# Vector Equations of Lines and Planes

> Vector equations of lines and planes for JEE Mathematics: r = a + lambda b, r.n = d, three-point planes, angles and Cartesian conversions.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/vector-equations-lines-planes
- Exam / course: JEE · Subject: Mathematics
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Vector Equations of Lines and Planes", PrepElephant, https://prepelephant.com/topics/jee/mathematics/vector-equations-lines-planes

## Direct answer

Points, lines and planes in three dimensions speak most fluently in vectors: a line through point a with direction b is r = a + λb for real λ; a line through two points a and b is r = a + λ(b − a); a plane with normal n through point a satisfies (r − a)·n = 0, written r·n = a·n. The plane through three non-collinear points a, b, c uses the normal (b − a) × (c − a), giving r·[(b − a) × (c − a)] = a·[(b − a) × (c − a)]. Angles follow from dot products: between two lines cos θ = |b₁·b₂|/(|b₁||b₂|), between line and plane sin θ = |b·n|/(|b||n|), between planes cos θ = |n₁·n₂|/(|n₁||n₂|) — the sine in the line-plane case is the detail exams harvest.

## What you must remember

- **Line forms:** r = a + λb (point-direction); r = a + λ(b − a) (two points); Cartesian (x − x₁)/a = (y − y₁)/b = (z − z₁)/c with direction ratios (a, b, c).
- **Plane forms:** r·n̂ = d (unit normal, d = distance from origin); r·n = d (any normal); intercept form x/a + y/b + z/c = 1 with intercepts a, b, c.
- **Three-point plane:** normal = (b − a) × (c − a); a plane through the line of intersection of two planes is S₁ + λS₂ = 0 (the family-of-planes trick).
- **Angle formulas:** line-line uses cos θ = |b₁·b₂|/(|b₁||b₂|); line-plane uses sin θ = |b·n|/(|b||n|); plane-plane uses cos θ = |n₁·n₂|/(|n₁||n₂|).
- **Perpendicularity/parallelism translations:** lines parallel iff b₁ ∥ b₂; line perpendicular to plane iff b ∥ n; planes parallel iff n₁ ∥ n₂; a line lies in a plane iff a satisfies the plane and b·n = 0.
- **Coplanarity test for two lines:** the lines r = a₁ + λb₁ and r = a₂ + μb₂ are coplanar iff (a₂ − a₁)·(b₁ × b₂) = 0, the scalar triple product condition.

## Building the forms

Construct the plane through A(1, 1, 0), B(2, 0, 1), C(3, 1, 1). Two in-plane vectors: AB = (1, −1, 1) and AC = (2, 0, 1). Their cross product: i-component (−1·1 − 1·0) = −1, j-component −(1·1 − 1·2) = 1, k-component (1·0 − (−1)·2) = 2, so n = (−1, 1, 2). The plane is −(x − 1) + (y − 1) + 2z = 0, i.e. −x + y + 2z = 0; the unused point C confirms it, since −3 + 1 + 2 = 0. One construction, whole toolkit: vectors from one vertex, cross product normal, point form, verification.

The family trick completes the picture: all planes through the line x + y + z = 1, 2x − y + 3z = 5 are (x + y + z − 1) + λ(2x − y + 3z − 5) = 0. To pick the member through (1, 0, 2): 1 + 0 + 2 − 1 + λ(2 − 0 + 6 − 5) = 2 + 3λ = 0, λ = −2/3, and expanding gives the specific plane. JEE asks exactly this — "the plane through the line of intersection and the point" — and the family method finishes it in three lines.

## Formula choice traps

The recurring error is the line-plane angle: students reflexively use cosine with the normal, landing on the complement — the sine formula exists because the line meets the plane, not the normal. The second trap is normalising too early: r·n̂ = d needs the unit normal — using an unnormalised n changes d — so either normalise first or use r·n = a·n. Third, converting vector to Cartesian form: from r = (2i − j + k) + λ(i + j − 2k), write x = 2 + λ, y = −1 + λ, z = 1 − 2λ and eliminate λ pairwise. Main-level questions convert forms and compute one angle; Advanced compose coplanarity with distances, where the scalar triple product is the entry ticket.

## Frequently asked questions

### What is the vector equation of a line through two given points?

r = a + λ(b − a), where a and b are the position vectors of the points; λ runs over all reals.

### How do you find the plane through three points?

Form vectors from one point to the other two, take their cross product as the normal n, then write (r − a)·n = 0 — checking first that the points are not collinear.

### Why does the line-plane angle use sine?

Because the angle between the line and the plane is the complement of the angle between the line's direction and the plane's normal, so sin θ = |b·n|/(|b||n|).

### What is the family of planes through a line of intersection?

S₁ + λS₂ = 0, where S₁ = 0 and S₂ = 0 are two planes through the line; λ selects the member satisfying one extra condition.

### How do you test whether two lines in space are coplanar?

Check the scalar triple product (a₂ − a₁)·(b₁ × b₂) = 0; coplanar lines then either intersect or are parallel.