A) \[\sqrt{2}\]
B) \[\sqrt{3}\]
C) \[1/\sqrt{2}\]
D) \[\sqrt{5}\]
Correct Answer: B
Solution :
Let \[{{\hat{n}}_{1}}\] and \[{{\hat{n}}_{2}}\] are the two unit vectors, then the sum is \[{{\overrightarrow{n}}_{s}}={{\hat{n}}_{1}}+{{\hat{n}}_{2}}\] or \[n_{s}^{2}=n_{1}^{2}+n_{2}^{2}+2{{n}_{1}}{{n}_{2}}\cos \theta \] \[=1+1+2\cos \theta \] Since it is given that \[{{n}_{s}}\] is also a unit vector, therefore \[1=1+1+2\cos \theta \]Þ \[\cos \theta =-\frac{1}{2}\] \\[\theta =120{}^\circ \] Now the difference vector is \[{{\hat{n}}_{d}}={{\hat{n}}_{1}}-{{\hat{n}}_{2}}\] or \[n_{d}^{2}=n_{1}^{2}+n_{2}^{2}-2{{n}_{1}}{{n}_{2}}\cos \theta \]\[=1+1-2\cos (120{}^\circ )\] \ \[n_{d}^{2}=2-2(-1/2)=2+1=3\]\[\Rightarrow \,\,{{n}_{d}}=\sqrt{3}\]
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