# Planes in 3D

> Planes in 3D for JEE Mathematics: general and intercept forms, plane through three points, angles between planes, point distance and the S1 + λS2 family.

- Canonical URL: https://prepelephant.com/topics/jee/mathematics/planes-in-3d
- 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: "Planes in 3D", PrepElephant, https://prepelephant.com/topics/jee/mathematics/planes-in-3d

## Direct answer

One equation, ax + by + cz + d = 0, fixes a plane with normal vector (a, b, c) — every direction lying in the plane is perpendicular to it. Intercepts p, q, r on the axes give the form x/p + y/q + z/r = 1; three non-collinear points fix a plane through a 3 × 3 determinant; and the family through the line of intersection of two planes is S1 + λS2 = 0. Angles and distances are one-liners: between planes, cos θ = |a1a2 + b1b2 + c1c2|/(|n1||n2|); between a line and a plane, sin θ = |b·n|/(|b||n|); and from a point, |ax1 + by1 + cz1 + d|/√(a^2 + b^2 + c^2).

## What you must remember

- **Normal vector:** the coefficients (a, b, c) carry the plane's entire orientation; d slides it along the normal.
- **Intercept form:** x/p + y/q + z/r = 1 reads the intercepts instantly; it requires all three intercepts nonzero.
- **Three-point plane:** the determinant with rows (x, y, z, 1), (x1, y1, z1, 1), (x2, y2, z2, 1), (x3, y3, z3, 1) vanishing — collinear points make it vanish identically, exposing the degenerate case.
- **Family:** the plane through the intersection line of S1 = 0 and S2 = 0 is S1 + λS2 = 0; S2 itself is the missing member.
- **Plane-plane angle:** cos θ = |n1·n2|/(|n1||n2|) — the modulus keeps the acute angle.
- **Line-plane angle:** sin θ = |b·n|/(|b||n|); a line perpendicular to the plane means b parallel to n.
- **Distances:** point to plane, |ax1 + by1 + cz1 + d|/√(a^2 + b^2 + c^2); between parallel planes, |d1 − d2|/√(a^2 + b^2 + c^2) after matching the normals.

## One plane through three intercepts

Find the plane through (1, 0, 0), (0, 2, 0) and (0, 0, 3), then read everything from it. The intercept form is immediate: x/1 + y/2 + z/3 = 1, which multiplies out to 6x + 3y + 2z = 6. The normal is (6, 3, 2) — and from this single vector the whole geometry follows. The distance from the origin is 6/√(36 + 9 + 4) = 6/7. The angle with the xy-plane (whose normal is (0, 0, 1)) satisfies cos θ = 2/7, so θ = cos^(-1)(2/7). Had a fourth condition arrived — the plane through that line of intersection passing through a point — the family form S1 + λS2 = 0 would absorb it by one substitution. The economy is the lesson: never solve for a plane with three simultaneous equations when the normal can be read or cross-produced directly, because distances and angles are dot products away once the normal is in hand.

## Where marks leak

JEE Main asks distance from a point, intercept form, and angle between planes — all normal-vector manipulations. JEE Advanced asks for the plane containing a given line and a point (family plus condition), the image of a point in a plane (foot of perpendicular then reflection), and the coplanarity of lines via scalar triple products. The recurring losses: dropping the modulus in the angle formula and reporting an obtuse angle the options never offered; feeding three collinear points into the determinant and reporting the resulting identity 0 = 0 as a plane; and computing the distance between parallel planes whose equations carry different normal magnitudes without normalising first — |d1 − d2|/√(...) is valid only when the normals match exactly. The line-plane angle formula with sine (not cosine) is a final memory worth over-drilling: the complement catches thousands of candidates every session.

## Frequently asked questions

### What is the normal to the plane ax + by + cz + d = 0?

The vector (a, b, c), perpendicular to every direction lying in the plane.

### What is the distance from (x1, y1, z1) to that plane?

The distance is |ax1 + by1 + cz1 + d|/√(a^2 + b^2 + c^2), with the modulus outside the entire numerator.

### How is the plane through three non-collinear points found?

Set the 4 × 4 determinant with rows (x, y, z, 1) and the three points to zero and expand.

### What equation runs through the intersection of two planes?

S1 + λS2 = 0 for a real parameter λ; each λ selects one plane of the pencil.

### What is the angle between a line and a plane?

sin θ = |b·n|/(|b||n|) — sine, because the angle is measured with the plane's surface, not its normal.