Let `A` and `B` be points with position vectors `veca` and `vecb` with respect to origin `O`. If the point `C` on `OA` is such that `2vec(AC)=vec(CO), vec(CD) ` is parallel to `vec(OB)` and `|vec(CD)|=3|vec(OB)|` then `vec(AD)` is (A) `vecb-veca/9` (B) `3vecb-veca/3` (C) `vecb-veca/3` (D) `vecb+veca/3`

Options

A. `3b-(a)/(2)`
B. `3b+(a)/(2)`
C. `3b-(a)/(3)`
D. `3b+(a)/(3)`

Explanation

To solve the problem step by step, we will analyze the given conditions and derive the required vector.

### Step 1: Understand the given vectors
Let the position vectors of points A and B be represented as:
- \( \vec{A} = \vec{a} \)
- \( \vec{B} = \vec{b} \)

The point C lies on the line segment OA.

### Step 2: Express the position vector of point C
We know that \( 2 \vec{AC} = \vec{CO} \). If we denote the position vector of point C as \( \vec{C} \), we can express this relationship as:
\[ \vec{AC} = \vec{C} - \vec{A} \]
\[ \vec{CO} = \vec{O} - \vec{C} = -\vec{C} \]
Thus, substituting these into the equation gives:
\[ 2(\vec{C} - \vec{A}) = -\vec{C} \]
This simplifies to:
\[ 2\vec{C} - 2\vec{A} = -\vec{C} \]
\[ 3\vec{C} = 2\vec{A} \]
\[ \vec{C} = \frac{2}{3} \vec{A} \]

### Step 3: Determine the position vector of point D
Since \( \vec{CD} \) is parallel to \( \vec{OB} \) and \( |\vec{CD}| = 3 |\vec{OB}| \), we can express \( \vec{CD} \) as:
\[ \vec{CD} = k \vec{b} \]
where \( k \) is a scalar. The magnitude condition gives:
\[ |\vec{CD}| = |k \vec{b}| = |k| |\vec{b}| = 3 |\vec{b}| \]
Thus, we can set \( |k| = 3 \), implying \( k = 3 \) (assuming the direction is the same).

### Step 4: Write the vector CD in terms of C and D
From the above, we have:
\[ \vec{CD} = \vec{D} - \vec{C} = 3\vec{b} \]
This leads to:
\[ \vec{D} = \vec{C} + 3\vec{b} \]

### Step 5: Substitute for C
Substituting \( \vec{C} = \frac{2}{3} \vec{A} \) into the equation for \( \vec{D} \):
\[ \vec{D} = \frac{2}{3} \vec{A} + 3\vec{b} \]

### Step 6: Find the vector AD
Now, we need to find \( \vec{AD} \):
\[ \vec{AD} = \vec{D} - \vec{A} \]
Substituting \( \vec{D} \):
\[ \vec{AD} = \left(\frac{2}{3} \vec{A} + 3\vec{b}\right) - \vec{A} \]
This simplifies to:
\[ \vec{AD} = \frac{2}{3} \vec{A} + 3\vec{b} - \frac{3}{3} \vec{A} \]
\[ \vec{AD} = \left(3\vec{b} - \frac{1}{3} \vec{A}\right) \]
\[ \vec{AD} = \frac{9\vec{b} - \vec{A}}{3} \]

### Final Result
Thus, we have:
\[ \vec{AD} = \frac{3\vec{b} - \vec{a}}{3} \]

### Conclusion
The answer corresponds to option (B):
\[ \vec{AD} = \frac{3\vec{b} - \vec{a}}{3} \]