Question

Find the angle between the diagonal of a cube of side length 15 and the diagonal of one of its faces, so that the two diagonals have a common vertex. The angle should be measured in radians. (Hint: we may assume that the cube is in the first octant, the origin is one of its vertices, and both diagonals start at the origin.)

Answer 1

the angle between the diagonal is 0.6155 RADIANS

Explanation:

Given the data in the question

as illustrated in the image below;

O is [0,0,0]

B is [15,0,15]

E is [15,15,15]

Now to find angle BOE

vector OB = [15,0,15]

vector OE = [15,15,15]

OB.OE = [15,0,15].[15,15,15] = 15 × 15 + 15 × 15 = 450 = |OB||OE|cos[angleBOE]

|OB| = √(15² + 15²) = √(225 + 225) = √450

|OE| = √(15² + 15² + 15²) = √(225 + 225 + 225) = √675

so

Angle BOE = cos⁻¹ ( 450 / ( √450 × √675 )

Angle BOE = cos⁻¹ ( 0.81649 )

Angle BOE = 35.265°

Angle BOE = 0.6155 RADIANS

Therefore, the angle between the diagonal is 0.6155 RADIANS