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 check if a triangle is a right triangle, compute the dot product of two sides. If the result is zero, the angle between the sides is \( \frac{\pi}{2} \).

Answer 1

Step 1: Find the vectors \( \overrightarrow{AB} \), \( \overrightarrow{BC} \), and \( \overrightarrow{AC} \).
\[ \overrightarrow{AB} = \vec{B} - \vec{A} = (\hat{i} - 3\hat{j} - 5\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = -\hat{i} - 2\hat{j} - 6\hat{k}. \] \[ \overrightarrow{AC} = \vec{C} - \vec{A} = (3\hat{i} - 4\hat{j} - 4\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = \hat{i} - 3\hat{j} - 5\hat{k}. \] \[ \overrightarrow{BC} = \vec{C} - \vec{B} = (3\hat{i} - 4\hat{j} - 4\hat{k}) - (\hat{i} - 3\hat{j} - 5\hat{k}) = 2\hat{i} - \hat{j} + \hat{k}. \] Step 2: Find the magnitudes of the vectors.
- Magnitude of \( \overrightarrow{AB} \): \[ \|\overrightarrow{AB}\| = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41}. \] - Magnitude of \( \overrightarrow{AC} \): \[ \|\overrightarrow{AC}\| = \sqrt{(1)^2 + (-3)^2 + (-5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35}. \] - Magnitude of \( \overrightarrow{BC} \): \[ \|\overrightarrow{BC}\| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}. \] Step 3: Use the dot product to find \( \cos C \).
To find \( \angle C \), use the vectors \( \overrightarrow{AC} \) and \( \overrightarrow{BC} \): \[ \overrightarrow{AC} \cdot \overrightarrow{BC} = (1)(2) + (-3)(-1) + (-5)(1) = 2 + 3 - 5 = 0. \] Thus: \[ \cos C = \frac{\overrightarrow{AC} \cdot \overrightarrow{BC}}{\|\overrightarrow{AC}\| \|\overrightarrow{BC}\|} = \frac{0}{\sqrt{35} \cdot \sqrt{6}} = 0. \] Step 4: Determine \( \angle C \).
Since \( \cos C = 0 \): \[ C = \frac{\pi}{2}. \] Step 5: Conclusion.
The angle \( C \) of \( \Delta ABC \) is: \[ \boxed{C = \frac{\pi}{2}} \quad (\text{a right angle}). \]