F3 - Angles involving planes

Angle between planes from normals

Can the same dot-product calculation describe two different geometric angles? Consider the planes $x+z=2$ and $y+z=5$. Their normals are $(1,0,1)$ and $(0,1,1)$. A scalar product compares those normals, but you need to understand why that gives an angle between the planes.

Imagine looking along the intersection of two nonparallel planes. In a cross-section perpendicular to that intersection, the planes appear as two intersecting lines, and their normals are perpendicular to those lines. Turning both lines through the same right angle preserves their smaller angle. Therefore the smaller angle between planes equals the smaller angle between their normal directions.

[DIAGRAM: asset_name: Plane angles from perpendicular normals; asset_slug: aqa7367-plane-angle; caption: In a perpendicular cross-section, the smaller plane angle equals the smaller normal angle; description: An equal-scale cross-section shows two traces of planes meeting at an angle of sixty degrees. A second panel shows a normal perpendicular to each trace, translated to a common origin. The normals also form sixty degrees. The blue plane and blue normal correspond, as do the clay plane and clay normal. Right-angle markers identify the perpendicular relationships.]

Neither plane has a preferred normal arrow. Reversing either normal must leave the plane angle unchanged, so use an absolute value just as for unoriented lines.

Constants in the plane equations locate the planes; their normals determine the angle.

Now multiply the second plane equation by $-1$. Only the chosen normal arrow changes.

Some questions explicitly request an obtuse angle between planes. First find the smaller angle, then take its supplement when required. For $0<\alpha<90^\circ$, the other dihedral angle is $180^\circ-\alpha$. The unqualified smaller angle includes $0^\circ$ and $90^\circ$ as boundary cases.

Line-plane angle as a complement

The angle between a line and a plane is measured from the line to its projection in the plane. Use the angle in $[0,90^\circ]$. When the line is perpendicular to the plane its projection has no direction, and the line-plane angle is defined as $90^\circ$.

The normal is perpendicular to the plane. In the cross-section containing the line direction and the normal, the smaller angle $\beta$ between those directions and the line-plane angle $\alpha$ therefore satisfy $\alpha+\beta=90^\circ$.

[DIAGRAM: asset_name: Complementary line and plane angles; asset_slug: aqa7367-line-plane-angle; caption: The line-plane angle and the smaller angle to the normal add to a right angle; description: An exact two-dimensional cross-section contains a horizontal trace of a plane, a vertical normal and a line direction at thirty degrees above the plane. The angle alpha from the plane to the line is thirty degrees and beta from the line to the normal is sixty degrees. A pale right-angle marker shows that the plane trace and normal are perpendicular. A note states alpha plus beta equals ninety degrees.]

If $\mathbf d\ne\mathbf0$ is the line direction and $\mathbf n\ne\mathbf0$ is the plane normal, the dot product gives $\cos\beta$, which equals $\sin\alpha$. This is why a direct line-plane calculation uses inverse sine, not inverse cosine.

The absolute value removes arbitrary choices of the line arrow or normal arrow.