A) \[\alpha =1,\text{ }\beta =-1\]
B) \[\alpha =1,\text{ }\beta =\pm 1\]
C) \[\alpha =-1,\text{ }\beta =\pm 1\]
D) \[\alpha =\pm 1,\text{ }\beta =1\]
Correct Answer: D
Solution :
Since, a, b and c are linearly dependent vectors. \[\therefore \] \[[abc]=0\] \[\Rightarrow \] \[\left| \begin{matrix} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \\ \end{matrix} \right|=0\] \[1(3\beta -4\alpha )-1(4\beta -4)+1(4\alpha -3)=0\] \[\Rightarrow \] \[-\beta +1=0\] \[\Rightarrow \] \[\beta =1\] Now, \[|c|=\sqrt{3}\] \[\Rightarrow \] \[\sqrt{1+{{\alpha }^{2}}+{{\beta }^{2}}}=\sqrt{3}\] \[\Rightarrow \] \[1+1+{{\alpha }^{2}}=3\] \[\Rightarrow \] \[{{\alpha }^{2}}=1\] \[\Rightarrow \] \[\alpha =\pm 1\]
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