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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.)

Respuesta :

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

[tex]OB\cdot OE=|OE|*|OB|cos(\theta)[/tex]

now, we can find values

[tex]OB\cdot OE=(9,0,9)\cdot (9,9,9) [/tex]

[tex]OB\cdot OE=9*9+0*9+9*9 [/tex]

[tex]OB\cdot OE=162 [/tex]

[tex]|OB|=\sqrt{9^2+0^2+9^2}[/tex]

[tex]|OB|=9\sqrt{2}[/tex]

[tex]|OE|=\sqrt{9^2+9^2+9^2}[/tex]

[tex]|OB|=9\sqrt{3}[/tex]

now, we can plug these values

[tex]162=9\sqrt{2}*9\sqrt{3}|cos(\theta)[/tex]

now, we can solve for theta

[tex]\theta=cos^{-1}(\frac{162}{81\sqrt{6} } )[/tex]

[tex]\theta=0.61548 radians[/tex]............Answer


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