Planes in 3DJEE Main

Normal, intercepts & distance — interactive Mathematics simulation for IIT-JEE.

Concept

A plane in 3D is fixed by a point on it and a perpendicular normal vector $\vec n = (A,B,C)$ . The scalar equation $Ax+By+Cz=D$ packs both: the coefficients are the normal, and $D$ locates the plane along that normal.

Key formula

Ax + By + Cz = D, \qquad d(O) = \frac{|D|}{\sqrt{A^2+B^2+C^2}}

Derivation

Every point $\vec r$ on the plane satisfies $\vec n\cdot\vec r = D$ , because $\vec n$ is perpendicular to every in-plane direction.

Setting two coordinates to zero gives the axis intercepts, e.g. $x = D/A$ .

The distance from the origin is the projection of any point's position onto the unit normal: $|D|/|\vec n|$ . For a general point $P$ replace $D$ with $D - \vec n\cdot\vec P$ .

Scenarios to explore

Planes in 3D — Plane equations, normals and intersections.

Real-world applications

Defining surfaces and clipping in 3D graphics.
Crystal lattice planes (Miller indices).
Constraint surfaces in optimisation.

JEE exam tips

Two planes are parallel iff their normals are proportional.
Angle between planes equals the angle between their normals: $\cos\theta = \frac{\vec n_1\cdot\vec n_2}{|\vec n_1||\vec n_2|}$ .

Common mistakes

Forgetting to normalise by $|\vec n|$ when computing distances.
Reading the intercept as $D$ instead of $D/A$ .
Thinking the normal must be a unit vector — any scalar multiple works.

Exam traps to avoid

A zero coefficient means the plane is parallel to that axis (no intercept).
The same plane has infinitely many equations (scale A, B, C, D together).