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Show that the angle between two diagonal...

Show that the angle between two diagonals of a cube is `cos^(-1)sqrt(1/3)dot`

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Show that the angle between two diagonals of a cube is cos^(-1)(1/3)dot

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In a quadrilateral A B C D , vec A C is the bisector of vec A Ba n d vec A D , angle between vec A Ba n d vec A D is 2pi//3 , 15| vec A C|=3| vec A B|=5| vec A D|dot Then the angle between vec B Aa n d vec C D is cos^(-1)(sqrt(14))/(7sqrt(2)) b. cos^(-1)(sqrt(21))/(7sqrt(3)) c. cos^(-1)2/(sqrt(7)) d. cos^(-1)(2sqrt(7))/(14)

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(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.

The angle between two planes x+2y+2z=3 and -5x+3y+4z=9 is (A) cos^-1(3sqrt(2))/10 (B) cos^-1 (19sqrt(2))/30 (C) cos^-1 (9sqrt(2))/20 (D) cos^-1 (3sqrt(2))/5

If veca=(3,1) and vecb=(1,2) represent the sides of a parallelogram then the angle theta between the diagonals of the paralelogram is given by (A) theta=cos^-1(1/sqrt(5)) (B) theta=cos^-1(2/sqrt(5)) (C) theta=cos^-1 (1/(2sqrt(5))) (D) theta = pi/2