In a `DeltaABC`, if 2AC=3CB, then 2OA+3OB is equal to

Options

A. 5OC
B. `-OC`
C. `OC`
D. none of these

Explanation

To solve the problem, we need to find the expression \(2OA + 3OB\) given that \(2AC = 3CB\) in triangle \(ABC\). Let's break this down step by step.

### Step 1: Understand the given relationship
We are given that \(2AC = 3CB\). This means that the vector \(AC\) is related to the vector \(CB\) in a specific ratio. We can express this relationship in terms of vectors:
\[ AC = \frac{3}{2} CB \]

### Step 2: Express the vectors \(OA\) and \(OB\)
Let \(O\) be the origin. We can express the vectors \(OA\) and \(OB\) in terms of the position vectors of points \(A\), \(B\), and \(C\):
- The vector \(OA\) can be expressed as:
\[ OA = OC + CA \]
- The vector \(OB\) can be expressed as:
\[ OB = OC + CB \]

### Step 3: Substitute \(OA\) and \(OB\) into the expression \(2OA + 3OB\)
Now we substitute the expressions for \(OA\) and \(OB\) into \(2OA + 3OB\):
\[ 2OA + 3OB = 2(OC + CA) + 3(OC + CB) \]
Expanding this gives:
\[ = 2OC + 2CA + 3OC + 3CB \]
Combining like terms, we have:
\[ = (2OC + 3OC) + 2CA + 3CB = 5OC + 2CA + 3CB \]

### Step 4: Use the relationship \(2AC = 3CB\)
From our earlier relationship, we know:
\[ 2AC = 3CB \implies 3CB = -2CA \]
Thus, we can replace \(3CB\) with \(-2CA\) in our expression:
\[ 2CA + 3CB = 2CA - 2CA = 0 \]

### Step 5: Finalize the expression
Substituting this back into our expression for \(2OA + 3OB\):
\[ 2OA + 3OB = 5OC + 0 = 5OC \]

### Conclusion
Thus, we find that:
\[ 2OA + 3OB = 5OC \]