finding vector in tetrahedron

Question

basically i have no idea how to even start doing this following question from last exam, so it goes like this:

In tetrahedron $OABC$, $$\overrightarrow{|OA|}=2, \overrightarrow{|OB|}= \overrightarrow{|OC|}= 1$$ $$\angle (\overrightarrow{OA},\overrightarrow{OB})=\angle (\overrightarrow{OA},\overrightarrow{OC})= \frac{4\pi}{3}, \angle (\overrightarrow{OB},\overrightarrow{OC}) = \frac{\pi}{2}$$ Let $T$ be a point of altitude(height) of side $ABC$ from point $C$. Find real number $\alpha,\beta,\gamma$ so that $$\overrightarrow{CT} = \alpha \overrightarrow{OA}+\beta \overrightarrow{OB}+\gamma \overrightarrow{OC}$$

Sorry, had to translate everything from native language, as much as i could i explained.. I have no idea even how to start it

Answer 1

HINT

Let assume wlog

• $OB=(1,0,0)$

• $OC=(0,1,0)$

then by symmetry

• $OA=(s,s,t)$

for some some $s,t$ such that

• $\sqrt{2s^2+t^2}=2$
• $\angle (\overrightarrow{OA},\overrightarrow{OB})=\angle (\overrightarrow{OA},\overrightarrow{OC})= \theta$