Vectors - Angle in a cube

li@mo

Can anybody help me.
I was looking on google but couldnt find this anywhere.

The problem is:

Determine the angle (either one) made between the diagonals of a cube using vector methods. (i,j,k)

Any help would be greatly appreciated.

Michael Collins

Can you find the vector representation for the two diagonals in question?

li@mo

Yes.
Represent both diagonals as vectors.
Would I be correct in saying that if both diagonals are represented as vectors that the angle can then be found by getting the dot product of them using the 2 methods and solve for angle using simultaneous equations?

Anyway, if anybody knows how to represent that diagonals as vectors, i would aprreciate the help!!

Delphi91

I'm a tad rusty on this but wouldn't the equation of the diagonal going from (0,0,0) to (x,x,x) [where x is the length of one side of the cube] be

d = x(cos 45)i + x(cos45)j + x(cos45)k?

where i, j and k are unit vectors in the x,y and z directions respectively.

If that's the case then a little bit of jiggery-pokery should enable you to find the other equation.

Michael Collins

li@mo wrote: »

Yes.
Represent both diagonals as vectors.

Yes, that's what I'm asking you to do!

Would I be correct in saying that if both diagonals are represented as vectors that the angle can then be found by getting the dot product of them using the 2 methods and solve for angle using simultaneous equations?

Exactly. So the problem is just one of finding the vector representation of those diagonals. This is an easy problem using vector methods. What are the coordinates of the vertices of such a cube?

Delphi91 wrote: »

I'm a tad rusty on this but wouldn't the equation of the diagonal going from (0,0,0) to (x,x,x) [where x is the length of one side of the cube] be

d = x(cos 45)i + x(cos45)j + x(cos45)k?

where i, j and k are unit vectors in the x,y and z directions respectively.

Not quite, there's actually no need for the cosines in there at all - leave them out and you'll have one diagonal vector! That's the beauty of vector algebra, it makes calculations such as these very simple.

d = x i + x j +x k, where x is the length of each side of the cube. (You can leave the x's out if you like since you'll get the same answer for any cube, regardless of its size.)

Can you see why this is the case lim@o? Maybe try and get the 2nd diagonal now youself, then try the dot product rule on both. Let us know if you get stuck...

li@mo

Thanks for the help.
I'm gonna have a go at figuring out the other vector and solving the problem.