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

We are given the sides of the triangle \( ABC \) 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: Centroid Calculation
The centroid \( G \) of a triangle is given by the average of the position vectors of the three vertices: \[ \vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3} \] Since \( AB = \vec{B} - \vec{A} \) and \( AC = \vec{A} - \vec{C} \), we can solve for the position vectors of \( \vec{A}, \vec{B}, \vec{C} \) and then calculate \( \vec{G} \).
Step 2: Compute \( AG \), \( BG \), and \( CG \)
From the centroid formula: \[ \vec{G} = \frac{(2, -1, 1) + (2, 1, 3) + (-1, 3, 5)}{3} = \left( \frac{3}{3}, \frac{3}{3}, \frac{9}{3} \right) = (1, 1, 3) \]
Thus, \( G = (1, 1, 3) \). Now, we find the squared distances from \( G \) to each point: - \( AG \): The distance from \( A = (2, -1, 1) \) to \( G = (1, 1, 3) \) is: \[ AG^2 = \left( \frac{1}{3} \right)^2 + \left( \frac{2}{3} \right)^2 + 6^2 = 41 \] - \( BG \): The distance from \( B = (2, 1, 3) \) to \( G = (1, 1, 3) \) is: \[ BG^2 = \left( \frac{1}{3} \right)^2 + \left( \frac{2}{3} \right)^2 + 2 \cdot 1^2 = 59 \] - \( CG \): The distance from \( C = (-1, 3, 5) \) to \( G = (1, 1, 3) \) is: \[ CG^2 = \left( \frac{2}{3} \right)^2 + \left( \frac{2}{3} \right)^2 + (3 - 5)^2 = 146 \]
Step 3: Calculate the Final Expression
Now, we calculate: \[ 6 \left( |\vec{AG}|^2 + |\vec{BG}|^2 + |\vec{CG}|^2 \right) = 6 \times \left[ 41 + 59 + 146 \right] = 6 \times 246 = 164 \]
Thus, the final value is \( 164 \).

Answer 2

Step 1: Given data: 

\[ \overrightarrow{AB} + \overrightarrow{AC} = \overrightarrow{CB} \]

Step 2: Let the position vectors of \( \overrightarrow{A} \) are \( \overrightarrow{A} \), then:

\[ \overrightarrow{AB} = \overrightarrow{B} - \overrightarrow{A} \] i.e., position vector of \( \overrightarrow{B} = 2i - j + k \)

Step 3: Now for \( \overrightarrow{C} \), position vector is:

\[ \overrightarrow{CA} = \overrightarrow{C} - \overrightarrow{A} \] i.e., position vector of \( \overrightarrow{C} = -i - 3j - 5k \)

Step 4: Now for centroid, the position vector \( \overrightarrow{G} \) is:

\[ \overrightarrow{G} = \frac{\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}}{3} \] Substitute the values of \( \overrightarrow{A}, \overrightarrow{B}, \overrightarrow{C} \): \[ \overrightarrow{G} = \frac{1}{3} (2i + 2j + 6k) \]

Step 5: Now calculate the distance from \( \overrightarrow{G} \) to \( \overrightarrow{A} \):

\[ \overrightarrow{AG} = \frac{1}{3} (i + 2j + 6k) \]

Step 6: Now calculate \( |AG|^2 \):

\[ |AG|^2 = \frac{1}{9} \times 41 \]

Step 7: Distance between \( B \) and \( G \) is:

\[ |BG|^2 = \frac{59}{9} \]

Step 8: Distance between \( C \) and \( G \) is:

\[ |CG|^2 = \frac{146}{9} \]

Step 9: Final calculation of the sum of squares of the distances:

\[ 6 \left[ |AG|^2 + |BG|^2 + |CG|^2 \right] = 6 \times \left( \frac{41}{9} + \frac{59}{9} + \frac{146}{9} \right) \]

Step 10: Final result:

\[ 6 \times \frac{246}{9} = 164 \]