TheoremProved

Distance from a point to a plane

Statement

In $Oxyz$ , the perpendicular distance from $M_0(x_0,y_0,z_0)$ to the plane $(P): Ax+By+Cz+D=0$ (with $A^2+B^2+C^2>0$ ) is $d(M_0,(P))=\frac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$ .

Why is it true?

Evaluating $Ax_0+By_0+Cz_0+D$ measures how far $M_0$ fails to satisfy the plane's equation, scaled by the length $|\vec{n}|=\sqrt{A^2+B^2+C^2}$ of the normal vector $\vec{n}=(A,B,C)$ . Dividing by $|\vec{n}|$ converts that algebraic residual into the true geometric distance along the normal direction.

Proof sketch

Pick any point M1(x1,y1,z1) on the plane (P). Because M1 lies on (P), its coordinates satisfy Ax1 + By1 + Cz1 + D = 0, so D = -(Ax1 + By1 + Cz1).

Let H be the orthogonal projection of M0 onto (P). The vector M1M0 = (x0-x1, y0-y1, z0-z1) decomposes into a component along the plane plus the normal component HM0, so the distance d(M0,(P)) = |HM0| is the absolute value of the scalar projection of M1M0 onto the normal vector n = (A,B,C).

That scalar projection has magnitude |n . M1M0| / |n| = |A(x0-x1) + B(y0-y1) + C(z0-z1)| / sqrt(A^2+B^2+C^2).

Replacing -(Ax1 + By1 + Cz1) with D inside the absolute value yields |Ax0 + By0 + Cz0 + D| / sqrt(A^2+B^2+C^2), completing the proof.

Topics that use this theorem

Coordinates in space

Step-by-step proofs

No step-by-step proof yet for this theorem.