• In the following problems we use a cube of edge 4 cm, with square base ABCD and top face EFGH, where E lies above A, F above B, G above C and H above D.
• We place A at the origin, so the coordinates of the vertices are:
• A(0, 0, 0), B(4, 0, 0), C(4, 4, 0), D(0, 4, 0), E(0, 0, 4), G(4, 4, 4).

Example (1): Positions of two lines

1. Determine the position of the lines AB and EH.
2. Solution:

1. Line AB consists of points (t, 0, 0), and line EH consists of points (0, s, 4).
2. If they met, there would be a common point. But every point on AB has third coordinate 0, while every point on EH has third coordinate 4, so there is no common point.
3. The direction of AB is (1, 0, 0) and the direction of EH is (0, 1, 0), so they are not parallel.
4. Therefore AB and EH are skew.

Example (2): A line perpendicular to a plane

1. Prove that AE is perpendicular to the plane ABCD.
2. Solution:

1. The line AE is vertical, with direction (0, 0, 4).
2. AB and AD are two intersecting lines in the plane ABCD, with directions (4, 0, 0) and (0, 4, 0).
3. AE is perpendicular to AB because the dot product of their directions is 0, and similarly AE is perpendicular to AD.
4. Hence AE is perpendicular to the plane ABCD, and the distance from E to this plane is the length AE = 4 cm.

Example (3): Length of the space diagonal

1. Find the length of the diagonal AG.
2. Solution:

• AG = √((4 − 0)² + (4 − 0)² + (4 − 0)²) = √48 = 4√3 ≈ 6.93 cm.

Example (4): Angle a diagonal makes with the horizontal base

• Find the angle that the diagonal AG makes with the plane ABCD.
• Solution:

1. G lies directly above C, so the perpendicular projection of AG on the plane ABCD is AC.
2. The length of AC = √(4² + 4²) = 4√2 cm.
3. tan θ = CG ÷ AC = 4 ÷ (4√2) = 1 ÷ √2 ≈ 0.7071.
4. So θ ≈ 35.26°.
5. sin θ = 4 ÷ (4√3) = 1 ÷ √3, which agrees with the right-angled triangle.