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 \[ \overrightarrow{AD}+\overrightarrow{OC}+\overrightarrow{DC}= \]

• A. \(\vec{d}-\vec{a}\)
• B. \(\frac{1}{2}\vec{a}+3\vec{d}\)
• C. \(\frac{1}{2}\vec{a}+2\vec{d}\)
• D. \(6\vec{d}\)

Hint:

In vector geometry, first express every required point as a position vector from the origin. Then use \(\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}\).

Answer 1

The Correct Option is D

Step 1: Record the given position vectors.
The position vector of A is a⃗, and the position vector of D is d⃗.

Step 2: Exploit the parallel condition OD ∥ AB.
Since OD is parallel to AB and their length ratio is OD/AB = 1/3, we have AB = 3·OD, giving AB⃗ = 3d⃗. Consequently, OB⃗ = OA⃗ + AB⃗ = a⃗ + 3d⃗.

Step 3: Exploit the parallel condition OA ∥ CB.
Given OA ∥ CB with OA/CB = 2, we obtain CB = OA/2, so CB⃗ = (1/2)a⃗ and thus BC⃗ = -(1/2)a⃗. Then OC⃗ = OB⃗ + BC⃗ = (a⃗ + 3d⃗) - (1/2)a⃗ = (1/2)a⃗ + 3d⃗.

Step 4: Compute each vector in the required sum.
AD⃗ = OD⃗ - OA⃗ = d⃗ - a⃗. DC⃗ = OC⃗ - OD⃗ = [(1/2)a⃗ + 3d⃗] - d⃗ = (1/2)a⃗ + 2d⃗.

Step 5: Add the three vectors together.
AD⃗ + OC⃗ + DC⃗ = (d⃗ - a⃗) + [(1/2)a⃗ + 3d⃗] + [(1/2)a⃗ + 2d⃗] = d⃗ + 3d⃗ + 2d⃗ - a⃗ + (1/2)a⃗ + (1/2)a⃗ = 6d⃗.

Step 6: Final conclusion.
The sum simplifies to 6d⃗.