A) \[\sqrt{2}\]
B) \[1\]
C) \[\sqrt{5}\]
D) \[\sqrt{3}\]
Correct Answer: D
Solution :
Given, \[|\vec{a}+\vec{b}|=1\] \[\therefore \] \[|\vec{a}+\vec{b}{{|}^{2}}={{1}^{2}}\] \[\Rightarrow \] \[|\vec{a}|+|\vec{b}{{|}^{2}}+2|\vec{a}||\vec{b}|=1\] \[\Rightarrow \] \[2|\vec{a}||\vec{b}|=1-(1+1)\] \[\Rightarrow \] \[2|\vec{a}||\vec{b}|=-1\] ?.(1) Now. \[|\vec{a}-\vec{b}{{|}^{2}}=|\vec{a}{{|}^{2}}+|\vec{b}{{|}^{2}}-2|\vec{a}||\vec{b}|\] \[={{1}^{2}}+{{1}^{2}}-(-1)=3\] [form Eq. (i)] \[\Rightarrow \] \[|\vec{a}-\vec{b}|=\sqrt{3}\]
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