A) \[\sqrt{3}\]
B) \[2\]
C) \[2\sqrt{3}\]
D) \[0\]
Correct Answer: C
Solution :
Given, \[(a+b)=c\] Squaring on both sides, \[{{(a+b)}^{2}}={{c}^{2}}\] \[\Rightarrow \] \[(a+b)\,.\,(a+b)-|c{{|}^{2}}\] \[\Rightarrow \] \[|a{{|}^{2}}+|b{{|}^{2}}+2a.b=|c{{|}^{2}}\] \[\Rightarrow \] \[4+4+2a.b=4\] \[(\because \,\,|a|=|b|=|c|=2\,given)\] \[\Rightarrow \] \[a\,.\,b=-2\] ?.(i) Now, we have \[|a+b{{|}^{2}}=|a{{|}^{2}}+|b{{|}^{2}}-2a.b\] \[=4+4-2(-2)\] [from Eq. (i)] \[\Rightarrow \] \[|a-b|=2\sqrt{3}\] \[\therefore \]Magnitude of \[(a-b)=2\sqrt{3}\]
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