A) \[{{a}^{2}}={{b}^{2}}+{{c}^{2}}\]
B) \[{{b}^{2}}={{c}^{2}}+{{a}^{2}}\]
C) \[{{c}^{2}}={{a}^{2}}+{{b}^{2}}\]
D) \[2{{a}^{2}}-{{b}^{2}}={{c}^{2}}\] (Note : Here \[a=\,\,|\mathbf{a}|,\,\,b=\,|\,\mathbf{b}|,\,\,c=\,|\mathbf{c}|)\]
Correct Answer: A
Solution :
Given that \[\Rightarrow \mathbf{a}\times \mathbf{b}=\mathbf{c}\] and angle between b and c is \[\frac{\pi }{2}\]. So, \[{{\mathbf{a}}^{2}}={{\mathbf{b}}^{2}}+{{\mathbf{c}}^{2}}+2\,\mathbf{b}\,\mathbf{.}\,\mathbf{c}\] or \[{{\mathbf{a}}^{2}}={{\mathbf{b}}^{2}}+{{\mathbf{c}}^{2}}+2|\mathbf{b}||\mathbf{c}|\,\cos \frac{\pi }{2}\] or \[{{\mathbf{a}}^{2}}={{\mathbf{b}}^{2}}+{{\mathbf{c}}^{2}}+0,\,\,\,\,\,\therefore \,\,{{\mathbf{a}}^{2}}={{\mathbf{b}}^{2}}+{{\mathbf{c}}^{2}}\] i.e., \[{{a}^{2}}={{b}^{2}}+{{c}^{2}}\].
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