## Answer :

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)

[tex]\frac{3a^2}{2}cos\theta=\frac{6a^2}{4}-2a^2\\\\\frac{3a^2}{2}cos\theta=\frac{6a^2}{4}-\frac{8a^2}{4}\\\\\frac{3a^2}{2}cos\theta=-\frac{2a^2}{4}\\\\\frac{3a^2}{2}cos\theta=-\frac{a^2}{2}\ \ \ \ \ /\cdot\frac{2}{3a^2}\\\\cos\theta=-\frac{1}{3}[/tex]