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If \( 2 \vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}| \) then the angle between \( \vec{a} \& \vec{b} \) is
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The correct answer is:
\( 60^{\circ} \)
Given that, \( 2 \vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}| \)
As we know, \( \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \)
So, \( 2|\vec{a}||\vec{b}| \cos \theta=|\vec{a}||\vec{b}| \)
\( \Rightarrow 2 \cos \theta=1 \)
\( \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=60^{\circ} \)
As we know, \( \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta \)
So, \( 2|\vec{a}||\vec{b}| \cos \theta=|\vec{a}||\vec{b}| \)
\( \Rightarrow 2 \cos \theta=1 \)
\( \Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=60^{\circ} \)
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