Find the value of `lamda` so that the points P, Q, R and S on the sides OA, OB, OC and AB, respectively, of a regular tetrahedron OABC are coplanar. It is given that `(OP)/(OA) = (1)/(3), (OQ)/(OB) = (1)/(2), (OR)/(OC) = (1)/(3) and (OS)/(AB)= lamda`.

Options

A. `lamda=1/2`
B. `lamda=-1`
C. `lamda=0`
D. fo no value of `lamda`

Answer

B

Explanation

Let `OA=a,OB=b and OC=c`.
then `AB=b-a and OP=(1)/(3)a`.
`OQ=(1)/(2)b,OR=(1)/(3)c`.
Since, P,Q,R and S are coplanar, then
`PS=alphaPQ+betaPR`
(PS can be written as a linear combination of PQ and PR)
`=alpha(OQ-OP)+beta(OR-OP)`
i.e., `OS-OP=-(alpha+beta)(a)/(3)+(alpha)/(2)b+(beta)/(3)c`
`implies OS=(1-alpha-beta)(a)/(3)+(alpha)/(2)b+(beta)/(3)c` . . (i)
Given `OS=lamdaAB=lamda(b-a)` . . . (ii)
From Eq. (i) and Eq. (ii), `beta=0,(1-alpha)/(3)=-lamda and (alpha)/(2)=lamda`.
`implies 2 lamda=1+3lamda` or `lamda=-1`.