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:
The problem asks for the position vector of the intersection of the medians of a triangle. The intersection point of the medians of any triangle is historically and mathematically known as the centroid of that triangle.

Step 2: Key Formula or Approach:
If a triangle has vertices with position vectors $\bar{a}$, $\bar{b}$, and $\bar{c}$, or coordinates $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, and $(x_3, y_3, z_3)$, the position vector of its centroid $\bar{g}$ is the arithmetic mean of its vertices: $$\bar{g} = \frac{\bar{a} + \bar{b} + \bar{c}}{3}$$

Step 3: Detailed Explanation:
From the coordinates given: Vector $\bar{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ Vector $\bar{b} = \hat{i} + 0\hat{j} + 3\hat{k}$ Vector $\bar{c} = 4\hat{i} + \hat{j} - 3\hat{k}$ Summing the vectors component by component: $$\bar{a} + \bar{b} + \bar{c} = (1 + 1 + 4)\hat{i} + (2 + 0 + 1)\hat{j} + (3 + 3 - 3)\hat{k}$$ $$\bar{a} + \bar{b} + \bar{c} = 6\hat{i} + 3\hat{j} + 3\hat{k}$$ Now divide by $3$ to find the centroid vector $\bar{g}$: $$\bar{g} = \frac{6\hat{i} + 3\hat{j} + 3\hat{k}}{3} = 2\hat{i} + \hat{j} + \hat{k}$$

Step 4: Final Answer:
The position vector of the point of intersection of the medians is $2\hat{i} + \hat{j} + \hat{k}$, which perfectly matches option (B).