A) \[0\]
B) \[\pm 1\]
C) \[\pm \,2\]
D) \[\pm \,3\]
Correct Answer: C
Solution :
Given that, \[\vec{a}.\vec{b}=\vec{a}.\vec{c}=0\] Now, \[|\vec{a}+\vec{b}+\vec{c}{{|}^{2}}=|\vec{a}{{|}^{2}}+|\vec{b}{{|}^{2}}+|\vec{c}{{|}^{2}}\] \[+2(\vec{a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a})\] \[={{(1)}^{2}}+{{(1)}^{2}}+{{(1)}^{2}}+2\left( 0+|\vec{b}||\vec{c}|\,\cos \frac{\pi }{3}+0 \right)\] \[=3+2\times 1\times 1\times \frac{1}{2}=4\] \[\Rightarrow \] \[|\vec{a}+\vec{b}+\vec{c}|=\pm 2\]
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