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Medium

The Normal Vector and Equation of a Plane

Simple Explanation

A plane can be described by any one point on it, plus a "normal vector" (a,b,c) β€” a vector perpendicular to the entire plane. The plane's equation is a(xβˆ’xβ‚€)+b(yβˆ’yβ‚€)+c(zβˆ’zβ‚€)=0, which simplifies to ax+by+cz=d.

Why Do We Need It?

This is the standard way to describe a flat surface in 3D space mathematically β€” essential for everything from computer graphics to structural engineering.

See It

A plane with its normal vector n
nPβ‚€

A schematic sketch of a tilted flat plane, shown as a shaded parallelogram, with an arrow labelled n pointing perpendicular to it

Formula

The Equation of a Plane from a Normal Vector

a(xβˆ’xβ‚€) + b(yβˆ’yβ‚€) + c(zβˆ’zβ‚€) = 0, equivalently ax+by+cz = d

A plane through a known point (xβ‚€,yβ‚€,zβ‚€), perpendicular to a normal vector (a,b,c), is described by this equation β€” the normal vector's components become the plane equation's coefficients.

(a,b,c)
β€” the normal vector β€” perpendicular to every line lying in the plane
(xβ‚€,yβ‚€,zβ‚€)
β€” a known point on the plane
d
β€” the constant axβ‚€+byβ‚€+czβ‚€, once the equation is expanded

When to use it: Whenever a plane needs to be described by an equation, from a known point and its normal direction.

Worked Example

Find the equation of a plane from a point and normal vector

Find the equation of the plane through Pβ‚€(1,2,3) with normal vector (4,βˆ’1,2).

    Why Does This Work?

    Every vector from the known point Pβ‚€ to any other point (x,y,z) IN the plane must be perpendicular to the normal vector (a,b,c) β€” and two vectors are perpendicular exactly when their dot product is zero, which is precisely what a(xβˆ’xβ‚€)+b(yβˆ’yβ‚€)+c(zβˆ’zβ‚€)=0 expresses.

    Real-Life Example

    Modeling a flat rooftop or solar panel surface

    An architect models a flat rooftop surface, needing both a reference point on it and the exact direction it is tilted (its normal direction).

    The normal vector and a single known point are exactly enough information to write down the plane's full equation.

    Practice

    Find d, for the plane through (2,1,5) with normal vector (3,2,βˆ’1), written as 3x+2yβˆ’z=d.

    Medium

    Common mistake

    Sign errors when distributing a(xβˆ’xβ‚€)+b(yβˆ’yβ‚€)+c(zβˆ’zβ‚€) β€” especially when a, b, or c is negative, it is easy to mishandle the resulting double-negative.

    Quick Review

    • a(xβˆ’xβ‚€) + b(yβˆ’yβ‚€) + c(zβˆ’zβ‚€) = 0, simplifying to ax+by+cz=d.
    • The normal vector (a,b,c) is perpendicular to every line lying in the plane.
    • d = axβ‚€+byβ‚€+czβ‚€, once the equation is expanded.