A) all values of\[\lambda \]
B) all except one value of \[\lambda \]
C) all except two values of \[\lambda \]
D) no value of\[\lambda \]
Correct Answer: C
Solution :
If\[a={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k},b={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}\]and \[c={{c}_{1}}\hat{i}+{{c}_{2}}\hat{j}+{{c}_{3}}\hat{k}\]are coplanar, then \[\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\ \end{matrix} \right|=0\] Given, three vectors are\[(a+2b+3c),(\lambda b+4c)\] and\[(2\lambda -1)c\]are coplanar, if \[\left| \begin{matrix} 1 & 2 & 3 \\ 0 & \lambda & 4 \\ 0 & 0 & 2\lambda -1 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[(2\lambda -1)(\lambda )=0\] \[\Rightarrow \] \[\lambda =0,\frac{1}{2}\] \[\therefore \]These three vectors are non-coplanar for all except two values of\[\left( i.e.,0,\frac{1}{2} \right)\].
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