Addition of Position Vectors of Reg. Tetrahedron - Spherical Geometry/Vectors (1 Viewer)

[Sy123]

I recently came across a STEP question involving vectors in 3D Space, and a tetrahedron bound by a Sphere.

If we consider the points on the sphere which are the vertices of the regular tetrahedron

And consider the position vectors from O the origin corresponding to the above points respectively.

Before it really began the question it sort of assumed



And I've been trying to prove it. So far I thought to express it in Spherical co-ordinate system.

As









(Forgive the notation if it is incorrect)

Firstly, is this correct?

Then I tried to put it into 'cartesian' form, by applying that



So I get 4 points in the form (x,y,z)

I add them all together and it doesn't give zero, what did I do wrong?

 

[braintic]

I recently came across a STEP question involving vectors in 3D Space, and a tetrahedron bound by a Sphere.

If we consider the points on the sphere which are the vertices of the regular tetrahedron

And consider the position vectors from O the origin corresponding to the above points respectively.

Before it really began the question it sort of assumed



And I've been trying to prove it. So far I thought to express it in Spherical co-ordinate system.

As









(Forgive the notation if it is incorrect)

Firstly, is this correct?

Then I tried to put it into 'cartesian' form, by applying that



So I get 4 points in the form (x,y,z)

I add them all together and it doesn't give zero, what did I do wrong?

I believe the problem is your assumption that the polar angle is 120 degrees.

Imagine the centre of the base of your tetrahedron. The lines joining this centre to the vertices of the base are separated by 120 degrees. Now lift this point up to your origin ... the lines joining THIS point to the same vertices must now be separated by LESS THAN 120 degrees. And by symmetry, this is the same as your polar angle.

In fact, I have set the calculation of this angle as one of the questions in my 3D-Trig worksheet for year 11. It is roughly 109.5 degrees.

 

[braintic]

It would be interesting to know whether this fixed up your solution.
I don't know much about the spherical coordinate system other than definitions.

 

[Sy123]

It would be interesting to know whether this fixed up your solution.
I don't know much about the spherical coordinate system other than definitions.

Well firstly my Cartesian form was wrong, it is supposed to be








But even then, no matter what angle alpha is there




Which is supposed to be the addition of all the z co-ordinates, so there is a direct problem with the method itself.

 

[braintic]

Assuming you have the apex at (0,0,1) and the base parallel to the x-y plane, shouldn't the other three vertices all have the same z-coordinate?
(But I don't really understand those formulae, so perhaps I am missing something)

 

[Trebla]

I recently came across a STEP question involving vectors in 3D Space, and a tetrahedron bound by a Sphere.

If we consider the points on the sphere which are the vertices of the regular tetrahedron

And consider the position vectors from O the origin corresponding to the above points respectively.

Before it really began the question it sort of assumed



And I've been trying to prove it. So far I thought to express it in Spherical co-ordinate system.

As









(Forgive the notation if it is incorrect)

Firstly, is this correct?

Then I tried to put it into 'cartesian' form, by applying that



So I get 4 points in the form (x,y,z)

I add them all together and it doesn't give zero, what did I do wrong?

The angle between two adjacent vectors is cos^-1(-1/3). The only way the vectors can add to zero is if their relative positions are exactly like that (which I suppose defines the tetrahedron shape).

Using that information, show that the addition of the horizontal components and the addition of the vertical components of the relevant vectors are zero.