Find the angle between the diagonal of the back and left faces of a cube with one vertex on the origin(adsbygoogle = window.adsbygoogle || []).push({});

A) 60 degrees

B) cos^-1(1/sqrt(6))

C) cos^-1(1/3*sqrt(2))

D) 90 degrees

E) 120 degrees

The two diagonals eminate from the vertex at the origin.

I will write to mean the "norm" of u and u*v to be u "dot" v

I know we have the formula cos(theta) = u*v/( [v]).

But we do not know know any points on the cube so we can't form any vectors...

But since these are the diagonals of a cube they should be the same size . It which case u = v

Then u*v = u^2 and = [v] which impies [v] = u^2. Then cos(theta) = 1 which implies theta = 0 which is definitely not the case...

My other thought is that if these are diagonals on the face of a cube, shouldnt they bisect the 90 degree angles of the cube into 45 degree angles? Then the angle between the two vectors would be 45 + 45 = 90 degrees choice D).

Is this correct?

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# Angle between two diagonals of a cube.

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