A) \[4{{a}^{2}}\]
B) \[2{{a}^{2}}\]
C) \[{{a}^{2}}\]
D) \[3{{a}^{2}}\]
Correct Answer: B
Solution :
Since\[(a\times \hat{i}).(a\times \hat{i})=\left| \begin{matrix} a.a & a.\hat{i} \\ \hat{i}.a & 1 \\ \end{matrix} \right|=|a{{|}^{2}}-a_{1}^{2}\] Similarly, \[{{(a\times \hat{j})}^{2}}=|{{a}^{2}}|-a_{2}^{2}\] and \[{{(a\times \hat{k})}^{2}}=|a{{|}^{2}}-a_{3}^{2}\] \[\therefore \] \[{{(a\times \hat{i})}^{2}}+{{(a\times \hat{j})}^{2}}+{{(a\times \hat{k})}^{2}}\] \[=3|a{{|}^{2}}-(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})\] \[=3|a{{|}^{2}}-|a{{|}^{2}}\] \[=2|a{{|}^{2}}=2{{a}^{2}}\] Alternate Solution Let\[a={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k},\]then \[|a{{|}^{2}}=a_{1}^{2}+a_{2}^{2}+a_{3}^{2}\] and\[a\times \hat{i}=({{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k})\times (\hat{i})\] \[=-{{a}_{2}}\hat{k}+{{a}_{3}}\hat{j}\] \[\Rightarrow \]\[{{(a\times \hat{i})}^{2}}=a_{2}^{2}+a_{3}^{2}\] Similarly, \[{{(a\times \hat{j})}^{2}}=a_{3}^{2}+a_{1}^{2}\] and \[{{(a\times \hat{k})}^{2}}=a_{1}^{2}+a_{2}^{2}\] \[\therefore \] \[{{(a\times \hat{i})}^{2}}+{{(a\times \hat{j})}^{2}}+{{(a\times \hat{k})}^{2}}\] \[=(a_{2}^{2}+a_{3}^{2}+a_{3}^{2}+a_{1}^{2}+a_{1}^{2}+a_{2}^{2})\] \[=2(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})=2{{(a)}^{2}}\]
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