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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
Options
\[\alpha = \frac{\pi}{4}\]
\[\alpha = \frac{\pi}{3}\]
\[\alpha = \frac{2\pi}{3}\]
\[\alpha = \frac{\pi}{2}\]
Solution
\[\alpha = \frac{2\pi}{3}\]
\[\vec{a} \text{ and } \vec{b} \text{ are unit vectors } . \]
\[ \Rightarrow \left| \vec{a} \right| = \left| \vec{b} \right| = 1................... \left( 1 \right)\]
\[\text{ Now }, \]
\[ \vec{a} . \vec{b} = \left| \vec{a} \right| \left| \vec{b} \right| \cos \alpha\]
\[ \Rightarrow \vec{a} . \vec{b} = \cos \alpha.................. \left( 2 \right)\] .................\[ \left[ \text{ Using } \left( 1 \right) \right]\]
\[\text{ Given that }\]
\[\left| \vec{a} + \vec{b} \right| = 1\]
\[\text{ Squaring both sides, we get }\]
\[ \left| \vec{a} + \vec{b} \right|^2 = 1\]
\[ \Rightarrow \left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 + 2 \vec{a} . \vec{b} = 1\]
\[ \Rightarrow 1 + 1 + 2 \cos \alpha = 1 ..........\left[ \text{ Using } \left( 1 \right) \text{ and } \left( 2 \right) \right]\]
\[ \Rightarrow 2 + 2 \cos \alpha = 1\]
\[ \Rightarrow 2 \cos \alpha = - 1\]
\[ \Rightarrow 2 \cos \alpha = - 1\]
\[ \Rightarrow \cos \alpha = \frac{- 1}{2}\]
\[ \Rightarrow \alpha = \frac{2\pi}{3}\]
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