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
\[
\overrightarrow{DE}=
\]
Show Hint
For midpoint problems in vectors, use the midpoint position vector formula:
\[
\overrightarrow{OM}=\frac{1}{2}\left(\overrightarrow{OP}+\overrightarrow{OQ}\right)
\]
Then subtract position vectors to find the required vector.
Updated On: Jun 22, 2026
- \(\dfrac{1}{2}\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right)\)
- \(\dfrac{1}{2}\left(\overrightarrow{OA}+\overrightarrow{OB}-\overrightarrow{OC}\right)\)
- \(\dfrac{1}{2}\left(\overrightarrow{OA}-\overrightarrow{OB}+\overrightarrow{OC}\right)\)
- \(\dfrac{1}{2}\left(-\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right)\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Write the position vector of \(D\).
Since \(D\) is the midpoint of \(OA\), we have \[ \overrightarrow{OD}=\frac{1}{2}\overrightarrow{OA} \]
Step 2: Write the position vector of \(E\).
Since \(E\) is the midpoint of \(BC\), we have \[ \overrightarrow{OE} = \frac{1}{2}\left(\overrightarrow{OB}+\overrightarrow{OC}\right) \]
Step 3: Find \(\overrightarrow{DE}\).
Using position vectors, \[ \overrightarrow{DE} = \overrightarrow{OE}-\overrightarrow{OD} \] Substituting the values, \[ \overrightarrow{DE} = \frac{1}{2}\left(\overrightarrow{OB}+\overrightarrow{OC}\right) - \frac{1}{2}\overrightarrow{OA} \] Taking \(\frac{1}{2}\) common, \[ \overrightarrow{DE} = \frac{1}{2}\left(-\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right) \]
Step 4: Final conclusion.
Hence, \[ \boxed{ \frac{1}{2}\left(-\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right) } \] which corresponds to option (4).
Since \(D\) is the midpoint of \(OA\), we have \[ \overrightarrow{OD}=\frac{1}{2}\overrightarrow{OA} \]
Step 2: Write the position vector of \(E\).
Since \(E\) is the midpoint of \(BC\), we have \[ \overrightarrow{OE} = \frac{1}{2}\left(\overrightarrow{OB}+\overrightarrow{OC}\right) \]
Step 3: Find \(\overrightarrow{DE}\).
Using position vectors, \[ \overrightarrow{DE} = \overrightarrow{OE}-\overrightarrow{OD} \] Substituting the values, \[ \overrightarrow{DE} = \frac{1}{2}\left(\overrightarrow{OB}+\overrightarrow{OC}\right) - \frac{1}{2}\overrightarrow{OA} \] Taking \(\frac{1}{2}\) common, \[ \overrightarrow{DE} = \frac{1}{2}\left(-\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right) \]
Step 4: Final conclusion.
Hence, \[ \boxed{ \frac{1}{2}\left(-\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right) } \] which corresponds to option (4).
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