A) \[\pm \,1,1\]
B) \[\pm \,2,1\]
C) \[0,\pm \,1\]
D) None of these
Correct Answer: A
Solution :
\[\because \]a, b and c are linearly dependent vectors. \[\Rightarrow \] \[[a\,b\,c]=0\] \[\Rightarrow \] \[\left| \begin{matrix} 1 & 1 & 2 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[1(3\beta -4\alpha )-1(4\beta -4)+1(4\alpha -3)=0\] \[\Rightarrow \] \[-\beta +1=0\]\[\Rightarrow \]\[\beta =1\] Now, \[|c|=\sqrt{3}\] \[\Rightarrow \] \[\sqrt{1+{{\alpha }^{2}}+{{\beta }^{2}}}=\sqrt{3}\] \[\Rightarrow \] \[1+1+{{\alpha }^{2}}=3\Rightarrow {{\alpha }^{2}}=1\] \[\therefore \] \[\alpha =\pm 1\]
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