show that acute angle between two diagonals of cube has cosine 1/3

Question

show acute angle between two diagonals of cube

 

Answer 1

Call the corners or vertices of the unit cube A, B, C, D, E, F, G, H. If A to D clockwise is the top face and E to H clockwise is the bottom face where AE form an edge of the cube, then two diagonals are AG and BH. They intersect at X, bisecting one another. The base diagonal EG has length sqrt(1+1)=sqrt(2) by Pythagoras, and this is the length of all face diagonals. The length of AG is sqrt(AE^2+EG^2)=sqrt(1+2)=sqrt(3). AX=BX=(1/2)AG=sqrt(3)/2. In triangle ABX, we apply the cosine rule: AB^2=AX^2+BX^2-2AX*BXcosX. So 1=3/4+3/4-2*3/4cosX. -(1/2)=-(3/2)cosX and cosX=(2/3)(1/2)=1/3.