Angles Between Planes

Why normals determine the angle between planes

To find the angle between two planes, it is useful to replace each plane with a vector that describes its orientation. A normal vector is perpendicular to every direction lying in its plane, so the angle between the planes is closely related to the angle between their normals.

There is one important qualification: a plane does not have a preferred choice between a normal vector and its negative. Reversing a normal changes the angle between the chosen vectors, but it does not change the plane. The angle between planes must therefore be defined so it does not depend on that choice.

Normal vectors and the angle between planes

A plane with equation $ax+by+cz=d$, where $a,b,c$ are not all zero, has normal vector $\mathbf{n}=(a,b,c)$. If a plane is given in point-normal form, its displayed normal vector serves the same purpose. Any nonzero scalar multiple of a plane’s normal is also a normal to that plane.

Definition: The angle between two planes is the smaller angle between their orientations. It is a non-obtuse angle, so it lies between $0$ and $\frac{\pi}{2}$, inclusive. If $\mathbf{n}_1$ and $\mathbf{n}_2$ are any nonzero normal vectors to the planes, that angle $\phi$ satisfies $$ \cos\phi = \frac{|\mathbf{n}_1\cdot\mathbf{n}_2|} {\|\mathbf{n}_1\|\,\|\mathbf{n}_2\|}, \qquad 0\leq \phi\leq \frac{\pi}{2}. $$

The absolute value is essential. Without it, the formula would give the angle between the particular vectors chosen as normals, which can be obtuse. The plane angle describes the planes’ orientations, not the arbitrary direction in which each normal points.

In degrees, the same definition uses an angle from $0^\circ$ to $90^\circ$. The planes are perpendicular exactly when their normals have dot product zero. They are parallel exactly when their normals are parallel, which gives a plane angle of $0^\circ$. The planes may be distinct and parallel, or they may be the same plane.

Deriving the formula

For nonzero vectors, the dot-product formula gives the angle $\theta$ between the chosen normals: $\cos\theta=(\mathbf{n}_1\cdot\mathbf{n}_2)/(\|\mathbf{n}_1\|\|\mathbf{n}_2\|)$. This vector angle is between $0$ and $\pi$. If the dot product is negative, $\theta$ is obtuse. But a normal can be replaced by its negative without changing its plane, and that replacement changes an obtuse vector angle to its supplement.

For a geometric view, consider a cross-section by a plane perpendicular to the line where two nonparallel planes intersect. Each original plane appears in the cross-section as a line. A normal to either original plane lies in this cross-sectional plane and is perpendicular to the corresponding cross-sectional line. Rotating both lines by $90^\circ$ to obtain their normals preserves the smaller angle between the lines. The angle between the planes is therefore the smaller angle between the normal directions.

Derivation: Let $\theta\in[0,\pi]$ be the angle between the chosen normals, and let $c=(\mathbf{n}_1\cdot\mathbf{n}_2)/(\|\mathbf{n}_1\|\|\mathbf{n}_2\|)=\cos\theta$. The smaller angle between the normal directions, allowing either normal to be reversed, is $\min(\theta,\pi-\theta)$. If $c\geq 0$, then $\theta\leq\pi/2$, so this smaller angle is $\theta$, with cosine $c=|c|$. If $c<0$, then $\theta>\pi/2$, so the smaller angle is $\pi-\theta$, whose cosine is $-\cos\theta=|c|$. Thus in either case its cosine is $|c|$, giving the stated formula.

This also shows why the formula is unaffected by choices of normal. Multiplying either normal by a nonzero scalar changes the numerator and denominator by the same absolute factor; reversing a normal may change the sign of the dot product, but the absolute value removes that sign.

Worked examples

Worked Example: Finding the angle from two plane equations

Find the angle between the planes $2x-y+2z=4$ and $x+2y+2z=1$.

The coefficients of $x,y,z$ give normals $\mathbf{n}_1=(2,-1,2)$ and $\mathbf{n}_2=(1,2,2)$. Their dot product is $2(1)+(-1)(2)+2(2)=2-2+4=4$.

Their lengths are $\|\mathbf{n}_1\|=\sqrt{2^2+(-1)^2+2^2}=\sqrt{9}=3$ and $\|\mathbf{n}_2\|=\sqrt{1^2+2^2+2^2}=\sqrt{9}=3$. Therefore, $\cos\phi=|4|/(3\cdot3)=4/9$, and $\phi=\arccos(4/9)\approx63.6^\circ$, rounded to one decimal place.

Check: both normals are nonzero, and the dot product is $4$, so the cosine of the angle between these chosen normals is $4/9$. Since $4/9$ is positive, their vector angle is already acute; the plane angle is the same $63.6^\circ$. Substitution into the defining formula gives $\cos\phi=4/9$, as required.

Worked Example: An obtuse angle between chosen normals

Find the angle between $x+2y+2z=0$ and $-2x+y-2z=5$, and compare it with the angle between the displayed normals.

Use $\mathbf{n}_1=(1,2,2)$ and $\mathbf{n}_2=(-2,1,-2)$. Their dot product is $1(-2)+2(1)+2(-2)=-2+2-4=-4$. Their lengths are $\|\mathbf{n}_1\|=\sqrt{1+4+4}=3$ and $\|\mathbf{n}_2\|=\sqrt{4+1+4}=3$.

The angle $\theta$ between these particular normals satisfies $\cos\theta=-4/(3\cdot3)=-4/9$, so $\theta=\arccos(-4/9)\approx116.4^\circ$, rounded to one decimal place. The plane angle instead satisfies $\cos\phi=|-4|/(3\cdot3)=4/9$, giving $\phi\approx63.6^\circ$.

