Question:
$OABC$ is a tetrahedron. If $D, E$ are the midpoints of $OA$ and $BC$ respectively, then $\overrightarrow{DE} =$
$OABC$ is a tetrahedron. If $D, E$ are the midpoints of $OA$ and $BC$ respectively, then $\overrightarrow{DE} =$
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When using midpoints in vectors, always express in terms of origin and apply vector subtraction carefully.
Updated On: May 19, 2025
- $\dfrac{1}{2} (\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC})$
- $\dfrac{1}{2} (\overrightarrow{OA} + \overrightarrow{OB} - \overrightarrow{OC})$
- $\dfrac{1}{2} (\overrightarrow{OA} - \overrightarrow{OB} + \overrightarrow{OC})$
- $\dfrac{1}{2} (-\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC})$
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The Correct Option is D
Solution and Explanation
$D$ is midpoint of $OA \Rightarrow \vec{OD} = \dfrac{1}{2} \vec{OA}$
$E$ is midpoint of $BC \Rightarrow \vec{OE} = \dfrac{1}{2} (\vec{OB} + \vec{OC})$
Then $\vec{DE} = \vec{OE} - \vec{OD} = \dfrac{1}{2} (\vec{OB} + \vec{OC}) - \dfrac{1}{2} \vec{OA}$
$= \dfrac{1}{2} (-\vec{OA} + \vec{OB} + \vec{OC})$
$E$ is midpoint of $BC \Rightarrow \vec{OE} = \dfrac{1}{2} (\vec{OB} + \vec{OC})$
Then $\vec{DE} = \vec{OE} - \vec{OD} = \dfrac{1}{2} (\vec{OB} + \vec{OC}) - \dfrac{1}{2} \vec{OA}$
$= \dfrac{1}{2} (-\vec{OA} + \vec{OB} + \vec{OC})$
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