Trigonometry in Three Dimensions

Two diagonals in a box

A box, or cuboid, measures 4 long, 3 wide and 2 high. Two diagonals in it are not edges. The floor diagonal runs across the bottom face from one corner to the opposite corner. The space diagonal runs from a bottom corner, through the inside of the box, to the top corner farthest from it.

Neither length can be measured straight off the box, and the space diagonal does not lie in any face. The method is to find right triangles that lie flat, in one plane, and to use Pythagoras and the trigonometric ratios inside them, one triangle at a time.

A box 4 by 3 by 2, with the floor diagonal dashed across the bottom and the space diagonal running from a bottom corner to the opposite top corner.

The floor diagonal

The floor is a flat rectangle, 4 by 3. Its diagonal cuts it into two right triangles, with the right angle at a corner of the rectangle, and the edges of 4 and 3 as the two shorter sides.

By Pythagoras, the floor diagonal is $\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$√(4² + 3²) = √(16 + 9) = √25 = 5.

The floor is a 4 by 3 rectangle, so its diagonal is $\sqrt{16+9}=5$√(16 + 9) = 5.

A triangle standing on the floor diagonal

At the end of the floor diagonal, an upright edge of 2 rises to the top corner. That edge is vertical, and the floor is horizontal, so the edge is at a right angle to every line on the floor that passes through its foot. One of those lines is the floor diagonal.

So the floor diagonal, the upright edge and the space diagonal make a right triangle, standing upright inside the box. Its base is 5, its height is 2, and its hypotenuse is the space diagonal.

By Pythagoras again, the space diagonal is $\sqrt{5^2+2^2}=\sqrt{25+4}=\sqrt{29}$√(5² + 2²) = √(25 + 4) = √29, which is 5.39 to 2 decimal places.

The shaded right triangle stands on the floor diagonal of 5, with the upright edge of 2. Its hypotenuse is the space diagonal, $\sqrt{29}$√29.

Both rounds at once

The first round gave the square of the floor diagonal as $4^2+3^2$4² + 3². The second round added the square of the height. So the square of the space diagonal is $4^2+3^2+2^2=16+9+4=29$4² + 3² + 2² = 16 + 9 + 4 = 29.

For any box with length l, width w and height h, the space diagonal d has $d^2=l^2+w^2+h^2$d² = l² + w² + h², so $d=\sqrt{l^2+w^2+h^2}$d = √(l² + w² + h²). A box 3 by 4 by 12 has a floor diagonal of $\sqrt{9+16}=5$√(9 + 16) = 5 and a space diagonal of $\sqrt{25+144}=\sqrt{169}=13$√(25 + 144) = √169 = 13; the formula gives $\sqrt{9+16+144}=13$√(9 + 16 + 144) = 13 directly.

The box 4 by 3 by 2, with the floor diagonal in gold and the upright edge in green. The space diagonal’s length is printed as |v|. Drag any of the three edges: the floor diagonal is always $\sqrt{x^2+y^2}$√(x² + y²), and the space diagonal is always $\sqrt{x^2+y^2+z^2}$√(x² + y² + z²). Set the box to 3 by 4 by 12 to see 5 and 13.

The angle between a line and a plane

The space diagonal rises from the floor at some angle. To measure the angle between a line and a plane, drop a perpendicular from the top of the line straight down to the plane. The line on the plane from the foot of the line to the foot of that perpendicular is the projection of the line, its shadow when a light shines straight down. The angle between the line and the plane is the angle between the line and its projection.

The projection of the space diagonal onto the floor is the floor diagonal, because the top corner sits straight above the far end of it. So the angle $\theta$θ is the angle at the foot of the upright triangle, between the space diagonal and the floor diagonal.

Seen from $\theta$θ, the height of 2 is opposite and the floor diagonal of 5 is adjacent, so $\tan \theta =2/5=0.4$tan θ = 2 / 5 = 0.4, and $\theta ={\tan }^{-1}\left(0.4\right)=21.8^{\circ }$θ = tan⁻¹(0.4) = 21.8°, to 1 decimal place.

The angle $\theta$θ between the space diagonal and the floor is in the upright triangle, between the space diagonal and its projection, the floor diagonal: $\tan \theta =2/5$tan θ = 2 / 5.

Why the projection, and not another line

The space diagonal makes a different angle with each line on the floor. With the edge of 4, for example, the angle $\varphi$φ has $\cos \varphi =4÷\sqrt{29}$cos φ = 4 ÷ √29, so $\varphi =42.0^{\circ }$φ = 42.0°. That edge is not straight below the diagonal, so it is not the right line to measure to.

Of all the lines on the floor through the foot of the diagonal, the projection makes the smallest angle with it, 21.8°. That smallest angle is what the angle between a line and a plane means.

The same triangle holds a second angle, at the top corner, between the space diagonal and the upright edge. There the floor diagonal is opposite and the height is adjacent, so its tangent is 5 ÷ 2 and the angle is 68.2°. The two angles add to 90°, as the two acute angles of a right triangle always do.

The usual mistakes

Adding or multiplying the sides of the floor. 4 + 3 = 7 is a walk round two edges, and 4 × 3 = 12 is the area. The diagonal needs Pythagoras: $\sqrt{16+9}=5$√(16 + 9) = 5.

Stopping at the floor. The floor diagonal, 5, is only the first round; the space diagonal still has to climb the height, so $d^2=25+4=29$d² = 25 + 4 = 29, not 25.

Taking the box’s three edges as one triangle. The length, width and height meet at a single corner and never close up into a triangle. The upright triangle is the floor diagonal, the height and the space diagonal.

Measuring the angle to the wrong line on the floor, such as an edge, instead of to the projection.

Turning the tangent upside down. 5 ÷ 2 is the tangent of the angle at the top, against the upright edge, not the angle with the floor.