Check: $116.4^\circ+63.6^\circ=180.0^\circ$ to the stated precision, so the two angles are supplements. Reversing $\mathbf{n}_2$ gives $(2,-1,2)$, whose dot product with $\mathbf{n}_1$ is $4$; this changes the chosen-normal angle to the acute $63.6^\circ$ without changing either plane.

Worked Example: Recognizing perpendicular planes

Determine the angle between $x+y+z=1$ and $x-y=0$.

Normals are $\mathbf{n}_1=(1,1,1)$ and $\mathbf{n}_2=(1,-1,0)$. Their dot product is $1(1)+1(-1)+1(0)=1-1+0=0$. Their lengths are $\|\mathbf{n}_1\|=\sqrt{1+1+1}=\sqrt{3}$ and $\|\mathbf{n}_2\|=\sqrt{1+1+0}=\sqrt{2}$. Hence $\cos\phi=|0|/(\sqrt{3}\sqrt{2})=0$, so $\phi=\arccos(0)=90^\circ$.

Check: the normal vectors have dot product zero, which is exactly the condition for the planes to be perpendicular. The formula also gives $\cos 90^\circ=0$, agreeing with the calculation.

Worked Example: Parallel planes with oppositely directed normals

Find the angle between $x-2y+2z=3$ and $-2x+4y-4z=7$.

The normals are $\mathbf{n}_1=(1,-2,2)$ and $\mathbf{n}_2=(-2,4,-4)=-2\mathbf{n}_1$. Their dot product is $1(-2)+(-2)(4)+2(-4)=-2-8-8=-18$. Their lengths are $\|\mathbf{n}_1\|=\sqrt{1+4+4}=3$ and $\|\mathbf{n}_2\|=\sqrt{4+16+16}=6$. Thus $\cos\phi=|-18|/(3\cdot6)=1$, so $\phi=0^\circ$.

Check: the normal vectors are scalar multiples, so the planes are parallel and have angle zero. The equations are not equivalent: multiplying the first plane’s left side by $-2$ would give $-2x+4y-4z=-6$, not $7$. They are distinct parallel planes, and the result $0^\circ$ is consistent with that geometry. The chosen normals point in opposite directions, so their vector angle is $180^\circ$; the plane angle remains $0^\circ$.

A reliable calculation process

For planes in general form, extract the normal vectors directly from the coefficients of $x,y,z$. Then compute the dot product and both vector lengths before applying the inverse cosine. Keeping the absolute value in the formula makes the answer the non-obtuse plane angle regardless of how the normals were written.

Read off the normals.
For $a_1x+b_1y+c_1z=d_1$ and $a_2x+b_2y+c_2z=d_2$, take $\mathbf{n}_1=(a_1,b_1,c_1)$ and $\mathbf{n}_2=(a_2,b_2,c_2)$.

Compute the dot product and lengths.
Find $\mathbf{n}_1\cdot\mathbf{n}_2$, $\|\mathbf{n}_1\|$, and $\|\mathbf{n}_2\|$. Both normals must be nonzero.

Apply the plane-angle formula.
Calculate $\phi=\arccos\left(|\mathbf{n}_1\cdot\mathbf{n}_2|/(\|\mathbf{n}_1\|\|\mathbf{n}_2\|)\right)$. The fraction is between zero and one, so $\phi$ is between $0$ and $\pi/2$.

Interpret the result.
A zero dot product means perpendicular planes; parallel normals mean parallel planes and plane angle zero. If the original dot product is negative, the chosen normals make an obtuse angle, but the plane angle is still non-obtuse.

Common mistake: forgetting the absolute value

A frequent error is to use the angle formula for two vectors without adjusting it for the plane-angle convention. For example, in the second worked example, using $\arccos(-4/9)$ directly produces $116.4^\circ$. That is the angle between the displayed normal vectors, not the angle between the planes.

The correction is to use the absolute value: $\arccos(|-4/9|)=\arccos(4/9)\approx63.6^\circ$. The obtuse answer is not meaningless; it correctly describes those particular directed normals. It is simply not the smaller angle between the planes.

Another useful check is that a plane angle cannot exceed $90^\circ$. If a calculation returns an obtuse angle as the angle between planes, check whether the absolute value was omitted or whether you have reported the chosen-normal angle instead.

Why this distinction matters

The angle between planes is a geometric property of the planes, so it must not depend on how their equations are scaled. Multiplying a plane equation by $-1$ reverses its displayed normal but leaves the plane unchanged. A formula without the absolute value could give different answers for two equivalent descriptions of the same plane; the non-obtuse formula avoids that inconsistency.

In practice, the method reduces a spatial geometry question to a small number of vector calculations. The key is to be clear about which angle is being requested: the angle between planes, or the angle between specific normal vectors. Both can be calculated, but only the former is independent of the chosen directions of the normals.

Check Your Understanding

For each question, use the plane-angle convention and explain how the normal vectors support your answer.

What normal vector can be read from the plane $3x-y+2z=6$?
Two plane normals have dot product $-5$ and lengths $2$ and $5$. Write the cosine of the angle between the planes.
Why can the angle between chosen normals be obtuse even though the angle between planes is non-obtuse?
What does a zero dot product between two plane normals tell you about the planes?
If two nonzero plane normals are scalar multiples, what is the angle between their planes?

What You'll Learn