A) 1
B) 4
C) \[\vec{0}\]
D) None of these
Correct Answer: C
Solution :
We have, \[{{\mathbf{\vec{V}}}_{1}}=\frac{1}{2}(\mathbf{\vec{a}}\times \mathbf{\vec{b}}),\,\,{{\mathbf{\vec{V}}}_{2}}=\frac{1}{2}(\mathbf{\vec{b}}\times \mathbf{\vec{c}}),\]\[{{\mathbf{\vec{V}}}_{3}}=\frac{1}{2}(\mathbf{\vec{c}}\times \mathbf{\vec{a}})\]and \[{{\mathbf{\vec{V}}}_{4}}=\frac{1}{2}\{(\mathbf{\vec{c}}-\mathbf{\vec{a}})\times (\mathbf{\vec{b}}-\mathbf{\vec{a}})\}\] \[\therefore \]\[{{\mathbf{\vec{V}}}_{1}}+{{\mathbf{\vec{V}}}_{2}}+{{\mathbf{\vec{V}}}_{3}}+{{\mathbf{\vec{V}}}_{4}}=\frac{1}{2}(\mathbf{\vec{a}}\times \mathbf{\vec{b}}+\mathbf{\vec{b}}\times \mathbf{\vec{c}}+\mathbf{\vec{c}}\times \mathbf{\vec{a}}+\] \[+\mathbf{\vec{c}}\times \mathbf{\vec{b}}-\mathbf{\vec{c}}\times \mathbf{\vec{a}}-\mathbf{\vec{a}}\times \mathbf{\vec{b}})=0\] \[|{{\mathbf{\vec{V}}}_{1}}+{{\mathbf{\vec{V}}}_{\mathbf{2}}}+{{\mathbf{\vec{V}}}_{3}}+{{\mathbf{\vec{V}}}_{4}}|\,\,=0\]You need to login to perform this action.
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