A) 0
B) 1
C) \[\frac{1}{4}(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})(b_{1}^{2}+b_{3}^{2})\]
D) \[\frac{3}{4}(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})(b_{1}^{2}+b_{2}^{2}+b_{3}^{2})(c_{1}^{2}+c_{3}^{2})\]
Correct Answer: C
Solution :
[c] \[{{\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\ \end{matrix} \right|}^{2}}={{[\overset{\to }{\mathop{A}}\,\overset{\to }{\mathop{B}}\,\overset{\to }{\mathop{C}}\,]}^{2}}={{((\overset{\to }{\mathop{A}}\,\times \overset{\to }{\mathop{B}}\,).\overset{\to }{\mathop{C}}\,)}^{2}}\] \[={{\left\{ |\overset{\to }{\mathop{A}}\,||\overset{\to }{\mathop{B}}\,|sin\frac{\pi }{6}(\overset{\to }{\mathop{C}}\,).\overset{\to }{\mathop{C}}\, \right\}}^{2}}\] \[=|\overset{\to }{\mathop{A}}\,{{|}^{2}}|\overset{\to }{\mathop{B}}\,{{|}^{2}}{{\left( \frac{1}{2} \right)}^{2}}|\overset{\to }{\mathop{C}}\,{{|}^{4}}=\frac{1}{4}|\overset{\to }{\mathop{A}}\,{{|}^{2}}|\overset{\to }{\mathop{B}}\,{{|}^{2}}\] \[=\frac{1}{4}(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})(b_{1}^{2}+b_{2}^{2}+b_{3}^{2})\]
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