Find the angel between any two diagonals of a cube.

Options

A. `sin^(-1)""2/3`
B. `cos^(-1)""1/2`
C. `cos^(-1)""1/sqrt3`
D. `cos^(-1)""1/3`

Answer

D

Explanation

Let OA,OB,and OC be the coteminous edges of a cube taken along the axers in such a way that
OA=OB=OC=a
Then, the coordinates of the vertices of the cube are
O(0,0,00, A(a,0,0), B(0,a,0), C(0,0,a)
P(a,a,a), L(0,a,a), M(a,0,a) and N(a,a,0)

The direction rations of the diagonals OP,AL,BM and CN are (a,a,a),(-a,a,a) and (a,a,-a)
Directions cosines of OP,AL ,BM and CN are
`(1/sqrt3,1/sqrt3,1/sqrt3),((-1)/sqrt3,1/sqrt3,1/sqrt3),(1/sqrt3,(-1)/sqrt3,1/sqrt3) and (1/sqrt3,1/sqrt3,(-1)/sqrt3)`, respectvely
If `theta` be the angle between OP and AL, then
`costheta={1/sqrt3.((-1)/sqrt3)+1/(sqrt3)"."1/sqrt3+1/sqrt3"."1/sqrt3}=1/3`
`therefore" "theta=cos^(-1)""(1/3)`