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प्रश्न
If \[\vec{a,} \vec{b,} \vec{c}\] are non-coplanar vectors, then \[\frac{\vec{a} \cdot \left( \vec{b} \times \vec{c} \right)}{\left( \vec{c} \times \vec{a} \right) \cdot \vec{b}} + \frac{\vec{b} \cdot \left( \vec{a} \times \vec{c} \right)}{\vec{c} \cdot \left( \vec{a} \times \vec{b} \right)}\] is equal to
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none of these
उत्तर
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We have
\[\frac{\vec{a} . \left( \vec{b} \times \vec{c} \right)}{\left( \vec{c} \times \vec{a} \right) . \vec{b}} + \frac{\vec{b} . \left( \vec{a} \times \vec{c} \right)}{\vec{c} . \left( \vec{a} \times \vec{b} \right)}\]
\[ \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \]
\[ = \frac{\left[ \vec{a} \vec{b} \vec{c} \right]}{\left[ \vec{c} \vec{a} \vec{b} \right]} + \frac{\left[ \vec{b} \vec{a} \vec{c} \right]}{\left[ \vec{c} \vec{a} \vec{b} \right]} \left( \text { By definition of scalar tiple product } \right)\]
\[ = \frac{\left[ \vec{a} \vec{b} \vec{c} \right]}{\left[ \vec{a} \vec{b} \vec{c} \right]} + \frac{- \left[ \vec{a} \vec{b} \vec{c} \right]}{\left[ \vec{a} \vec{b} \vec{c} \right]} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \hspace{0.167em} \left( \text { Change in cyclic order of vectors changes the sign of the scalar triple product } \right)\]
\[ = 1 - 1 \hspace{0.167em} \hspace{0.167em} \]
\[ = 0\]
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