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If \( \vec{a}, \vec{b}, \vec{c} \) are unit vectors such that \( \vec{a}+\vec{b}+\vec{c}=\overrightarrow{0} \) then the value of \( \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a} \)
is equal to
Options:
is equal to
Solution:
2475 Upvotes
Verified Answer
The correct answer is:
\( -\frac{3}{2} \)
For unit vectors, we know that
\[
|\vec{a}|=|\vec{b}|=|\vec{c}|=1 \rightarrow(1)
\]
\[
\text { Given that, } \vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}
\]
Squaring on both sides of Eq. (1),
\[
\begin{array}{l}
|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0 \\
\Rightarrow 1+1+1+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0 \\
\Rightarrow \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=\frac{-3}{2}
\end{array}
\]
\[
|\vec{a}|=|\vec{b}|=|\vec{c}|=1 \rightarrow(1)
\]
\[
\text { Given that, } \vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}
\]
Squaring on both sides of Eq. (1),
\[
\begin{array}{l}
|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0 \\
\Rightarrow 1+1+1+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0 \\
\Rightarrow \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=\frac{-3}{2}
\end{array}
\]
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