Question:
\(OABCD\) is a pentagon in which the sides \(OA\) and \(CB\) are parallel and the sides \(OD\) and \(AB\) are parallel. Also, it is given that
\[
\frac{OA}{CB}=2,\qquad \frac{OD}{AB}=\frac{1}{3}.
\]
If \(\overrightarrow{OA}=\vec{a}\), \(\overrightarrow{OD}=\vec{d}\), then
\[
\overrightarrow{AD}+\overrightarrow{OC}+\overrightarrow{DC}=
\]
\(OABCD\) is a pentagon in which the sides \(OA\) and \(CB\) are parallel and the sides \(OD\) and \(AB\) are parallel. Also, it is given that
\[
\frac{OA}{CB}=2,\qquad \frac{OD}{AB}=\frac{1}{3}.
\]
If \(\overrightarrow{OA}=\vec{a}\), \(\overrightarrow{OD}=\vec{d}\), then
\[
\overrightarrow{AD}+\overrightarrow{OC}+\overrightarrow{DC}=
\]
Show Hint
In vector geometry, first express every required point as a position vector from the origin. Then use \(\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}\).
Updated On: Jun 18, 2026
- \(\vec{d}-\vec{a}\)
- \(\frac{1}{2}\vec{a}+3\vec{d}\)
- \(\frac{1}{2}\vec{a}+2\vec{d}\)
- \(6\vec{d}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Write the given position vectors.
Given, \[ \overrightarrow{OA}=\vec{a} \] and \[ \overrightarrow{OD}=\vec{d}. \] So the position vector of \(A\) is \(\vec{a}\), and the position vector of \(D\) is \(\vec{d}\).
Step 2: Use \(OD\parallel AB\).
Since \[ OD\parallel AB \] and \[ \frac{OD}{AB}=\frac{1}{3}, \] we get \[ AB=3OD. \] Therefore, \[ \overrightarrow{AB}=3\vec{d}. \] Hence, \[ \overrightarrow{OB} = \overrightarrow{OA}+\overrightarrow{AB} = \vec{a}+3\vec{d}. \]
Step 3: Use \(OA\parallel CB\).
Since \[ OA\parallel CB \] and \[ \frac{OA}{CB}=2, \] we get \[ CB=\frac{1}{2}OA. \] Thus, \[ \overrightarrow{CB}=\frac{1}{2}\vec{a}. \] So, \[ \overrightarrow{BC}=-\frac{1}{2}\vec{a}. \] Therefore, \[ \overrightarrow{OC} = \overrightarrow{OB}+\overrightarrow{BC}. \] \[ \overrightarrow{OC} = \vec{a}+3\vec{d}-\frac{1}{2}\vec{a}. \] \[ \overrightarrow{OC} = \frac{1}{2}\vec{a}+3\vec{d}. \]
Step 4: Find the required vector sum.
Now, \[ \overrightarrow{AD} = \overrightarrow{OD}-\overrightarrow{OA} = \vec{d}-\vec{a}. \] Also, \[ \overrightarrow{DC} = \overrightarrow{OC}-\overrightarrow{OD} = \left(\frac{1}{2}\vec{a}+3\vec{d}\right)-\vec{d}. \] \[ \overrightarrow{DC} = \frac{1}{2}\vec{a}+2\vec{d}. \] Therefore, \[ \overrightarrow{AD}+\overrightarrow{OC}+\overrightarrow{DC} = (\vec{d}-\vec{a}) + \left(\frac{1}{2}\vec{a}+3\vec{d}\right) + \left(\frac{1}{2}\vec{a}+2\vec{d}\right). \] \[ = \vec{d}+3\vec{d}+2\vec{d} -\vec{a}+\frac{1}{2}\vec{a}+\frac{1}{2}\vec{a}. \] \[ = 6\vec{d}. \]
Step 5: Final conclusion.
Hence, \[ \boxed{6\vec{d}} \]
Given, \[ \overrightarrow{OA}=\vec{a} \] and \[ \overrightarrow{OD}=\vec{d}. \] So the position vector of \(A\) is \(\vec{a}\), and the position vector of \(D\) is \(\vec{d}\).
Step 2: Use \(OD\parallel AB\).
Since \[ OD\parallel AB \] and \[ \frac{OD}{AB}=\frac{1}{3}, \] we get \[ AB=3OD. \] Therefore, \[ \overrightarrow{AB}=3\vec{d}. \] Hence, \[ \overrightarrow{OB} = \overrightarrow{OA}+\overrightarrow{AB} = \vec{a}+3\vec{d}. \]
Step 3: Use \(OA\parallel CB\).
Since \[ OA\parallel CB \] and \[ \frac{OA}{CB}=2, \] we get \[ CB=\frac{1}{2}OA. \] Thus, \[ \overrightarrow{CB}=\frac{1}{2}\vec{a}. \] So, \[ \overrightarrow{BC}=-\frac{1}{2}\vec{a}. \] Therefore, \[ \overrightarrow{OC} = \overrightarrow{OB}+\overrightarrow{BC}. \] \[ \overrightarrow{OC} = \vec{a}+3\vec{d}-\frac{1}{2}\vec{a}. \] \[ \overrightarrow{OC} = \frac{1}{2}\vec{a}+3\vec{d}. \]
Step 4: Find the required vector sum.
Now, \[ \overrightarrow{AD} = \overrightarrow{OD}-\overrightarrow{OA} = \vec{d}-\vec{a}. \] Also, \[ \overrightarrow{DC} = \overrightarrow{OC}-\overrightarrow{OD} = \left(\frac{1}{2}\vec{a}+3\vec{d}\right)-\vec{d}. \] \[ \overrightarrow{DC} = \frac{1}{2}\vec{a}+2\vec{d}. \] Therefore, \[ \overrightarrow{AD}+\overrightarrow{OC}+\overrightarrow{DC} = (\vec{d}-\vec{a}) + \left(\frac{1}{2}\vec{a}+3\vec{d}\right) + \left(\frac{1}{2}\vec{a}+2\vec{d}\right). \] \[ = \vec{d}+3\vec{d}+2\vec{d} -\vec{a}+\frac{1}{2}\vec{a}+\frac{1}{2}\vec{a}. \] \[ = 6\vec{d}. \]
Step 5: Final conclusion.
Hence, \[ \boxed{6\vec{d}} \]
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