bĵ + ck̂, then a, b, c are direction ratios of the line, and conversely (Remark, page 382). NCERT warns that the number b here is not |b|.

Parametric and Cartesian equations

With A(x_₁, y_₁, z_₁), r = xî + yĵ + zk̂ and direction ratios a, b, c, equating components gives the parametric equations, and eliminating λ gives the Cartesian equation (page 382):

With direction cosines l, m, n in place of a, b, c, the same form holds (NCERT’s Note, page 382). NCERT Example 6 (pages 382–383): through (5, 2, −4) parallel to 3î + 2ĵ − 8k̂, the line is r = 5î + 2ĵ − 4k̂ + λ(3î + 2ĵ − 8k̂), or (x − 5)/3 = (y − 2)/2 = (z + 4)/(−8).

Reading a Cartesian equation. The numerators must be x − x_₁ with x alone, coefficient 1: then the point is (x_₁, y_₁, z_₁) and the denominators are direction ratios. The fraction form is literal only for nonzero denominators. A printed zero denominator is shorthand for a constant-coordinate condition, never division by zero: x/1 = y/0 = z/0 is the x-axis, y = 0 and z = 0.

Two distinct points. Page 381 lists “two given points” as the second way to fix a line. The 2026–27 text does not derive the equation, but its Summary (page 392) gives the line through A(a) and B(b): r = a + λ(b − a), since b − a is a direction.

NCERT slip (page 377): the chapter’s introduction still promises “the equations of lines and planes”, angles involving planes, and the distance of a point from a plane. The 2026–27 chapter covers lines only, Sections 11.2 to 11.5; planes are no longer in the book. In an exam, answer from Sections 11.2–11.5: the chapter’s exercises use only lines.

Solved examples

Common mistakes to avoid

Quick revision summary

• Vector form: r = a + λb, through the point a and parallel to b; each λ gives one point of the line.
• Parametric form: x = x_₁ + λa, y = y_₁ + λb, z = z_₁ + λc.
• Cartesian form: (x − x_₁)/a = (y − y_₁)/b = (z − z_₁)/c, with a, b, c direction ratios; a zero denominator means that coordinate is constant.

Questions students ask

What do the letters r and λ (or t) mean in the vector equation of a line?

In the vector equation of a line, r = a + λb, r is the position vector of a general point P(x, y, z) of the line and λ is a real number, the parameter, that picks out one point: λ = 0 gives the point A with position vector a, and positive and negative λ give points on either side of A. Since AP = λb, the distance AP is |λ| |b|, so λ counts steps of length |b|, and it is the signed distance from A only when b is a unit vector. The letter does not matter: NCERT itself uses t and s in Exercise 11.2, Q15 (page 390), and a second line needs its own letter, such as μ.

How do you find the equation of a line through two points in 3D?

To find the line through distinct points A(x_₁, y_₁, z_₁) and B(x_₂, y_₂, z_₂), with position vectors a and b, take A as the point and b − a, with direction ratios x_₂ − x_₁, y_₂ − y_₁, z_₂ − z_₁, as the direction: r = a + λ(b − a), as in NCERT’s Summary (page 392). NCERT’s 2022 list of rationalised content (page 45) dropped Section 11.3.2, Equation of a Line Passing through Two Given Points, so the Cartesian form is no longer printed, but it follows in the same way: (x − x_₁)/(x_₂ − x_₁) equals the matching y and z quotients. For A(3, 1, −2) and B(0, 5, −1), the direction ratios are −3, 4, 1, and λ = 1 gives B as a check.

How do you write a line segment AB in vector form?

Beyond the NCERT text, the segment AB is the part of the line r = a + λ(b − a) with 0 ≤ λ ≤ 1: λ = 0 gives A, λ = 1 gives B, and λ < 0 or λ > 1 gives points of the line beyond A or beyond B respectively. Each λ in between gives the point dividing AB internally in the ratio λ : (1 − λ), since r = (1 − λ)a + λb is the section formula in vectors in another form, and λ = 1/2 gives the mid-point. For A(2, 0, −3) and B(−4, 6, 3), λ = 1/3 gives the point (0, 2, −1), one third of the way from A to B.

What is the difference between the equation of a line and of a plane in 3D?

Beyond the 2026–27 NCERT book, which no longer covers planes, a plane in space is given by one linear equation Ax + By + Cz = D (A, B, C not all 0), while a line needs two at once, as the line where two non-parallel planes meet. So a line’s Cartesian equation, a chain of three equal fractions, is really two equations, which is also why each coordinate axis needs two (coordinate axes and planes). In vector form a line, r = a + λb, has one parameter, and a plane, r = a + λb + μc with b and c not parallel, has two.

How do you find the direction ratios of a line written as 2x = 3y = 4z?

To read the direction ratios of a line such as 2x = 3y = 4z, first put each of x, y and z alone on top: dividing all three parts by 12 gives x/6 = y/4 = z/3, so the direction ratios are 6, 4, 3 and the line passes through the origin. Reading 2, 3, 4 is a slip: the direction ratios are proportional to the reciprocals ½, ⅓ and ¼ of the coefficients, not to the coefficients themselves. With √(36 + 16 + 9) = √61, the direction cosines are 6/√61, 4/√61 and 3/√61, or all three negated.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Find the point of the line (x − 1)/2 = (y + 3)/(−1) = z/4 given by λ = 2.

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   x = 1 + 2λ, y = −3 − λ and z = 4λ, so λ = 2 gives (5, −5, 8).

2. Q2. Is the point (7, 2, −3) on the line r = î + 4ĵ + k̂ + λ(3î − ĵ − 2k̂)?

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   From x: 1 + 3λ = 7 gives λ = 2. Then y = 4 − 2 = 2 and z = 1 − 4 = −3, which match. Yes, it is on the line.

3. Q3. Write (2x − 1)/4 = (y + 3)/(−2) = (3 − z)/1 in standard form, and give a point and direction ratios.

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   (x − ½)/2 = (y + 3)/(−2) = (z − 3)/(−1). A point is (½, −3, 3), and the direction ratios are 2, −2, −1.

4. Q4. Find the vector equation of the line through the origin and (2, −1, 3).

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   r = λ(2î − ĵ + 3k̂).