Let O be an interior point of triangle ABC, such that `2vec(OA)+3vec(OB)+4vec(OC)=0`, then the ratio of the area of `DeltaABC` to the area of `DeltaAOC` is

Options

A. `3:1`
B. `3:2`
C. `2:1`
D. `4:3`

Answer

To solve the problem, we need to find the ratio of the area of triangle ABC to the area of triangle AOC given the equation \(2\vec{OA} + 3\vec{OB} + 4\vec{OC} = 0\).

### Step-by-Step Solution:

1. **Understanding the given condition**:
We have the equation \(2\vec{OA} + 3\vec{OB} + 4\vec{OC} = 0\). This can be rewritten in terms of the position vectors of points A, B, C, and O.
\[ 2\vec{OA} + 3\vec{OB} + 4\vec{OC} = 0 \implies 2(\vec{O} - \vec{A}) + 3(\vec{O} - \vec{B}) + 4(\vec{O} - \vec{C}) = 0 \]
Rearranging gives:
\[ (2 + 3 + 4)\vec{O} - (2\vec{A} + 3\vec{B} + 4\vec{C}) = 0 \implies 9\vec{O} = 2\vec{A} + 3\vec{B} + 4\vec{C} \]
Thus, we can express \(\vec{O}\) as:
\[ \vec{O} = \frac{2\vec{A} + 3\vec{B} + 4\vec{C}}{9} \]

2. **Finding the area of triangle AOC**:
The area of triangle AOC can be calculated using the formula for the area of a triangle formed by vectors. The area of triangle AOC is given by:
\[ \text{Area of } \Delta AOC = \frac{1}{2} |\vec{OA} \times \vec{OC}| \]

3. **Finding the area of triangle ABC**:
Similarly, the area of triangle ABC can be calculated as:
\[ \text{Area of } \Delta ABC = \frac{1}{2} |\vec{AB} \times \vec{AC}| \]

4. **Using the ratios of the areas**:
The ratio of the areas of triangles ABC and AOC can be derived from the coefficients in the expression for \(\vec{O}\). The coefficients of the vectors give us a way to relate the areas:
\[ \text{Area of } \Delta ABC = \text{Area of } \Delta AOB + \text{Area of } \Delta AOC + \text{Area of } \Delta BOC \]
From the coefficients, we can see that:
\[ \text{Area of } \Delta AOB : \text{Area of } \Delta AOC : \text{Area of } \Delta BOC = 2 : 4 : 3 \]
Thus, the area of triangle ABC can be expressed in terms of the area of triangle AOC:
\[ \text{Area of } \Delta ABC = \text{Area of } \Delta AOC + \text{Area of } \Delta AOB + \text{Area of } \Delta BOC = 4k + 2k + 3k = 9k \]
where \(k\) is the area of triangle AOC.

5. **Calculating the ratio**:
Now we can find the ratio:
\[ \frac{\text{Area of } \Delta ABC}{\text{Area of } \Delta AOC} = \frac{9k}{4k} = \frac{9}{4} \]

### Final Answer:
The ratio of the area of triangle ABC to the area of triangle AOC is \( \frac{9}{4} \).