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Angle between two diagonals of a cube. 
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#1
Apr510, 09:18 PM

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Find the angle between the diagonal of the back and left faces of a cube with one vertex on the origin
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 [u] to mean the "norm" of u and u*v to be u "dot" v I know we have the formula cos(theta) = u*v/([u] [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 [u] = [v] which impies [u][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? 


#2
Apr510, 11:05 PM

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Just because [itex] \textbf{u} = \textbf{v} [/itex], you still have to accaount for the angle which is unknown, so[itex] \textbf{u} \bullet \textbf{v} = \textbf{u}\textbf{v} cos(\theta) =\textbf{u}^2 cos(\theta) [/itex] so if I'm getting it correctly, try considering the triangle created by joining the ends of the 2 diagonals across the top face how are the lengths of each side of the traingle related? 


#3
Apr510, 11:08 PM

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otherwise do it the brute force way & write out the vectors explicitlye eg. (1,1,0) and find their magnitudes & dot product then use the results to find the angle



#4
Jan112, 10:15 AM

P: 10

Angle between two diagonals of a cube.
There are 4 diagonals in a cube. Section the cube such that you have 6 equal square based pyramids, formed by the 6 faces and the 4 diagonals. Now consider one of the pyramids. Let ABCD denote the 4 corners of the square base and P denote the apex of the pyramid, which is at the centre of the cube. The angle required is surely given by, say, angle APB. This is trivial to find: let the cube have sides of length say x; drop a perpendicular from P to centre of square base and call this point G; length PG is, trivially, x/2; length GA is half the diagonal of square base (or sqr(2)*x/2). By Pythagoras theorem length PA (=PB) can be found:
PA^2 = PG^2 + GA^2, which yields PA =sqr(3)*x/2. Hence we have 3 sides of the triangle PAB. Angle BPA is required and this is found by the cosine rule. The calculation yields: Cos(angle BPA) = 1/3 or angle BPA = 70.52877937 degrees. This is the angle between 2 diagonals of a cube. 


#5
Jan112, 01:48 PM

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@Bill Crean: The problem asks the angle between face diagonals. Read the original post.
@d.almont: I suggest to draw a picture. Something like the one I attached. You can place that cube as you like in the coordinate system and can choose unit edge length. ehild 


#6
Jan112, 01:55 PM

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this one is probably old enough to be considered closed, but just for completeness, consider the dot product method
by taking the difference of the corner point positions of the cube of unit length positioned with a corner at the origin, one vector representing a diagonal is (1,1,1) = (1,1,1)(0,0,0) similarly for another (1,1,1) = (1,1,0)(0,0,1) both have magnitude sqrt(3) taking the dot product gives: 3.cos(theta)=1 


#7
Jan112, 02:00 PM

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as mentioned though this is reopening a pretty old thread 


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