A) \[\overset{\to }{\mathop{\mathbf{a}}}\,\cdot \overset{\to }{\mathop{\mathbf{b}}}\,+\overset{\to }{\mathop{\mathbf{b}}}\,\cdot \overset{\to }{\mathop{\mathbf{c}}}\,+\overset{\to }{\mathop{\mathbf{c}}}\,\cdot \overset{\to }{\mathop{\mathbf{b}}}\,=0\]
B) \[\overset{\to }{\mathop{\mathbf{a}}}\,\times \overset{\to }{\mathop{\mathbf{b}}}\,=\overset{\to }{\mathop{\mathbf{b}}}\,\times \overset{\to }{\mathop{\mathbf{c}}}\,=\overset{\to }{\mathop{\mathbf{c}}}\,\times \overset{\to }{\mathop{\mathbf{a}}}\,=0\]
C) \[\overset{\to }{\mathop{\mathbf{a}}}\,\cdot \overset{\to }{\mathop{\mathbf{b}}}\,=\overset{\to }{\mathop{\mathbf{b}}}\,\cdot \overset{\to }{\mathop{\mathbf{c}}}\,=\overset{\to }{\mathop{\mathbf{c}}}\,\cdot \overset{\to }{\mathop{\mathbf{a}}}\,=0\]
D) \[\overset{\to }{\mathop{\mathbf{a}}}\,\times \overset{\to }{\mathop{\mathbf{a}}}\,+\overset{\to }{\mathop{\mathbf{a}}}\,\times \overset{\to }{\mathop{\mathbf{c}}}\,+\overset{\to }{\mathop{\mathbf{c}}}\,\times \overset{\to }{\mathop{\mathbf{a}}}\,=0\]
Correct Answer: B
Solution :
We have,\[\overset{\to }{\mathop{\mathbf{a}}}\,+\overset{\to }{\mathop{\mathbf{b}}}\,+\overset{\to }{\mathop{\mathbf{c}}}\,=0\] \[\therefore \] \[\overset{\to }{\mathop{\mathbf{a}}}\,+\overset{\to }{\mathop{\mathbf{b}}}\,=-\overset{\to }{\mathop{\mathbf{c}}}\,\] \[\Rightarrow \] \[\overset{\to }{\mathop{\mathbf{a}}}\,\times \overset{\to }{\mathop{\mathbf{c}}}\,+\overset{\to }{\mathop{\mathbf{b}}}\,\times \overset{\to }{\mathop{\mathbf{c}}}\,=0\] \[\Rightarrow \] \[\overset{\to }{\mathop{\mathbf{b}}}\,\times \overset{\to }{\mathop{\mathbf{c}}}\,=\overset{\to }{\mathop{\mathbf{c}}}\,\times \overset{\to }{\mathop{\mathbf{a}}}\,\] Similarly, \[\overset{\to }{\mathop{\mathbf{a}}}\,\times \overset{\to }{\mathop{\mathbf{b}}}\,=\overset{\to }{\mathop{\mathbf{b}}}\,\times \overset{\to }{\mathop{\mathbf{c}}}\,=\overset{\to }{\mathop{\mathbf{c}}}\,\times \overset{\to }{\mathop{\mathbf{a}}}\,\]
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