A) \[\vec{b}\times \vec{a}\]
B) \[\overset{\to }{\mathop{a}}\,\]
C) \[\overset{\to }{\mathop{a}}\,\times \overset{\to }{\mathop{b}}\,\]
D) \[\overset{\to }{\mathop{b}}\,\]
Correct Answer: C
Solution :
\[(\hat{i}\times a.b)\hat{i}+(\hat{j}\times a.b)\hat{j}+(\hat{k}\times a.b)\hat{k}\]\[=[\,\hat{i}\,ab\,]\hat{i}+[\hat{j}\,a\,b]\hat{j}+[\hat{k}\,a\,b]\hat{k}\] Let \[a={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}\] \[b={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}\] \[\therefore \] \[[\hat{i}\cdot \,ab]=\left| \begin{matrix} 1 & 0 & 0 \\ {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ \end{matrix} \right|=({{a}_{2}}{{b}_{3}}-{{b}_{2}}{{a}_{3}})\] and \[[\hat{j}\,ab]=\left| \begin{matrix} 1 & 0 & 0 \\ {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\ \end{matrix} \right|=({{a}_{1}}{{b}_{2}}-{{b}_{2}}{{a}_{1}})\] \[\therefore \] \[[i\,ab]\hat{i}+[\hat{j}\,ab]\hat{j}\,[\hat{k}\,ab]\hat{k}\] \[=({{a}_{2}}{{b}_{3}}-{{b}_{2}}{{a}_{3}})\hat{i}+({{b}_{1}}{{a}_{3}}-{{a}_{1}}{{b}_{3}})\hat{j}+({{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}})\hat{k}\] \[=a\times b\]
You need to login to perform this action.
You will be redirected in
3 sec
