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.)

the angle between the diagonal is 0.6155 RADIANS

Step-by-step 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  

true

step-by-step explanation:

i think the correct option is the second option.