3D DistanceJEE Main

Point-to-plane perpendicular distance — interactive Mathematics simulation for IIT-JEE.

Concept

The distance from a point to a plane is measured along the plane's normal. Substituting the point into the plane expression and normalising by $|\vec n|$ gives a signed value whose magnitude is the distance and whose sign tells you which side of the plane the point is on.

Key formula

d = \frac{|A x_0 + B y_0 + C z_0 - D|}{\sqrt{A^2 + B^2 + C^2}}

Derivation

Take the unit normal $\hat n = (A,B,C)/|\vec n|$ . For any point $Q$ on the plane, $\vec n\cdot\vec Q = D$ .

The distance is the projection of $\vec{QP} = \vec P - \vec Q$ onto $\hat n$ : $(\vec n\cdot\vec P - D)/|\vec n|$ .

Dropping a perpendicular from $P$ a signed distance along $\hat n$ lands on the foot $F = P - \big(\tfrac{\vec n\cdot P - D}{|\vec n|}\big)\hat n$ .

Scenarios to explore

3D Distance — Point-to-plane and point-to-line distances.

Real-world applications

Collision and clearance checks in 3D engines.
Fitting planes to point clouds (residual distances).
Computing layer thickness in CAD and manufacturing.

JEE exam tips

Two parallel planes $\vec n\cdot\vec r = D_1$ and $D_2$ are $|D_1 - D_2|/|\vec n|$ apart.
A point lies on the plane exactly when the numerator is zero.

Common mistakes

Forgetting to divide by $|\vec n|$ — that gives the scaled value, not the distance.
Dropping the constant $D$ when substituting the point.
Reporting the signed value when the question wants the unsigned distance.

Exam traps to avoid

The sign of the distance flips depending on the orientation chosen for $\vec n$ .
Use the same form of the equation ( $=D$ vs $-D=0$ ) consistently.