A) \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=bc+ca+ab\]
B) \[5\sqrt{2}\]
C) \[({{a}^{2}}+ab,\,{{b}^{2}}+ab,\,-ab)\]
D) \[(-bc,\,{{b}^{2}}+bc,\,{{c}^{2}}+bc),\]
Correct Answer: A
Solution :
\[A=\frac{1}{2}|{{\mathbf{d}}_{1}}\times {{\mathbf{d}}_{2}}|\,=\frac{1}{2}\left| \,\begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{3}{2} & \frac{1}{2} & -1 \\ 2 & -6 & 8 \\ \end{matrix}\, \right|\] \[=\frac{1}{2}|-2\mathbf{i}-14\mathbf{j}-10\mathbf{k}|\] \[A=\frac{1}{2}\sqrt{4+{{(14)}^{2}}+100}=\frac{1}{2}\sqrt{300}\]\[=\frac{1}{2}\,.\,10\sqrt{3}=5\sqrt{3}\].
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