# 3D Trigonometry/GeometryWatch

#1
In chemistry a common shape of a molecule is with 4 atoms on the points of a tetrahedron and 1 dead in the middle, with each outer atom joined to the middle one. We get told that the angle between any 2 of these joins is approximately 109°, but we have never been shown any reason for this value. I'm kinda stuck as to how to go about finding this, anyone know how, or even got any suggestions on where to start? It seems the sort of thing that SHOULD be pretty easy, but I don't have a clue.
http://www.elmhurst.edu/~chm/vchembo...204methane.gif

Thanks, Chris.
0
12 years ago
#2
I spent ages trying this in various ways and as far as I'm aware, I didn't get very far [maybe I did and just didn't spot it... hmm, likely...]. The main points I kept in mind were that all the bonds would be at the same angle to each other, and that this angle would be as big as possible [although those two points seem to fit together].
Lots of drawings seemed helpful, but they weren't that great since I could't work out why it's 109.5 degrees.
Perhaps vectors would be useful?
0
12 years ago
#3
Oho, vectors seemed to be looking very promising... and then I got 105.257 degrees, which is wrong. Blast it. I'll try again and let you know if [and how] it works.
0
12 years ago
#4
...yes, vectors give 109.471220635 degress [when you do the vectors correctly, whoops], according to my calculator.
Basically you need to construct a tetrahedron using vectors [which can be done using basic trig and the modulus of the vectors - it's probably easiest to let this be 1], and find the central point in a similar way. Then you find the direction vectors from the centre to two of the outer points. Then you can use a.b = |a||b|cos(theta) to find the angle.
Wahey, I feel happy now
Anyway, I can try writing that out and can send it to you. I don't think I can face the thought of typing it in an ordinary post, it would take years.
0
#5
Hehe, well done. No point trying to write up the whole thing, I don't know anything about 3D vectors. Not surprising I wasn't getting anywhere then. Thanks anyway.
0
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