A) \[\lambda \overrightarrow{a}\]
B) \[\lambda \overrightarrow{b}\]
C) \[\lambda \overrightarrow{c}\]
D) 0
Correct Answer: D
Solution :
If\[\overrightarrow{a}+2\overrightarrow{b}\]is collinear with\[\overrightarrow{c}\],then \[\overrightarrow{a}+2\overrightarrow{b}=t\overrightarrow{c}\] ...(i) Also, if\[\overrightarrow{b}+3\overrightarrow{c}\]is collinear with\[\overrightarrow{a},\]then \[\overrightarrow{b}+3\overrightarrow{c}=\lambda \overrightarrow{a}\] ...(ii) \[\Rightarrow \] \[\overrightarrow{b}=\lambda \overrightarrow{a}-3\overrightarrow{c}\] On putting this value in Eq. (i) \[\overrightarrow{a}+2(\lambda \overrightarrow{a}-3\overrightarrow{c})=t\overrightarrow{c}\] \[\Rightarrow \] \[\overrightarrow{a}+2\lambda \overrightarrow{a}-6\overrightarrow{c}=t\overrightarrow{c}\] \[\Rightarrow \] \[(\overrightarrow{a}-6\overrightarrow{c})=t\overrightarrow{c}-2\lambda \overrightarrow{a}\] On comparing, we get \[1=-2\lambda \Rightarrow \lambda =-\frac{1}{2}\] and \[-6=t\Rightarrow t=-6\] From Eq. (i) \[\overrightarrow{a}+2\overrightarrow{b}=-6\overrightarrow{c}\] \[\Rightarrow \] \[\overrightarrow{a}+2\overrightarrow{b}+6\overrightarrow{c}=0\]
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