Question

Let the three sides of a triangle ABC be given by the vectors $$ 2\hat{i} - \hat{j} + \hat{k}, \quad \hat{i} - 3\hat{j} - 5\hat{k}, \quad \text{and} \quad 3\hat{i} - 4\hat{j} - 4\hat{k}. $$ Let G be the centroid of the triangle ABC. Then $$ 6 \left( |\vec{AG}|^2 + |\vec{BG}|^2 + |\vec{CG}|^2 \right) \text{ is equal to}$$ _______

Hint:

To calculate the sum of squared distances in a triangle, use the centroid formula and the properties of vectors.

Answer 1

Correct Answer: 164

The sides of triangle \( ABC \) are provided as vectors:\[AB = 2\hat{i} - \hat{j} + \hat{k}, \quad AC = \hat{i} - 3\hat{j} - 5\hat{k}, \quad BC = 3\hat{i} - 4\hat{j} - 4\hat{k}\]
Step 1: Determine Vertex Position Vectors
The centroid \( G \) of a triangle is the average of its vertex position vectors:\[\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3}\]Given \( AB = \vec{B} - \vec{A} \) and \( AC = \vec{C} - \vec{A} \), we derive the position vectors of vertices \( A, B, \) and \( C \) and subsequently compute \( \vec{G} \).
Step 2: Calculate Squared Distances from Centroid
The centroid \( G \) is calculated as:\[\vec{G} = \frac{(2, -1, 1) + (2, 1, 3) + (-1, 3, 5)}{3} = \left( \frac{3}{3}, \frac{3}{3}, \frac{9}{3} \right) = (1, 1, 3)\]Thus, \( G = (1, 1, 3) \). The squared distances from \( G \) to each vertex are:- \( AG^2 \): Distance from \( A = (2, -1, 1) \) to \( G = (1, 1, 3) \):\[AG^2 = \left( \frac{1}{3} \right)^2 + \left( \frac{2}{3} \right)^2 + 6^2 = 41\]- \( BG^2 \): Distance from \( B = (2, 1, 3) \) to \( G = (1, 1, 3) \):\[BG^2 = \left( \frac{1}{3} \right)^2 + \left( \frac{2}{3} \right)^2 + 2 \cdot 1^2 = 59\]- \( CG^2 \): Distance from \( C = (-1, 3, 5) \) to \( G = (1, 1, 3) \):\[CG^2 = \left( \frac{2}{3} \right)^2 + \left( \frac{2}{3} \right)^2 + (3 - 5)^2 = 146\]
Step 3: Compute the Final Value
The final expression is evaluated as:\[6 \left( |\vec{AG}|^2 + |\vec{BG}|^2 + |\vec{CG}|^2 \right) = 6 \times \left[ 41 + 59 + 146 \right] = 6 \times 246 = 164\]
The resulting value is \( 164 \).