Question:

Let $\vec{OA}=2\hat{i}+3\hat{j}-5\hat{k}$, $\vec{OB}=3\hat{i}+\hat{j}-2\hat{k}$, $\vec{OC}=6\hat{i}-5\hat{j}+7\hat{k}$ be position vectors of A, B and C. Then ________.

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Collinear points satisfy $\vec{AC} = k\vec{AB}$.
Updated On: Jun 26, 2026
  • $\vec{AC}=3\vec{AB}$
  • $\vec{AB}=3\vec{BC}$
  • $\vec{AC}=2\vec{AB}$
  • $\vec{BC}=3\vec{AB}$
  • $\vec{AC}=4\vec{AB}$
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The Correct Option is

Solution and Explanation

Step 1: Concept
Calculate vectors $\vec{AB}$ and $\vec{AC}$ by subtracting position vectors.

Step 2: Meaning

$\vec{AB} = \vec{OB} - \vec{OA} = (3-2)\hat{i} + (1-3)\hat{j} + (-2+5)\hat{k} = \hat{i} - 2\hat{j} + 3\hat{k}$.

Step 3: Analysis

$\vec{AC} = \vec{OC} - \vec{OA} = (6-2)\hat{i} + (-5-3)\hat{j} + (7+5)\hat{k} = 4\hat{i} - 8\hat{j} + 12\hat{k}$.
Observing: $\vec{AC} = 4(\hat{i} - 2\hat{j} + 3\hat{k})$.

Step 4: Conclusion

$\vec{AC} = 4\vec{AB}$. Final Answer: (E)
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