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Question

Find the value of λ 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 OPOA=13, OQOB=12 and OSAB=λ.

A
λ=12
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B
λ=1
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C
λ=0
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D
For no value of λ
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Solution

The correct option is B λ=−1Let −−→OA=→a, −−→OB=→b and −−→OC=→c, then, −−→AB=→b−→a and −−→OP=13→a, −−→OQ=13→b, −−→OR=13→c Since, P, Q, R and S are coplanar, then −→PS=α−−→PQ+β−−→PR (−→PS can be written as a linear combination of −−→PQ and −−→PR) ⇒−→PS=α(−−→OQ−−−→OP)+β(−−→OR−−−→OP) i.e.−−→OS−−−→OP=−(α+β)→a3+α2→b+β3→c ⇒−−→OS=(1−α−β)→a3+α2→b+β3→c …[1] Given −−→OS=λ−−→AB=λ(→b−→a) …[2] From [1] and [2], β=0, 1−α3=−λ and α2=λ ⇒2λ=1+3λ or, λ=−1

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