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Angle between two diagonals of a cube.

  1. Apr 5, 2010 #1
    Find the angle between the diagonals of the back and left faces of a cube with one vertex on the origin.

    A) 60 degrees
    B) arcos(1/sqrt(6))
    C) arcos(1/(3sqrt(2))
    D) 90 degrees
    E) 120 degrees.

    I think the answer should be D) 90 degrees. Since these vectors are diagonals on the face of the cube, they bisect the angles of the cube. The since it is a cube every vertex forms a 90 degree angle.

    Can someone tell me if I am correct.


    I know the formula for the cos(theta) using the dot product but the problem is i do not know the coordinates of this cube all I know one vertex is on the origin. But these vectors must be the same legnth right? if they lie on the faces of the cube.

    but then if you use a.b = abcosθ, a=b... So the norm(a) * norm(b) is really a^2 and a dot b is really a^2 as well..

    This leaves you with cos(theta) = 1... which mean theta is 0 which is not possible.. can someone help please
     
    Last edited: Apr 5, 2010
  2. jcsd
  3. Apr 5, 2010 #2

    tiny-tim

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    welcome to pf!

    hi mary ! welcome to pf! :smile:

    (have a square-root: √ and a degree: º :wink:)
    hint: use a.b = abcosθ (or draw the triangle, and then look at it!) :smile:
     
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