Question

The position vector of the point of intersection of the medians of a triangle, whose vertices are $A(1, 2, 3)$, $B(1, 0, 3)$ and $C(4, 1, -3)$ is

• A. $6\hat{i} + 3\hat{j} + 3\hat{k}$
• B. $2\hat{i} + \hat{j} + \hat{k}$
• C. $3\hat{i} + \hat{j} + \hat{k}$
• D. $\hat{i} + \hat{j} + \hat{k}$

Hint:

To solve this in your head, just average the coordinates individually: $$x = \frac{1+1+4}{3} = 2, \quad y = \frac{2+0+1}{3} = 1, \quad z = \frac{3+3-3}{3} = 1$$ This gives the coordinate position $(2, 1, 1)$, which translates to $2\hat{i} + \hat{j} + \hat{k}$ directly.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Find the position vector of the intersection of medians (centroid) of a triangle with given vertex position vectors.

Step 2: Key Formula or Approach:
The centroid's position vector is the arithmetic mean of the three vertex vectors: ḡ = (ā + b̄ + c̄)/3.

Step 3: Detailed Explanation:
Summing: (î + 2ĵ + 3k̂) + (î + 0ĵ + 3k̂) + (4î + ĵ - 3k̂) = (1+1+4)î + (2+0+1)ĵ + (3+3-3)k̂ = 6î + 3ĵ + 3k̂. Dividing by 3: ḡ = 2î + ĵ + k̂.

Step 4: Final Answer:
The centroid position vector is 2î + ĵ + k̂, option (B).