Question

If \(\vec{r}\) is the position vector of the centroid \(G\) of triangle \(ABC\) and \[ 2\hat{i} + \hat{j} - \hat{k} \quad \text{and} \quad 2\hat{i} + 4\hat{j} - 4\hat{k} \] are respectively the position vectors of its vertices \(A\) and \(B\), then \[ AG^2 + BG^2 + CG^2 =\ ? \]

• A. \(77\)
• B. \(\mathbf{74}\)
• C. \(86\)
• D. \(83\)

Hint:

The sum of squares of distances from centroid to vertices equals: \[ AG^2 + BG^2 + CG^2 = \frac{1}{3}(AB^2 + BC^2 + CA^2) \] or can be computed directly using vector subtraction and magnitude squared.

Answer 1

The Correct Option is B

Let: \[ \vec{A} = 2\hat{i} + \hat{j} - \hat{k}, \quad \vec{B} = 2\hat{i} + 4\hat{j} - 4\hat{k}, \quad \vec{C} = \vec{c} \] Let centroid: \[ \vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3} \Rightarrow \vec{C} = 3\vec{G} - \vec{A} - \vec{B} \] We know: \[ AG^2 + BG^2 + CG^2 = \frac{1}{3}(AB^2 + BC^2 + CA^2) \quad \text{(Using the identity for sum of squares of distances from centroid)} \] But using vectors: \[ AG = \vec{A} - \vec{G}, \quad BG = \vec{B} - \vec{G}, \quad CG = \vec{C} - \vec{G} \] Then using identities or direct vector calculation (from solution steps), the result simplifies to: \[ AG^2 + BG^2 + CG^2 = 74 \]