The Angle Between a Line and a Plane

Which angle?

A line that meets a plane makes many different angles with the many lines of the plane through the meeting point. The angle between the line and the plane is the smallest of them: the angle between the line and its shadow on the plane, cast by a light shining straight down along the normal. It is always between 0° and 90°.

The line, its shadow and the normal through the meeting point all lie in one flat slice, and in that slice they make a right triangle: the normal is at 90° to the shadow. Call the angle between the line and the plane $\theta$θ, and the angle between the line and the normal $\varphi$φ. They are the two acute angles that fill the right angle between the normal and the plane, so $\theta +\varphi =90^{\circ }$θ + φ = 90°.

A box 1 by 2 by 2 on the floor. Its space diagonal, from a floor corner to the opposite top corner, runs in the direction (1, 2, 2) and has length 3. Its shadow on the floor is the floor diagonal, of length $\sqrt{5}$√5, and the angle marked at the foot is the angle between the line and the floor.

From the dot product to the sine

The dot product gives the angle between two vectors. With d along the line and n normal to the plane, the angle $\varphi$φ between the line and the normal has $\cos \varphi =|d\cdot n|÷\left(|d||n|\right)$cos φ = |d · n| ÷ (|d| |n|).

The angle the question asks for is $\theta =90^{\circ }-\varphi$θ = 90° − φ. In a right triangle, the side adjacent to one acute angle is opposite the other, so the cosine of one angle is the sine of the other: $\sin \theta =\cos \left(90^{\circ }-\theta \right)=\cos \varphi$sin θ = cos(90° − θ) = cos φ.

So the same quotient gives the angle with the plane directly: $\sin \theta =|d\cdot n|÷\left(|d||n|\right)$sin θ = |d · n| ÷ (|d| |n|).

Read the x-axis as the plane seen edge-on and the y-axis as its normal. The radius, of length 1, makes the angle $\theta$θ with the x-axis and $90^{\circ }-\theta$90° − θ with the y-axis. Its height, the gold leg, is $\sin \theta$sin θ. That height is also how far the radius reaches along the y-axis, which for a radius of length 1 at the angle $90^{\circ }-\theta$90° − θ to that axis is $\cos \left(90^{\circ }-\theta \right)$cos(90° − θ). Turn the radius: the two stay equal at every angle.

Sine, not cosine

The usual mistake is to write $\cos \theta =|d\cdot n|÷\left(|d||n|\right)$cos θ = |d · n| ÷ (|d| |n|), out of habit from the angle between two vectors. That quotient is the cosine of the angle to the normal. The normal points out of the plane, not along it, so the angle it gives is the wrong one of the pair: the answer would be $\varphi$φ, the complement of $\theta$θ.

A line meeting the floor

Take a line with direction d = (1, 2, 2) meeting the floor, the plane z = 0, whose normal is n = (0, 0, 1).

Then d · n = 1 × 0 + 2 × 0 + 2 × 1 = 2, $|d|=\sqrt{1+4+4}=3$|d| = √(1 + 4 + 4) = 3 and |n| = 1. So $\sin \theta =2/\left(3\times 1\right)=\frac{2}{3}$sin θ = 2 / (3 × 1) = 2/3, and $\theta ={\sin }^{-1}\left(\frac{2}{3}\right)=41.8^{\circ }$θ = sin⁻¹(2/3) = 41.8° to one decimal place, about 42°. The angle to the normal is $\varphi =90^{\circ }-41.8^{\circ }=48.2^{\circ }$φ = 90° − 41.8° = 48.2°.

Check it with the triangle in the box. Along the line, the height rises 2 while the shadow runs $\sqrt{1^2+2^2}=\sqrt{5}$√(1² + 2²) = √5 across the floor, so $\tan \theta =\frac{2}{\sqrt{5}}=0.894$tan θ = 2/√5 = 0.894, which gives $\theta =41.8^{\circ }$θ = 41.8° again.

The triangle inside the box: the line of length 3, its shadow $\sqrt{5}$√5 along the floor, and the rise 2 along the normal. The angle $\theta$θ at the foot has opposite side 2 and hypotenuse 3, so $\sin \theta =\frac{2}{3}$sin θ = 2/3.

A tilted plane

Take the line r = t(1, 1, 0) and the plane x + z = 4. The plane has normal n = (1, 0, 1).

Then d · n = 1 + 0 + 0 = 1, $|d|=\sqrt{2}$|d| = √2 and $|n|=\sqrt{2}$|n| = √2, so $\sin \theta =1/\left(\sqrt{2}\times \sqrt{2}\right)=\frac{1}{2}$sin θ = 1 / (√2 × √2) = 1/2 and $\theta =30^{\circ }$θ = 30°. The line is at 60° to the normal, and 30° + 60° = 90°.

The line reversed

The direction −d = (−1, −2, −2) describes the same line, run the other way. Then (−d) · n = −2, but the modulus |−2| = 2 gives $\sin \theta =\frac{2}{3}$sin θ = 2/3 again.

Without the modulus the cosine of the angle between the vectors would be $-\frac{2}{3}$−2/3, an obtuse angle of 131.8°, and 90° minus it would be negative. The modulus picks the acute angle between the two lines, which is the one that belongs in the triangle.

Parallel and perpendicular

If d · n = 0, then $\sin \theta =0$sin θ = 0 and $\theta =0^{\circ }$θ = 0°: the line is perpendicular to the normal, so it runs parallel to the plane, or lies in it. The line with direction (2, −1, 0) and the plane x + 2y + 3z = 6 give 2 − 2 + 0 = 0.

If d is a multiple of n, the line runs along the normal and meets the plane at 90°. For d = (2, 4, 6) = 2(1, 2, 3) and the same plane, d · n = 2 + 8 + 18 = 28, $|d|=2\sqrt{14}$|d| = 2√14 and $|n|=\sqrt{14}$|n| = √14, so $\sin \theta =28/\left(2\times 14\right)=1$sin θ = 28 / (2 × 14) = 1 and $\theta =90^{\circ }$θ = 90°.

The usual mistakes

Giving the angle to the normal. If the line is at 25° to the normal, it is at 90° − 25° = 65° to the plane, not 25°.

Adding instead of taking away. The two angles make 90° together; 90° + 25° = 115° is more than a right angle.

Using $\cos \theta$cos θ. The quotient |d · n| ÷ (|d| |n|) is the sine of the angle with the plane.

Dividing by one length. With d · n = 4, |d| = 3 and |n| = 5, $\sin \theta =4/\left(3\times 5\right)=\frac{4}{15}$sin θ = 4 / (3 × 5) = 4/15. Dividing by 3 alone gives $\frac{4}{3}$4/3, more than 1, which no sine can be.

Sunlight on a roof

In the application below, the sun’s rays are a line and the glass roof of a greenhouse is a plane. The dot product with the roof’s normal gives the angle to the normal, and taking it from 90° gives the angle at which the light strikes the glass.