Question

\(OABC\) is a tetrahedron. If \(D,E\) are the midpoints of \(OA\) and \(BC\) respectively, then \[ \overrightarrow{DE}= \]

• A. \(\dfrac{1}{2}\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right)\)
• B. \(\dfrac{1}{2}\left(\overrightarrow{OA}+\overrightarrow{OB}-\overrightarrow{OC}\right)\)
• C. \(\dfrac{1}{2}\left(\overrightarrow{OA}-\overrightarrow{OB}+\overrightarrow{OC}\right)\)
• D. \(\dfrac{1}{2}\left(-\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right)\)

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.

Answer 1

The Correct Option is D

Step 1: Fix the reference point.
Take $O$ as the origin so positions of $A$, $B$, $C$ are $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$.
Step 2: Find the midpoint $D$ of $OA$.
Since $D$ is the midpoint of $OA$, \[ \overrightarrow{OD}=\frac{1}{2}\overrightarrow{OA}. \] Step 3: Find the midpoint $E$ of $BC$.
Since $E$ is the midpoint of $BC$, \[ \overrightarrow{OE}=\frac{1}{2}\left(\overrightarrow{OB}+\overrightarrow{OC}\right). \] Step 4: Express $\overrightarrow{DE}$.
Using $\overrightarrow{DE}=\overrightarrow{OE}-\overrightarrow{OD}$, \[ \overrightarrow{DE}=\frac{1}{2}\left(\overrightarrow{OB}+\overrightarrow{OC}\right)-\frac{1}{2}\overrightarrow{OA}. \] Step 5: Collect the terms.
Factoring out $\dfrac{1}{2}$, \[ \overrightarrow{DE}=\frac{1}{2}\left(-\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right). \] Step 6: State the result.
This matches the required form.
\[ \boxed{\dfrac{1}{2}\left(-\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right)} \]