Vector Algebra Question 210
Question: Find the value of $ \lambda $ 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 $ \frac{OP}{OA}=\frac{1}{3},\frac{OQ}{OB}=\frac{1}{2},\frac{QR}{OC}=\frac{1}{3} $ and $ \frac{OS}{AB}=\lambda $ .
Options:
A) $ \lambda =\frac{1}{2} $
B) $ \lambda =-1 $
C) $ \lambda =0 $
D) for no value of $ \lambda $
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Answer:
Correct Answer: B
Solution:
- [b] Let $ \overrightarrow{OA}=\overrightarrow{a},\overrightarrow{OB}=\overrightarrow{b} $ and $ \overrightarrow{OC}=\overrightarrow{c} $ , Then $ \overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a} $ and $ \overrightarrow{OP}=\frac{1}{3}\overrightarrow{a,} $ $ \overrightarrow{OQ}=\frac{1}{2}\overrightarrow{b,}\overrightarrow{OR}=\frac{1}{3}\overrightarrow{c.} $ Since P, Q, R and S are coplanar, then $ \overrightarrow{PS}=\alpha \overrightarrow{PQ}+\beta \overrightarrow{PR} $ ( $ \overrightarrow{PS} $ can be written as a linear combination of $ \overrightarrow{PQ} $ and $ \overrightarrow{PR} $ ) $ \alpha (\overrightarrow{OQ}-\overrightarrow{OP})+\beta (\overrightarrow{OR}-\overrightarrow{OP}) $ $ i.e.,\overrightarrow{OS}-\overrightarrow{OP}=-(\alpha +\beta )\frac{{\vec{a}}}{3}+\frac{\alpha }{2}\vec{b}+\frac{\beta }{3}\vec{c} $
$ \Rightarrow \overrightarrow{OS}=(1-\alpha -\beta )\frac{{\vec{a}}}{3}+\frac{\alpha }{2}\vec{b}+\frac{\beta }{3}\vec{c} $ …(i) Given $ \overrightarrow{OS}=\lambda \overrightarrow{AB}=\lambda (\vec{b}-\vec{a}) $ …(ii) From (i) and (ii), $ \beta =0,\frac{1-\alpha }{3}=-\lambda $ and $ \frac{\alpha }{2}=\lambda $
$ \Rightarrow 2\lambda =1+3\lambda $ Or $ \lambda =-1 $
BETA
