Angle between two planes: definition and coordinate formula

2026-10-22 huynhvietkhanh Comments: 0

Use nonzero normal vectors and an absolute-value dot product to find the prescribed angle from 0 to 90 degrees.

High school3D Coordinate Geometry / Solid Geometry & Vectors

Updated: October 22, 2026

The angle between two planes is obtained from the angle between their normal vectors. The absolute value makes the result independent of the chosen normal orientations.

Learning goals: determine an angle from 0 to 90 degrees; use the coordinate formula; distinguish the plane angle from the angle between oriented vectors.

Prerequisites: nonzero normal vectors, the dot product, vector lengths, and the angle between lines.

On this page: Definition · Method · Examples · Practice · Further study

1. How is the angle between planes chosen?

If $(P)$ and $(Q)$ intersect along $\Delta$, choose in each plane a line perpendicular to $\Delta$ at the same point. The angle between these lines is the angle between the planes. Choose the angle no greater than $90^\circ$. Parallel or coincident planes have angle $0^\circ$.

Let $\varphi\in[0^\circ;90^\circ]$ be the required angle, and $\alpha\in[0^\circ;180^\circ]$ the angle between the normal vectors $\vec n_P,\vec n_Q$. Then $\varphi=\min\{\alpha,180^\circ-\alpha\}$. They agree if $\alpha\le90^\circ$ and are supplementary if $\alpha>90^\circ$.

New perpendicular cross-section: the traces of the planes and chosen normals give the same acute angle.

New schematic with one normal reversed: the plane angle remains acute while the normal angle is supplementary.

2. Formula using normal vectors

For two normal vectors different from $\vec0$,

$$\cos\varphi=|\cos\alpha|=\frac{|\vec n_P\cdot\vec n_Q|}{\|\vec n_P\|\,\|\vec n_Q\|}.$$

Let $(P):a_1x+b_1y+c_1z+d_1=0$ and $(Q):a_2x+b_2y+c_2z+d_2=0$, with each coefficient triple of $x,y,z$ different from $(0;0;0)$. Choose $\vec n_P=(a_1;b_1;c_1)$ and $\vec n_Q=(a_2;b_2;c_2)$, and calculate

$$\cos\varphi=\frac{|a_1a_2+b_1b_2+c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}.$$

Calculate the dot product and the two lengths separately, then find $\varphi$ in the prescribed interval. The Cauchy–Schwarz inequality ensures that the ratio belongs to $[0;1]$. It is $0$ when the planes are perpendicular and $1$ when the normals are parallel, meaning that the planes are parallel or coincident.

The constants $d_1,d_2$ affect the plane positions but do not appear in the angle formula. Multiplying a plane equation by a number different from $0$ does not change the angle: numerator and denominator are both multiplied by its absolute value.

Reversing one normal may replace $\alpha$ by its supplementary angle, but leaves $|\cos\alpha|$ unchanged. Retain the absolute value even when a particular example has a positive dot product. Do not substitute the sine formula for a line–plane angle into a two-plane problem.

3. Worked examples

Example 1. Two planes in Cartesian space

Find the angle between $(P):x+2y+z+4=0$ and $(Q):-x+y+2z+3=0$. Choose $\vec n_P=(1;2;1)$ and $\vec n_Q=(-1;1;2)$. Then

$$\vec n_P\cdot\vec n_Q=-1+2+2=3,\qquad\|\vec n_P\|=\|\vec n_Q\|=\sqrt6.$$

$$\cos\varphi=\frac{|3|}{\sqrt6\sqrt6}=\frac12.$$

Since $\varphi\in[0^\circ;90^\circ]$, we conclude $\varphi=60^\circ$.

Example 2. A negative dot product

For $(P):x=0$ and $(Q):-x+y-1=0$, the normals $(1;0;0)$ and $(-1;1;0)$ have dot product $-1$. Thus $\cos\varphi=1/\sqrt2$ and $\varphi=45^\circ$. The angle between the chosen normals is $135^\circ$; it is not the prescribed plane angle.

Example 3. Two special cases

The planes $z=0$ and $z=3$ have parallel normals, so their angle is $0^\circ$, although they have no common point. The planes $z=0$ and $x=0$ have perpendicular normals, so their angle is $90^\circ$. If two equations differ only by a multiplier different from $0$, they represent the same plane and still have angle $0^\circ$.

4. Practice

Find the angle between each pair of planes:

$x+y=0$ and $x+z=1$.
$2x-y+z=3$ and $-4x+2y-2z=5$.
$x+y+z=1$ and $x-y=2$.

Hints and answers

The normals $(1;1;0)$ and $(1;0;1)$ give $\cos\varphi=1/2$, so $\varphi=60^\circ$.
The normals are parallel, so the angle is $0^\circ$. These planes are distinct and parallel: multiplying the first equation by $-2$ gives right-hand constant $-6$, different from $5$.
Their dot product is $1-1=0$, so the angle is $90^\circ$.

Common mistakes: choosing normal $\vec0$; answering with an obtuse normal angle; using squared lengths in the denominator; omitting the absolute value; confusing angle $0^\circ$ with coincidence of the planes.

5. Further study and resources

Normal vectors to a plane (English lesson) · Equations of planes (English lesson) · Relative positions of two planes (English lesson) · Angle between a line and a plane: definition and calculation (English lesson).

Original resource on Scribd (external link).