A) \[\lambda =\frac{1}{2}\]
B) \[\lambda =-1\]
C) \[\lambda =0\]
D) for no value of \[\lambda \]
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\[\overrightarrow{PR}\]) \[\alpha (\overrightarrow{OQ}-\overrightarrow{OP})+\beta (\overrightarrow{OR}-\overrightarrow{OP})\] \[i.e.,\text{ }\overrightarrow{OS}-\overrightarrow{OP}=-(\alpha +\beta )\frac{{\vec{a}}}{3}+\frac{\alpha }{2}\vec{b}+\frac{\beta }{3}\vec{c}\] \[\Rightarrow \overrightarrow{OS}=(1-\alpha -\beta )\frac{{\vec{a}}}{3}+\frac{\alpha }{2}\vec{b}+\frac{\beta }{3}\vec{c}\] ...(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 \[\lambda =-1\]
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