Question:
If \( |\vec{a}|=3, |\vec{b}|=1, |\vec{c}|=4 \) and \( \vec{a}+\vec{b}+\vec{c}=0 \), then the value of \( \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} \) is
If \( |\vec{a}|=3, |\vec{b}|=1, |\vec{c}|=4 \) and \( \vec{a}+\vec{b}+\vec{c}=0 \), then the value of \( \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} \) is
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Always square vector sum when given zero vector condition.
Updated On: May 1, 2026
- \( 13 \)
- \( 26 \)
- \( -29 \)
- \( -13 \)
- \( -26 \)
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The Correct Option is D
Solution and Explanation
Concept:
Use identity:
\[
(\vec{a}+\vec{b}+\vec{c})^2 = 0
\]
Step 1: Expand square.
\[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0 \]
Step 2: Substitute values.
\[ 9 + 1 + 16 + 2S = 0 \]
Step 3: Simplify.
\[ 26 + 2S = 0 \]
Step 4: Solve for \( S \).
\[ 2S = -26 \]
Step 5: Final result.
\[ S = -13 \]
Step 1: Expand square.
\[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0 \]
Step 2: Substitute values.
\[ 9 + 1 + 16 + 2S = 0 \]
Step 3: Simplify.
\[ 26 + 2S = 0 \]
Step 4: Solve for \( S \).
\[ 2S = -26 \]
Step 5: Final result.
\[ S = -13 \]
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