Question:

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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Solution and Explanation

Concept:
• Centroid: \[ \vec{Q} = \frac{\vec{a}+\vec{b}+\vec{c}}{3} \]

Step 1:
Write vectors.
\[ \overrightarrow{QA} = \vec{a} - \vec{Q} \] \[ \overrightarrow{QB} = \vec{b} - \vec{Q} \] \[ \overrightarrow{QC} = \vec{c} - \vec{Q} \]

Step 2:
Add them.
\[ (\vec{a} - \vec{Q}) + (\vec{b} - \vec{Q}) + (\vec{c} - \vec{Q}) \] \[ = \vec{a} + \vec{b} + \vec{c} - 3\vec{Q} \]

Step 3:
Substitute centroid.
\[ 3\vec{Q} = \vec{a} + \vec{b} + \vec{c} \]

Step 4:
Simplify.
\[ = 0 \]

Step 5:
Final Answer.
\[ \boxed{\vec{0}} \]
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