A) \[\frac{1}{2}\]
B) \[\frac{1}{4}\]
C) 1
D) 2
Correct Answer: A
Solution :
| Hint: \[[\vec{a}\,\vec{b}\,\vec{c}]=\vec{a}.(\vec{b}\,\times \,\vec{c})=\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\ \end{matrix} \right|\] |
| \[[\vec{a}\,\,\vec{b}\,\,\vec{c}]=(\vec{a}\,\,\times \,\,\vec{b}\,).\vec{c}\] |
| \[[\vec{a}\,\,\vec{b}\,\,\vec{c}]=\pm \,\,\left| \,\vec{a}\,\,\times \,\,\vec{b}\, \right|\,\,\left| \,\vec{c}\, \right|\cos 0\] |
| \[\because \vec{c}\,||\,\vec{a}\,\,\times \,\,\vec{b}\] \[{{[\vec{a}\,\,\vec{b}\,\,\vec{c}]}^{2}}=\,{{\left| \,\overrightarrow{a}\, \right|}^{2}}{{\left| \,\vec{b}\, \right|}^{2}}\,\,\,\,{{\sin }^{2}}\frac{\pi }{2}\] |
| \[{{[\vec{a}\,\,\vec{b}\,\,\vec{c}]}^{2}}=\frac{1}{2}{{\left| \,\overrightarrow{a}\, \right|}^{2}}{{\left| \,\vec{b}\, \right|}^{2}}\] |
| \[\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} \\ \end{matrix} \right|=\frac{1}{2}(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})(b_{1}^{2}+b_{2}^{2}+b_{3}^{2})\] |
| \[\lambda =\frac{1}{2}\] |
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