A) \[1\]
B) \[2\]
C) \[\sqrt{2}\]
D) \[\sqrt{3}\]
Correct Answer: D
Solution :
Given, \[|\vec{a}|=|\vec{b}|=|\vec{a}-\vec{b}|=1\] \[\Rightarrow \] \[|\vec{a}-\vec{b}{{|}^{2}}={{1}^{2}}\] \[\Rightarrow \] \[|\vec{a}{{|}^{2}}+|\vec{b}{{|}^{2}}-2|\vec{a}|.|\vec{b}|=1\] \[\Rightarrow \] \[2|\vec{a}||\vec{b}|=1\] \[\Rightarrow \] \[2|\vec{a}||\vec{b}|=1\] Now, \[|\vec{a}+\vec{b}{{|}^{2}}=|\vec{a}-\vec{b}{{|}^{2}}+4|\vec{a}|.|\vec{b}|\] \[=1+4\left( \frac{1}{2} \right)=3\] \[\Rightarrow \] \[|\vec{a}+\vec{b}|=\sqrt{3}\]
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