Find the angle between the diagonal of a cube of side length 9 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.)

Firstly, we can sketch points as

O as (0,0,0)

B as (9,0,9)

E as (9,9,9)

now, we can find vectors

OB=(9,0,9)-(0,0,0)

OB=(9,0,9)

OE=(9,9,9,)-(0,0,0)

OE=(9,9,9)

we can use dot product formula to find angle

now, we can find values

now, we can plug these values

now, we can solve for theta

............Answer

answer: the common difference is -0.10

y=mx+