A) 1
B) 3
C) ? 3/2
D) 3/2
Correct Answer: C
Solution :
Squaring \[(\mathbf{a}+\mathbf{b}+\mathbf{c})=\mathbf{0},\] we get \[{{\mathbf{a}}^{2}}+{{\mathbf{b}}^{2}}+{{\mathbf{c}}^{2}}+2\mathbf{a}.\mathbf{b}+2\mathbf{b}.\mathbf{c}+2\mathbf{c}.\mathbf{a}=0\] Þ \[|\mathbf{a}{{|}^{2}}+|\mathbf{b}{{|}^{2}}+|\mathbf{c}{{|}^{2}}+2(\mathbf{a}.\mathbf{b}+\mathbf{b}.\mathbf{c}+\mathbf{c}.\mathbf{a})=0\] Þ \[2(\mathbf{a}.\mathbf{b}+\mathbf{b}.\mathbf{c}+\mathbf{c}.\mathbf{a})=-3\] \[\Rightarrow \mathbf{a}.\mathbf{b}+\mathbf{b}.\mathbf{c}+\mathbf{c}.\mathbf{a}=-\frac{3}{2}\].
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