A) \[\overset{{{\to }^{2}}}{\mathop{a}}\,+\overset{{{\to }^{2}}}{\mathop{b}}\,-(\overrightarrow{a}.\overrightarrow{b})\]
B) \[\overset{{{\to }^{2}}}{\mathop{a}}\,+\overset{{{\to }^{2}}}{\mathop{b}}\,-{{(\overrightarrow{a}.\overrightarrow{b})}^{2}}\]
C) \[\overset{{{\to }^{2}}}{\mathop{a}}\,+\overset{{{\to }^{2}}}{\mathop{b}}\,-2\,\,\overrightarrow{a}\,\,.\,\,\overrightarrow{b}\]
D) \[\overset{{{\to }^{2}}}{\mathop{a}}\,+\overset{{{\to }^{2}}}{\mathop{b}}\,-2\,\,\overrightarrow{a}\,\,.\,\,\overrightarrow{b}\]
E) none of the above
Correct Answer: B
Solution :
\[\therefore \]\[{{(\overrightarrow{a}\times \overrightarrow{b})}^{2}}+{{(\overrightarrow{a}.\overrightarrow{b})}^{2}}\] \[=({{\overrightarrow{a}}^{2}}{{\overrightarrow{b}}^{2}}{{\sin }^{2}}\theta )+({{\overrightarrow{a}}^{2}}{{\overrightarrow{b}}^{2}}{{\cos }^{2}}\theta )\] \[={{\overrightarrow{a}}^{2}}{{\overrightarrow{b}}^{2}}({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )={{\overrightarrow{a}}^{2}}{{\overrightarrow{b}}^{2}}\] \[\therefore \] \[{{(\overrightarrow{a}\times \overrightarrow{b})}^{2}}={{\overrightarrow{a}}^{2}}\,\,\,{{\overrightarrow{b}}^{2}}-{{(\overrightarrow{a}\,\,.\,\,\,\overrightarrow{b})}^{2}}\]
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