If `veca, vecb` and `vecc` are position vectors of A,B, and C respectively of `DeltaABC` and `if|veca-vecb|=4,|vecb-vec(c)|=2, |vecc-veca|=3`, then the distance between the centroid and incenter of `triangleABC` is

Options

A. 1
B. `1/2`
C. `1/3`
D. `2/3`

Explanation

To find the distance between the centroid and the incenter of triangle ABC given the position vectors of points A, B, and C, we can follow these steps:

### Step 1: Define the position vectors
Let:
- \(\vec{a}\) be the position vector of point A
- \(\vec{b}\) be the position vector of point B
- \(\vec{c}\) be the position vector of point C

### Step 2: Use the given distances
We are given:
- \(|\vec{a} - \vec{b}| = 4\)
- \(|\vec{b} - \vec{c}| = 2\)
- \(|\vec{c} - \vec{a}| = 3\)

### Step 3: Find the centroid (G)
The centroid \(G\) of triangle ABC is given by the formula:
\[ \vec{G} = \frac{\vec{a} + \vec{b} + \vec{c}}{3} \]

### Step 4: Find the incenter (I)
The incenter \(I\) can be calculated using the formula:
\[ \vec{I} = \frac{a \vec{a} + b \vec{b} + c \vec{c}}{a + b + c} \]
where \(a\), \(b\), and \(c\) are the lengths of the sides opposite to vertices A, B, and C respectively.

### Step 5: Calculate the lengths of the sides
From the given distances, we can assign:
- Let \(a = |\vec{b} - \vec{c}| = 2\)
- Let \(b = |\vec{c} - \vec{a}| = 3\)
- Let \(c = |\vec{a} - \vec{b}| = 4\)

### Step 6: Substitute the values into the incenter formula
Now substituting the values into the incenter formula:
\[ \vec{I} = \frac{2\vec{a} + 3\vec{b} + 4\vec{c}}{2 + 3 + 4} = \frac{2\vec{a} + 3\vec{b} + 4\vec{c}}{9} \]

### Step 7: Calculate the distance \(|\vec{I} - \vec{G}|\)
Now we need to find the distance between the centroid and incenter:
\[ |\vec{I} - \vec{G}| = \left| \frac{2\vec{a} + 3\vec{b} + 4\vec{c}}{9} - \frac{\vec{a} + \vec{b} + \vec{c}}{3} \right| \]

### Step 8: Simplify the expression
To simplify, we can express \(\frac{\vec{a} + \vec{b} + \vec{c}}{3}\) in terms of a common denominator:
\[ |\vec{I} - \vec{G}| = \left| \frac{2\vec{a} + 3\vec{b} + 4\vec{c}}{9} - \frac{3(\vec{a} + \vec{b} + \vec{c})}{9} \right| \]
\[ = \left| \frac{(2\vec{a} + 3\vec{b} + 4\vec{c}) - (3\vec{a} + 3\vec{b} + 3\vec{c})}{9} \right| \]
\[ = \left| \frac{-\vec{a} + 0\vec{b} + \vec{c}}{9} \right| \]
\[ = \frac{|\vec{c} - \vec{a}|}{9} \]

### Step 9: Substitute the known distance
From the problem, we know \(|\vec{c} - \vec{a}| = 3\):
\[ |\vec{I} - \vec{G}| = \frac{3}{9} = \frac{1}{3} \]

### Conclusion
Thus, the distance between the centroid and the incenter of triangle ABC is:
\[ \frac{1}{3} \]