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