A) \[\alpha =\sqrt{2},\beta =1\]
B) \[\alpha =1,\beta =\pm 1\]
C) \[\alpha =\pm 1,\beta =1\]
D) \[\alpha =\pm 1,\beta =-1\]
Correct Answer: C
Solution :
[c] Since \[\vec{a},\,\vec{b}\] and \[\vec{c}\] are coplanar therefore \[\left| \begin{matrix} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \\ \end{matrix} \right|=0\Rightarrow \beta =1;\left| {\vec{c}} \right|=\sqrt{1+{{\alpha }^{2}}+{{\beta }^{2}}}=\sqrt{3}\] \[\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}=2\Rightarrow {{\alpha }^{2}}=1\therefore \alpha =\pm 1\]
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