# Bisectors in tetrahedron

• Nov 26th 2010, 09:04 AM
francois
Bisectors in tetrahedron
I've got such a problem. In tetrahedron we're considering the three bisectors of angles which are having common vertex. How to proof that if two of this bisectors are perpendicular that implicates that the third one must be also perpendicular to this two bisectors??
• Nov 27th 2010, 12:27 AM
Opalg
Quote:

Originally Posted by francois
I've got such a problem. In tetrahedron we're considering the three bisectors of angles which are having common vertex. How to proof that if two of this bisectors are perpendicular that implicates that the third one must be also perpendicular to this two bisectors??

Take that vertex to be the origin, and let p, q, r be unit vectors along the three edges which meet at the origin. Then p+q is a vector along the bisector of the angle between p and q. Suppose we know that this bisector is perpendicular to the bisector of the angle between p and r. Then the scalar product (p+q).(p+r) is 0. Therefore p.q + p.r + q.r = –1. But that equation is symmetric in p, q and r. Consequently, the other two pairs of bisectors are also perpendicular to each other.
• Nov 27th 2010, 12:48 AM
francois
Sorry but I don't understand your proof. If you could explain a little bit more precisely
• Nov 27th 2010, 05:08 AM
Opalg
Quote:

Originally Posted by francois
Sorry but I don't understand your proof. If you could explain a little bit more precisely

If p and q are unit vectors (or more generally vectors with the same length) then the points represented by the vectors 0, p, p+q and q form the vertices of a rhombus. The diagonals of a rhombus bisect the angles at the vertices. Thus the vector p+q bisects the angle between p and q.

The other thing I used in the proof is that the condition for two vectors to be perpendicular is that their scalar product ("dot product") should be zero.

Does that help to make things clearer? If not, you'll have to give a bit more detail about what it is that you don't understand.
• Nov 27th 2010, 11:23 AM
francois
I understand. But if you could make some visualisation of this problem and write on the drawing all the vectors and write the equations. Thank you for help
• Dec 3rd 2010, 09:59 AM
randomuser
small question
Quote:

Originally Posted by Opalg
If p and q are unit vectors (or more generally vectors with the same length) then the points represented by the vectors 0, p, p+q and q form the vertices of a rhombus. The diagonals of a rhombus bisect the angles at the vertices. Thus the vector p+q bisects the angle between p and q.

The other thing I used in the proof is that the condition for two vectors to be perpendicular is that their scalar product ("dot product") should be zero.

Does that help to make things clearer? If not, you'll have to give a bit more detail about what it is that you don't understand.

hello, i am also interested in this proof. I understand the whole idea, but could you explain how "(p+q).(p+r) is 0" implies "p.q + p.r + q.r = –1" ?
• Dec 3rd 2010, 10:38 AM
Opalg
Quote:

Originally Posted by randomuser
could you explain how "(p+q).(p+r) is 0" implies "p.q + p.r + q.r = –1" ?

Multiply out the brackets (using what Americans call the FOIL procedure) to get (p+q).(p+r) = p.p + p.r + q.p + q.r. Then use the fact the p.p = 1 (because p is a unit vector).
• Dec 3rd 2010, 10:44 AM
randomuser
Quote:

Originally Posted by Opalg
Multiply out the brackets (using what Americans call the FOIL procedure) to get (p+q).(p+r) = p.p + p.r + q.p + q.r. Then use the fact the p.p = 1 (because p is a unit vector).

thanks, I just didn't know that this operation can be used on vectors too :)
• Dec 3rd 2010, 02:25 PM
atreyyu
Seems so easy with all the vector geometry instead of tedious elementary proofs...