Let p be the position vector of orthocentre and g is the position vector of the centroid of `DeltaABC`, where circumcentre is the origin. If `p=kg`, then the value of k is

Explanation

To solve the problem, we need to find the value of \( k \) such that the position vector of the orthocenter \( P \) is equal to \( k \) times the position vector of the centroid \( G \) of triangle \( ABC \), given that the circumcenter is at the origin.

### Step-by-Step Solution:

1. **Understanding the Position Vectors**:
- Let \( O \) be the position vector of the circumcenter, which is given as the origin \( O = 0 \).
- Let \( H \) be the position vector of the orthocenter, denoted as \( P \).
- Let \( G \) be the position vector of the centroid.

2. **Using the Relationship Between the Points**:
- In triangle geometry, the centroid \( G \), orthocenter \( H \), and circumcenter \( O \) are related by the following ratio:
\[ G = \frac{2H + O}{3} \]
- Since \( O = 0 \), we can simplify this to:
\[ G = \frac{2H}{3} \]

3. **Expressing \( H \) in Terms of \( G \)**:
- Rearranging the equation gives:
\[ 3G = 2H \]
- This can be rewritten as:
\[ H = \frac{3}{2} G \]

4. **Relating \( P \) and \( G \)**:
- Since \( P \) represents the position vector of the orthocenter \( H \), we can substitute:
\[ P = \frac{3}{2} G \]

5. **Finding the Value of \( k \)**:
- From the equation \( P = kG \), we can compare:
\[ kG = \frac{3}{2} G \]
- Therefore, we can conclude that:
\[ k = \frac{3}{2} \]

### Final Answer:
The value of \( k \) is \( \frac{3}{2} \).

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