1.6 - Angles and distances involving planes

1.6 - Angles and distances involving planes

Normals turn spatial angle and distance questions into scalar products. You will distinguish a line-plane angle from a normal angle, find the angle between planes, and calculate the shortest distance from a point to a plane.

Angle between a line and a plane

A line points upwards through a horizontal plane. Is the angle found by taking its scalar product with a vertical vector the angle with the plane? It is the angle with the normal, so we need its complement.

For a line with nonzero direction d\mathbf d and a plane with nonzero normal n\mathbf n, let β\beta be the smaller angle between their directions, taking it between 0∘0^\circ and 90∘90^\circ. The line-plane angle is α=90∘−β\alpha=90^\circ-\beta. This angle measures the line's inclination above the plane and also lies between 0∘0^\circ and 90∘90^\circ.

[DIAGRAM: asset_name: A line-plane angle is a complement; asset_slug: cam9231-line-plane-angle; caption: In a normal section, the line-plane angle and the line-normal angle add to 90 degrees; description: Exact cross-section through a plane normal and an intersecting line. The plane appears as a horizontal grey line, its normal as a vertical sage arrow, and the line direction as a blue arrow at 35 degrees above the plane. The 35 degree angle alpha and 55 degree angle beta are drawn in distinct arcs. A right-angle marker identifies the plane-normal angle.]
In a normal section, the line-plane angle and the line-normal angle add to 90 degrees

Line-plane angle

sin⁡α=∣d⋅n∣∣d∣ ∣n∣,0≤α≤90∘.\sin\alpha=\frac{|\mathbf d\cdot\mathbf n|}{|\mathbf d|\,|\mathbf n|}, \qquad 0\le\alpha\le90^\circ.

The absolute value makes the result independent of which sense of the line or normal you choose. This follows from sin⁡α=cos⁡β\sin\alpha=\cos\beta.

Not in the formula booklet: learn it. The scalar-product angle formula used to derive it is in the booklet.

When d⋅n=0\mathbf d\cdot\mathbf n=0, the angle is 0∘0^\circ: the line is parallel to, or contained in, the plane. When the directions d\mathbf d and n\mathbf n are parallel, the angle is 90∘90^\circ: the line is perpendicular to the plane. At this boundary its projection onto the plane has zero length, so the formula supplies the angle directly.

Worked example 1: inclination to a horizontal plane

Question. Find the acute angle between r=(2,1,0)+t(1,2,2)\mathbf r=(2,1,0)+t(1,2,2) and the plane z=4z=4.

Plan. Use the direction of the line and the plane's normal, then take an inverse sine.

Here d=(1,2,2)\mathbf d=(1,2,2), n=(0,0,1)\mathbf n=(0,0,1), so

sin⁡α=21+4+4 (1)=23.\sin\alpha=\frac{2}{\sqrt{1+4+4}\,(1)}=\frac23.

Answer. α=sin⁡−1(2/3)=41.8∘\alpha=\sin^{-1}(2/3)=41.8^\circ, to one decimal place.

Check. The line rises 22 units for a horizontal displacement of length 12+22=5\sqrt{1^2+2^2}=\sqrt5. Thus tan⁡α=2/5\tan\alpha=2/\sqrt5, which gives the same angle independently.

Reverse only the line's direction. A geometric line has no preferred sense, so its inclination must remain the same.

Worked example 2: a negative scalar product

Question. Find the acute angle between r=(2,1,0)+s(−1,−2,−2)\mathbf r=(2,1,0)+s(-1,-2,-2) and z=4z=4.

Plan. Take the magnitude of the scalar product before applying the line-plane formula.

sin⁡α=∣(−1,−2,−2)⋅(0,0,1)∣3(1)=∣−2∣3=23.\sin\alpha=\frac{|(-1,-2,-2)\cdot(0,0,1)|}{3(1)} =\frac{|-2|}{3}=\frac23.

Answer. The angle is again 41.8∘41.8^\circ.

Check. The line's direction is the negative of the previous direction, so both equations describe the same geometric line. Its ratio of vertical to horizontal displacement magnitudes is unchanged.

A cosine applied directly to the positive ratio would give 48.2∘48.2^\circ, the smaller angle with the normal. Decide which angle is requested before pressing an inverse-trigonometric key.

Watch out

For a line and a plane, use sine of the angle with the plane, or find the normal angle with cosine and subtract it from 90∘90^\circ.

Use a tilted plane in the next calculation.

Angle between two planes

The smaller angle between two planes equals the smaller angle between their normals. In a cross-section perpendicular to the planes' intersection, each normal is a quarter-turn from its own plane trace; rotating both traces by a quarter-turn preserves their angle.

Unlike the line-plane calculation, this uses cosine of the required angle directly. Unless a different dihedral angle is specified, take the smaller angle, including the limiting values 0∘0^\circ and 90∘90^\circ.

Angle between planes

For nonzero normals n1,n2\mathbf n_1,\mathbf n_2, the smaller plane angle θ\theta satisfies

cos⁡θ=∣n1⋅n2∣∣n1∣ ∣n2∣,0≤θ≤90∘.\cos\theta=\frac{|\mathbf n_1\cdot\mathbf n_2|}{|\mathbf n_1|\,|\mathbf n_2|}, \qquad 0\le\theta\le90^\circ.

Parallel or coincident planes have angle 0∘0^\circ; perpendicular planes have orthogonal normals and angle 90∘90^\circ.

Not in the formula booklet: learn it. The underlying scalar-product formula is in the booklet.

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