Planes in 3D
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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.