Question

Let \[\vec{a} \text{ and } \vec{b}\]  be two unit vectors and α be the angle between them. Then, \[\vec{a} + \vec{b}\] is a unit vector if 

पर्याय

• \[\alpha = \frac{\pi}{4}\] 

• \[\alpha = \frac{\pi}{3}\] 

•  \[\alpha = \frac{2\pi}{3}\] 

   

•  \[\alpha = \frac{\pi}{2}\]

Answer 1

  \[\alpha = \frac{2\pi}{3}\]  

\[\vec{a} \text{ and } \vec{b} \text{ are unit vectors } . \]

\[ \Rightarrow \left| \vec{a} \right| = \left| \vec{b} \right| = 1................... \left( 1 \right)\]

\[\text{ Now }, \]

\[ \vec{a} . \vec{b} = \left| \vec{a} \right| \left| \vec{b} \right| \cos \alpha\]

\[ \Rightarrow \vec{a} . \vec{b} = \cos \alpha.................. \left( 2 \right)\]     .................\[ \left[ \text{ Using } \left( 1 \right) \right]\] 

\[\text{ Given that }\]

\[\left| \vec{a} + \vec{b} \right| = 1\]

\[\text{ Squaring both sides, we get }\]

\[ \left| \vec{a} + \vec{b} \right|^2 = 1\]

\[ \Rightarrow \left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 + 2 \vec{a} . \vec{b} = 1\]

\[ \Rightarrow 1 + 1 + 2 \cos \alpha = 1 ..........\left[ \text{ Using } \left( 1 \right) \text{ and } \left( 2 \right) \right]\]

\[ \Rightarrow 2 + 2 \cos \alpha = 1\]

\[ \Rightarrow 2 \cos \alpha = - 1\]

\[ \Rightarrow 2 \cos \alpha = - 1\]

\[ \Rightarrow \cos \alpha = \frac{- 1}{2}\]

\[ \Rightarrow \alpha = \frac{2\pi}{3}\]