Question

Let the vectors \( \overrightarrow{AB} = 2\hat{i} + 2\hat{j} + \hat{k} \) and \( \overrightarrow{AC} = 2\hat{i} + 4\hat{j} + 4\hat{k} \) be two sides of a triangle ABC. If \( G \) is the centroid of \( \triangle ABC \), then \( \frac{22}{7} |\overrightarrow{AG}|^2 + 5 = \): (a) 25

• A. 25
• B. 38
• C. 47
• D. 52

Hint:

The centroid of a triangle is found using the formula \( \overrightarrow{G} = \frac{\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}}{3} \). Always check vector components carefully to ensure correct calculations.

Answer 1

The Correct Option is B

The centroid \( G \) of a triangle is given by the formula: \[ \overrightarrow{AG} = \frac{\overrightarrow{O} + \overrightarrow{B} + \overrightarrow{C}}{3} \] where \( A \) is taken as the origin, \( B \) has the position vector \( \overrightarrow{B} = 2\hat{i} + 2\hat{j} + \hat{k} \), and \( C \) has the position vector \( \overrightarrow{C} = 2\hat{i} + 4\hat{j} + 4\hat{k} \). Substituting these values: \[ \overrightarrow{AG} = \frac{0 + (2\hat{i} + 2\hat{j} + \hat{k}) + (2\hat{i} + 4\hat{j} + 4\hat{k})}{3} \] \[ = \frac{4\hat{i} + 6\hat{j} + 5\hat{k}}{3} \] Now, computing \( |\overrightarrow{AG}|^2 \): \[ |\overrightarrow{AG}|^2 = \left( \frac{4}{3} \right)^2 + \left( \frac{6}{3} \right)^2 + \left( \frac{5}{3} \right)^2 \] \[ = \frac{16}{9} + \frac{36}{9} + \frac{25}{9} = \frac{77}{9} \] Multiplying by \( \frac{22}{7} \): \[ \frac{22}{7} \times \frac{77}{9} = \frac{1694}{63} = 33. \] Adding 5: \[ 33 + 5 = 38. \] Thus, the final result is: Final Answer: \( \mathbf{38} \).