Question

SHOW THAT THE ANGEL BETWEEN ANY 2 DIAGONALS OF A CUBE IS cos-1 (1/3)

Answer 1

Vector for one diagnol (1,1,1) - (0,0,0) = (1,1,1)
Vector for another diagnol (0,1,1) - (1,0,0) = (-1,1,1)

Do cos product of two vectors , that is

sqrt(3) sqrt(3) cos(theta) = (1,1,1).(-1,1,1)
3cos(theta) = 1.(-1) + 1.(1) + 1.(1) = 1

therefore cos(theta) = 1/3
angle betwwen diagnols = cos-1 (1/3)

Answer 2

`Look\ at\ the\ picture.\\\\|BD|=a\sqrt2\\\\|BD_1|=|DB_1|=a\sqrt3\\\\|BE|=|DE|=(a\sqrt3)/(2)\\\\Use\ law\ of\ cosine:\\\\(a\sqrt2)^2=\left((a\sqrt3)/(2)\right)^2+\left((a\sqrt3)/(2)\right)^2-2\cdot(a\sqrt3)/(2)\cdot(a\sqrt3)/(2)\cdot cos\theta\\\\2a^2=(3a^2)/(4)+(3a^2)/(4)-(3a^2)/(2)cos\theta`


`(3a^2)/(2)cos\theta=(6a^2)/(4)-2a^2\\\\(3a^2)/(2)cos\theta=(6a^2)/(4)-(8a^2)/(4)\\\\(3a^2)/(2)cos\theta=-(2a^2)/(4)\\\\(3a^2)/(2)cos\theta=-(a^2)/(2)\ \ \ \ \ /\cdot(2)/(3a^2)\\\\cos\theta=-(1)/(3)`