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If \( \vec{a} \) and \( \vec{b} \) are two unit vectors inclined at an angle \( \frac{\Pi}{3} \), then the value of \( |\vec{a}+\vec{b}| \) is
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Verified Answer
The correct answer is:
greater than \( 1 \)
Given that \( |\vec{a}|=|\vec{b}|=1 \)
\(\begin{array}{l}
\text { and } \theta=\frac{\Pi}{3} \\
|\vec{a}+ \vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+2(|\vec{a}| \cdot|\vec{b}| \cdot \cos \theta) \\
=1+1+2 \cdot 1 \cdot 1 \cdot \cos \frac{\Pi}{3} \\
=1+1+2 \cdot \frac{1}{2}=3 \\
\text { So }|\vec{a}+\vec{b}|>1
\end{array}\)
\(\begin{array}{l}
\text { and } \theta=\frac{\Pi}{3} \\
|\vec{a}+ \vec{b}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+2(|\vec{a}| \cdot|\vec{b}| \cdot \cos \theta) \\
=1+1+2 \cdot 1 \cdot 1 \cdot \cos \frac{\Pi}{3} \\
=1+1+2 \cdot \frac{1}{2}=3 \\
\text { So }|\vec{a}+\vec{b}|>1
\end{array}\)
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