question_answer

If the sum of two unit vectors is a unit vector, show that the magnitude of their difference is \[\sqrt{3}.\]   

Answer:

Let \[\overrightarrow{a}\]and \[\overrightarrow{b}\]are the two unit vectors. According to the question, \[\overrightarrow{a}+\overrightarrow{b}\]is also a unit vector. \[\therefore \] \[|\overrightarrow{a}+\overrightarrow{b}{{|}^{2}}={{(\overrightarrow{a})}^{2}}+{{(\overrightarrow{b})}^{2}}+2\overrightarrow{a}\cdot \overrightarrow{b}\] \[\Rightarrow \]   \[1=1+1+2\,(1)\,\,(1)\,\cos \theta \] \[\Rightarrow \]   \[\cos \theta =\frac{-\,1}{2}\] Now, \[|\overrightarrow{a}-\overrightarrow{b}{{|}^{2}}={{(\overrightarrow{a})}^{2}}+{{(\overrightarrow{b})}^{2}}-2\overrightarrow{a}\cdot \overrightarrow{b}\] \[=1+1-2\cdot 1\cdot 1\,\cos \theta \] \[=2-2\cdot \left( \frac{-\,1}{2} \right)\] \[=2+1=3\] \[\therefore \]      \[|\overrightarrow{a}-\overrightarrow{b}|\,\,=\sqrt{3}\]           Hence proved.