Question

The position vectors of vertices of \( \Delta ABC \) are \( A(2\hat{i} - \hat{j} + \hat{k}) \), \( B(\hat{i} - 3\hat{j} - 5\hat{k}) \), and \( C(3\hat{i} - 4\hat{j} - 4\hat{k}) \). Find all the angles of \( \Delta ABC \).

Hint:

To determine if a triangle is a right triangle, check the dot products of all pairs of side vectors. If one dot product is zero, the corresponding angle is \( \frac{\pi}{2} \).

Answer 1

Step 1: Compute vectors \( \overrightarrow{AB} \), \( \overrightarrow{AC} \), and \( \overrightarrow{BC} \).
Given: \[ \vec{A} = 2\hat{i} - \hat{j} + \hat{k}, \quad \vec{B} = \hat{i} - 3\hat{j} - 5\hat{k}, \quad \vec{C} = 3\hat{i} - 4\hat{j} - 4\hat{k}. \] - Vector \( \overrightarrow{AB} \): \[ \overrightarrow{AB} = \vec{B} - \vec{A} = (\hat{i} - 3\hat{j} - 5\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = -\hat{i} - 2\hat{j} - 6\hat{k}. \] - Vector \( \overrightarrow{AC} \): \[ \overrightarrow{AC} = \vec{C} - \vec{A} = (3\hat{i} - 4\hat{j} - 4\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = \hat{i} - 3\hat{j} - 5\hat{k}. \] - Vector \( \overrightarrow{BC} \): \[ \overrightarrow{BC} = \vec{C} - \vec{B} = (3\hat{i} - 4\hat{j} - 4\hat{k}) - (\hat{i} - 3\hat{j} - 5\hat{k}) = 2\hat{i} - \hat{j} + \hat{k}. \] Step 2: Use the dot product to check orthogonality.
To determine if \( \Delta ABC \) has a right angle, compute the dot products between pairs of vectors. - Compute \( \overrightarrow{AB} \cdot \overrightarrow{AC} \): \[ \overrightarrow{AB} \cdot \overrightarrow{AC} = (-1)(1) + (-2)(-3) + (-6)(-5) = -1 + 6 + 30 = 35 \quad (\text{not orthogonal}). \] - Compute \( \overrightarrow{AC} \cdot \overrightarrow{BC} \): \[ \overrightarrow{AC} \cdot \overrightarrow{BC} = (1)(2) + (-3)(-1) + (-5)(1) = 2 + 3 - 5 = 0 \quad (\text{orthogonal}). \] - Compute \( \overrightarrow{AB} \cdot \overrightarrow{BC} \): \[ \overrightarrow{AB} \cdot \overrightarrow{BC} = (-1)(2) + (-2)(-1) + (-6)(1) = -2 + 2 - 6 = -6 \quad (\text{not orthogonal}). \] Since \( \overrightarrow{AC} \cdot \overrightarrow{BC} = 0 \), \( \angle C = \frac{\pi}{2} \), and \( \Delta ABC \) is a right triangle. Step 3: Conclusion.
The triangle \( \Delta ABC \) has a right angle at \( C \), and the angle is: \[ \boxed{C = \frac{\pi}{2}}. \]