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If \( \vec{a} \) and \( \vec{b} \) are unit vectors then what is the angle between \( \vec{a} \) and \( \vec{b} \) for \( \sqrt{3} \vec{a}-\vec{b} \) to be unit
vector?
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vector?
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Verified Answer
The correct answer is:
\( 30^{\circ} \)
Given that, $|\vec{a}|=|\vec{b}|=1$
and and $(\sqrt{3} \vec{a}-\vec{b}) \cdot(\sqrt{3} \vec{a}-\vec{b})=1$
$\Rightarrow 3|\vec{a}|^{2}+|\vec{b}|^{2}-2 \sqrt{3} \vec{a} \cdot \vec{b}=1$
$\Rightarrow 3+1-2 \sqrt{3} \cos \theta=1$
$\Rightarrow \cos \theta=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2}$
$\Rightarrow \theta=30^{\circ}$
and and $(\sqrt{3} \vec{a}-\vec{b}) \cdot(\sqrt{3} \vec{a}-\vec{b})=1$
$\Rightarrow 3|\vec{a}|^{2}+|\vec{b}|^{2}-2 \sqrt{3} \vec{a} \cdot \vec{b}=1$
$\Rightarrow 3+1-2 \sqrt{3} \cos \theta=1$
$\Rightarrow \cos \theta=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2}$
$\Rightarrow \theta=30^{\circ}$
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