Question

Let $O$ be any point inside a tetrahedron $A B C D$ . The line joining $O$ to the vertices meet the opposite faces in $P , \, Q , \, R , \, S$ respectively. If $\frac{O P}{A P} + \frac{O Q}{B Q} + \frac{O R}{C R} + \frac{O S}{D S} = k$ , then the value of $k$

Answer 1

Let $\frac{O P}{A P} = \frac{\lambda }{1}$

$\overset{ \rightarrow }{p} = \left(\frac{\lambda }{\lambda - 1}\right) \overset{ \rightarrow }{a} \, \ldots \left(i\right)$
Now $\overset{ \rightarrow }{B C} , \overset{ \rightarrow }{B D} , \overset{ \rightarrow }{B P}$ are coplanar
$\Rightarrow \left(\overset{ \rightarrow }{c} - \overset{ \rightarrow }{b}\right) \times \left(\overset{ \rightarrow }{d} - \overset{ \rightarrow }{b}\right) \cdot \left(\overset{ \rightarrow }{p} - \overset{ \rightarrow }{b}\right) = 0$
$\frac{\lambda }{\lambda - 1} = \frac{\left[\overset{ \rightarrow }{c} \overset{ \rightarrow }{d} \overset{ \rightarrow }{b}\right]}{\left[\overset{ \rightarrow }{c} \overset{ \rightarrow }{d} \overset{ \rightarrow }{a} \left]-\right[ \overset{ \rightarrow }{b} \overset{ \rightarrow }{d} \overset{ \rightarrow }{a} \left]-\right[ \overset{ \rightarrow }{c} \overset{ \rightarrow }{b} \overset{ \rightarrow }{a}\right]}$
$\lambda = \frac{\left[\overset{ \rightarrow }{b} \overset{ \rightarrow }{c} d\right]}{\left[\right. \overset{ \rightarrow }{b} \overset{ \rightarrow }{c} \overset{ \rightarrow }{d} \left]+\right[ \overset{ \rightarrow }{a} \overset{ \rightarrow }{b} \overset{ \rightarrow }{d} \left]+\right[ \overset{ \rightarrow }{a} \overset{ \rightarrow }{d} \overset{ \rightarrow }{c} \left]+\right[ \overset{ \rightarrow }{a} \overset{ \rightarrow }{c}} \, \ldots \left(2\right)$
Similarly
If $\frac{O Q}{B Q} = \mu , \frac{O R}{C R} = \alpha , \frac{O S}{D S} = \beta $
$\Rightarrow \mu = \frac{\left[\overset{ \rightarrow }{a} \overset{ \rightarrow }{d} \overset{ \rightarrow }{c}\right]}{\left[\overset{ \rightarrow }{a} \overset{ \rightarrow }{d} \overset{ \rightarrow }{c} \left]+\right[ \overset{ \rightarrow }{a} \overset{ \rightarrow }{b} \overset{ \rightarrow }{d} \left]+\right[ \overset{ \rightarrow }{b} \overset{ \rightarrow }{c} \overset{ \rightarrow }{d} \left]+\right[ \overset{ \rightarrow }{a} \overset{ \rightarrow }{c} \overset{ \rightarrow }{d}\right]}$
and so an
$\Rightarrow \frac{O D}{A P} + \frac{O Q}{B Q} + \frac{O R}{C R} + \frac{O S}{D S} = 1$