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Question
If \[\vec{a}\] and \[\vec{b}\] are unit vectors, then find the angle between \[\vec{a}\] and \[\vec{b}\] given that \[\left( \sqrt{3} \vec{a} - \vec{b} \right)\] is a unit vector.
Solution
Let the angle between \[\vec{a}\] and \[\vec{b}\] be \[\theta\] It is given that \[\left| \vec{a} \right| = \left| \vec{b} \right| = \left| \sqrt{3} \vec{a} - \vec{b} \right| = 1\]
\[\left| \sqrt{3} \vec{a} - \vec{b} \right| = 1\]
\[ \Rightarrow \left| \sqrt{3} \vec{a} - \vec{b} \right|^2 = 1\]
\[ \Rightarrow \left| \sqrt{3} \vec{a} \right|^2 - 2\sqrt{3} \vec{a} . \vec{b} + \left| \vec{b} \right|^2 = 1\]
\[ \Rightarrow 3 \left| \vec{a} \right|^2 - 2\sqrt{3}\left| \vec{a} \right|\left| \vec{b} \right|\cos\theta + \left| \vec{b} \right|^2 = 1\]
\[\Rightarrow 3 \times 1 - 2\sqrt{3} \times 1 \times 1 \times \cos\theta + 1 = 1\]
\[ \Rightarrow 2\sqrt{3}\cos\theta = 3\]
\[ \Rightarrow \cos\theta = \frac{\sqrt{3}}{2} = \cos\frac{\pi}{6}\]
\[ \Rightarrow \theta = \frac{\pi}{6}\]
Thus, the angle between \[\vec{a}\] and \[\vec{b}\] is \[\frac{\pi}{6}\]
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