(i) Find the angle bewteen the lines whose direction ratios are 1, 2, 3 and - 3 , 2 , 1
(ii) Find the angle between two diagonals of a cube.

Explanation

(i) Let `theta` be the required angle , themn
`cos theta = (1 xx - 3 + 2 xx2 + 3 xx 1 )/(sqrt(1 + 4 + 9) sqrt(1 + 4 + 9 )) = (4)/(14) = (2)/(7)`
`rArr theta = cos^(-1)((2)/(7))`
(ii) From the adjoining figure, the direction rations of the diagonals OP and CD of a given cube ltbgt are given by
` a - 0 , a - 0 , a - 0 ` and
` a - 0 , a - 0, 0 - a`

and hence their respective direction cosines are
`- (a)/(sqrt(a^(2)+ a^(2)+a^(2))), (a)/(sqrt(a^(2)+ a^(2)+a^(2))),(a)/(sqrt(a^(2)+ a^(2)+a^(2))) i.e., (1)/(sqrt(3)), (1)/(sqrt(3)), (1)/(sqrt(3))`
and ` (a)/(sqrt(a^(2)+ a^(2)+a^(2))), (a)/(sqrt(a^(2)+ a^(2)+a^(2))),(-a)/(sqrt(a^(2)+ a^(2)+a^(2))) i.e., (1)/(sqrt(3)), (1)/(sqrt(3)), (-1)/(sqrt(3))`
Let `theta (1)/(sqrt(3)), (1)/(sqrt(3))+(1)/(sqrt(3)), (1)/(sqrt(3)),+(1)/(sqrt(3)), (-1)/(sqrt(3))= (1)/(3)+(1)/(3) - (1)/(3) = (1)/(3)`
`rArr theta cos^(-1) ((1)/(3))`