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

Sagot :

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

[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]
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