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If\,\overrightarrow{A}={{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k},\,then\,\widehat{i}\times \left( \widehat{i}\times \overrightarrow{A} \right)\,is
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If\,\overrightarrow{A}={{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k},\,then\,\widehat{i}\times \left( \widehat{i}\times \overrightarrow{A} \right)\,is

Solution: Option 3 is correct

We\,have:\,\,\overrightarrow{A}={{a}_{1}}\widehat{i}+{{a}_{2}}\widehat{j}+{{a}_{3}}\widehat{k},\,

\,\widehat{i}\times \left( \widehat{i}\times \overrightarrow{A} \right)\,=?

\left( \widehat{i}\times \overrightarrow{A} \right)={{a}_{1}}\left( \widehat{i}\times \widehat{i} \right)+{{a}_{2}}\left( \widehat{i}\times \widehat{j} \right)+{{a}_{3}}\left( \widehat{i}\times \widehat{k} \right)

\left( \widehat{i}\times \overrightarrow{A} \right)={{a}_{2}}\widehat{k}-{{a}_{3}}\widehat{j}

Now,\,\,\widehat{i}\times \left( \widehat{i}\times \overrightarrow{A} \right)\,=\,\widehat{i}\times \left( {{a}_{2}}\widehat{k}-{{a}_{3}}\widehat{j} \right)

\,\widehat{i}\times \left( \widehat{i}\times \overrightarrow{A} \right)\,={{a}_{2}}\left( \widehat{i}\times \widehat{k} \right)-{{a}_{3}}\left( \widehat{i}\times \widehat{j} \right)

\therefore \,\widehat{i}\times \left( \widehat{i}\times \overrightarrow{A} \right)\,=-{{a}_{2}}\widehat{j}-{{a}_{3}}\widehat{k}

i

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