Find the Angle Between Two Planes from Three Points

To find the angle between two planes, compare their normal vectors: each normal is perpendicular to its plane, so the angle between the planes is the acute angle between the normals. For the three given points, a cross product gives a normal to the plane they determine.

The plane $z=0$ is the $xy$-plane, whose normal vector is $(0,0,1)$. The plane through $(0,0,0)$, $(2,3,0)$, and $(0,3,3)$ has normal vector $(3,-2,2)$. Using these normals gives an acute angle of about $60.9^\circ$.

Find two vectors in the plane

Use one of the points as a base point and subtract it from the other two. Since the first point is the origin, the vectors are

$$\mathbf{u}=(2,3,0), \qquad \mathbf{v}=(0,3,3).$$

Both vectors lie in the plane through the three points. They are not parallel, so they span that plane.

Compute a normal vector

A cross product of two nonparallel vectors in a plane is perpendicular to that plane. Take

$$\begin{aligned} \mathbf{n}_2 &= \mathbf{u}\times\mathbf{v} \\ &= (3\cdot 3-0\cdot 3,\;0\cdot 0-2\cdot 3,\;2\cdot 3-3\cdot 0) \\ &= (9,-6,6) \\ &= 3(3,-2,2). \end{aligned}$$

A normal vector can be multiplied by any nonzero scalar without changing its direction, so use the simpler normal $(3,-2,2)$.

The other plane is $z=0$. Its normal is

$$\mathbf{n}_1=(0,0,1),$$

because the equation has coefficient vector $(0,0,1)$.

Apply the angle formula

For nonzero vectors $\mathbf{a}$ and $\mathbf{b}$, the cosine of the angle between them is their dot product divided by the product of their lengths. For the standard angle between planes, take the absolute value in the numerator so the result is the acute angle:

$$\cos\theta=\frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{\|\mathbf{n}_1\|\,\|\mathbf{n}_2\|}.$$

Here,

$$\mathbf{n}_1\cdot\mathbf{n}_2 = (0)(3)+(0)(-2)+(1)(2)=2,$$

and

$$\|\mathbf{n}_1\|=1, \qquad \|\mathbf{n}_2\|=\sqrt{3^2+(-2)^2+2^2}=\sqrt{17}.$$

Therefore,

$$\theta=\arccos\left(\frac{2}{\sqrt{17}}\right)\approx 60.9^\circ.$$

In radians, this is approximately $1.0637$. Thus, the acute angle between the planes is about $60.9^\circ$.

Check by writing the plane equation

The plane through the points contains the origin, so its equation has the form $ax+by+cz=0$. Its normal $(3,-2,2)$ gives

$$3x-2y+2z=0.$$

The listed points satisfy this equation: substituting $(2,3,0)$ gives $6-6=0$, and substituting $(0,3,3)$ gives $-6+6=0$. This confirms the normal vector used in the angle calculation.

Common mistakes

• Using a point instead of a normal: The coordinates of a point on a plane do not, by themselves, give its orientation. First construct two direction vectors, then take their cross product.
• Forgetting the absolute value: Without it, the dot-product formula can return an obtuse angle if one normal points in the opposite direction. A plane has no preferred normal direction, so its standard angle with another plane is acute or right.
• Confusing the angle with the angle between normals: The acute angle between two planes equals the acute angle between their normals. If you use a signed or oriented angle convention, the result can differ; the usual geometry question asks for the smaller, non-oriented angle.
• Using the wrong normal for $z=0$: The $xy$-plane is horizontal, so its normal points in the $z$ direction: $(0,0,1)$, not a vector lying in the plane.