Planes in Space

The General Equation of a Plane

The Geometric Meaning of the Normal Vector

The normal vector $\vec{n} = (A, B, C)$ is perpendicular to every vector in the plane, and so it is perpendicular to the plane itself.

Relations between the Coefficients

• $A$, $B$ and $C$ cannot all be zero
• $D$ is determined by the position of the plane
• multiplying both sides of the equation by a nonzero constant does not change the plane

The Point-Normal Form of a Plane

The Features of the Point-Normal Form

• it gives the normal vector and a point on the plane directly
• it makes it easy to build an equation from geometric conditions
• the form is intuitive and easy to understand

The Intercept Form of a Plane

The Geometric Meaning of the Intercepts

• $a$: the $x$-coordinate of the point where the plane meets the $x$-axis
• $b$: the $y$-coordinate of the point where the plane meets the $y$-axis
• $c$: the $z$-coordinate of the point where the plane meets the $z$-axis

The Three-Point Form of a Plane

How the Three-Point Form Is Derived

It uses the property that the scalar triple product is zero: the necessary and sufficient condition for three vectors to be coplanar is that their scalar triple product vanishes.

Converting between the Forms

The General Form → the Point-Normal Form

Convert $Ax + By + Cz + D = 0$ into:

$A(x - x_0) + B(y - y_0) + C(z - z_0) = 0$

where $P_0(x_0, y_0, z_0)$ is a point on the plane.

The Point-Normal Form → the General Form

Simply expand the point-normal form to get the general form.

The Intercept Form → the General Form

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \Rightarrow \frac{x}{a} + \frac{y}{b} + \frac{z}{c} - 1 = 0$

---

Exercises

Exercise 1

Find the equation of the plane passing through the point $A(1, 2, 3)$ with normal vector $(2, -1, 1)$.

Exercise 2

Find the equation of the plane passing through the three points $A(1, 0, 0)$, $B(0, 1, 0)$ and $C(0, 0, 1)$.

---

Summary

Symbols Used in This Article