Question

Let $O$ be any point inside a tetrahedron $ABCD$ . The line joining $O$ to the vertices meet the opposite faces in $P,Q,R,S$ respectively. If $\frac{OP}{AP}+\frac{OQ}{BQ}+\frac{OR}{CR}+\frac{OS}{DS}=k$ , then the value of $k$

Answer 1

Answer: 1

Let $\frac{OP}{AP}=\frac{\lambda }{1}$

$\overrightarrow{p}=\left(\frac{\lambda }{\lambda -1}\right)\overrightarrow{a}\dots \left(i\right)$
Now $\overrightarrow{\mathrm{BC}},\overrightarrow{\mathrm{BD}},\overrightarrow{\mathrm{BP}}$ are coplanar
$\Rightarrow \left(\overrightarrow{c}-\overrightarrow{b}\right)\times \left(\overrightarrow{d}-\overrightarrow{b}\right)\cdot \left(\overrightarrow{p}-\overrightarrow{b}\right)=0$
$\frac{\lambda }{\lambda -1}=\frac{\left[\overrightarrow{c}\overrightarrow{d}\overrightarrow{b}\right]}{\left[\overrightarrow{c}\overrightarrow{d}\overrightarrow{a}\left]-\right[\overrightarrow{b}\overrightarrow{d}\overrightarrow{a}\left]-\right[\overrightarrow{c}\overrightarrow{b}\overrightarrow{a}\right]}$
$\lambda =\frac{\left[\overrightarrow{b}\overrightarrow{c}d\right]}{[\overrightarrow{b}\overrightarrow{c}\overrightarrow{d}\left]+\right[\overrightarrow{a}\overrightarrow{b}\overrightarrow{d}\left]+\right[\overrightarrow{a}\overrightarrow{d}\overrightarrow{c}\left]+\right[\overrightarrow{a}\overrightarrow{c}}\dots \left(2\right)$
Similarly
If $\frac{OQ}{BQ}=\mu ,\frac{OR}{CR}=\alpha ,\frac{OS}{DS}=\beta$
$\Rightarrow \mu =\frac{\left[\overrightarrow{a}\overrightarrow{d}\overrightarrow{c}\right]}{\left[\overrightarrow{a}\overrightarrow{d}\overrightarrow{c}\left]+\right[\overrightarrow{a}\overrightarrow{b}\overrightarrow{d}\left]+\right[\overrightarrow{b}\overrightarrow{c}\overrightarrow{d}\left]+\right[\overrightarrow{a}\overrightarrow{c}\overrightarrow{d}\right]}$
and so an
$\Rightarrow \frac{OD}{AP}+\frac{OQ}{BQ}+\frac{OR}{CR}+\frac{OS}{DS}=1$