Planes JEE adds

A plane is fixed by a point on it and a normal (a direction perpendicular to it). Through a point with position vector a and normal n: (r − a)·n = 0, that is r·n = a·n.

Cartesian form: Ax + By + Cz = D, where A, B, C are direction ratios of the normal. The plane through (x₁, y₁, z₁) with normal ratios A, B, C is A(x − x₁) + B(y − y₁) + C(z − z₁) = 0.

Intercept form: x/p + y/q + z/r = 1 cuts the axes at (p, 0, 0), (0, q, 0), (0, 0, r). For example 2x + 3y + 6z = 12 cuts them at 6, 4 and 2.

Distance of the point (x₁, y₁, z₁) from Ax + By + Cz = D is |Ax₁ + By₁ + Cz₁ − D| / √(A² + B² + C²). From the origin to 2x + 3y + 6z = 12 it is 12/7; from (1, 2, 0) it is |2 + 6 − 12|/7 = 4/7.

Angle between two planes = angle between their normals: cos θ = |n₁·n₂|/(|n₁||n₂|). The floor z = 0 and 2x + 3y + 6z = 12 meet at cos θ = 6/7.

Angle φ between a line (direction b) and a plane (normal n) is the complement of the angle between b and n: sin φ = |b·n|/(|b||n|). A line along 2î + 3ĵ + 6k̂ meets the floor at sin φ = 6/7, φ ≈ 59.0°.

A line is parallel to a plane when b·n = 0, and perpendicular to it when b is parallel to n.