Problem with Tetrahedron

[pephiroth]

Hey all, there is an Tetrahedron SABC, SO - height. I have to prove that SO vector = 1/3 (SA + SB + SC) vectors. Any ideas how to do that?
[h=1]Thanks for helping [/h]

 

[soroban]

Hello, pephiroth!

\(\displaystyle \text{There is a tetrahedron }S\!ABC\text{ with }SO = \text{height.}\)

\(\displaystyle \text{Prove that: }\:\overrightarrow{SO} \;=\;\frac{1}{3}\left(\overrightarrow{SA} + \overrightarrow{SB} + \overrightarrow{SC}\right) \)

Sketch the tetrahedron.
Base triangle \(\displaystyle ABC\) is on the "floor".
Vertex \(\displaystyle S\) is above it.
Draw segments \(\displaystyle SA, SB, SC, AB, BC, CA.\)
\(\displaystyle SO\) is the altitude to \(\displaystyle \Delta ABC.\)
Draw segments \(\displaystyle AO, BO, CO.\)

We have: .\(\displaystyle \begin{Bmatrix}\overrightarrow{SO} &=& \overrightarrow{SA} + \overrightarrow{AO} & [1] \\ \overrightarrow{SO} &=& \overrightarrow{SB} + \overrightarrow{BO} & [2] \\ \overrightarrow{SO} &=& \overrightarrow{SC} + \overrightarrow{CO} & [3]\end{Bmatrix}\)


\(\displaystyle \text{Add [1], [2] and [3]: }\:3\cdot\overrightarrow{SO} \;=\;(\overrightarrow{SA} + \overrightarrow{SB} + \overrightarrow{SC}) + \underbrace{(\overrightarrow{AO} + \overrightarrow{BO} + \overrightarrow{CO})}_{\text{This is }\vec 0} \)
. . \(\displaystyle 3\cdot\overrightarrow{SO} \;=\;\overrightarrow{SA} + \overrightarrow{SB} + \overrightarrow{SC} \)

. . . . \(\displaystyle \overrightarrow{SO} \;=\;\frac{1}{3}\left(\overrightarrow{SA} + \overrightarrow{SB} + \overrightarrow{SC}\right)\)