Question:medium
Let the position vectors of points \(A, B, C\) be \( \vec{a}, \vec{b}, \vec{c} \) respectively. Let \(Q\) be the centroid. Then \( \overrightarrow{QA} + \overrightarrow{QB} + \overrightarrow{QC} = \)
Let the position vectors of points \(A, B, C\) be \( \vec{a}, \vec{b}, \vec{c} \) respectively. Let \(Q\) be the centroid. Then \( \overrightarrow{QA} + \overrightarrow{QB} + \overrightarrow{QC} = \)
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Sum of vectors from centroid to vertices of triangle is always zero.
Updated On: May 8, 2026
- \( \frac{\vec{a}+\vec{b}+\vec{c}}{2} \)
- \( 2\vec{a}+\vec{b}+\vec{c} \)
- \( \vec{a}+\vec{b}+\vec{c} \)
- \( \frac{\vec{a}+\vec{b}+\vec{c}}{3} \)
- \( \vec{0} \)
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The Correct Option is
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