Prove that the vectgors `2hati-hatj+hatk, hati-3hatj-5hatk and 3hati-4hatj-4hatk` form a righat angled triangle.

Options

A. Equilateral triangle
B. Isosceles triangle
C. Right triangle
D. Right Isosceles triangle

Answer

To prove that the vectors \( \mathbf{A} = 2\hat{i} - \hat{j} + \hat{k} \), \( \mathbf{B} = \hat{i} - 3\hat{j} - 5\hat{k} \), and \( \mathbf{C} = 3\hat{i} - 4\hat{j} - 4\hat{k} \) form a right-angled triangle, we will follow these steps:

### Step 1: Find the position vectors
Let:
- \( \mathbf{A} = 2\hat{i} - \hat{j} + \hat{k} \)
- \( \mathbf{B} = \hat{i} - 3\hat{j} - 5\hat{k} \)
- \( \mathbf{C} = 3\hat{i} - 4\hat{j} - 4\hat{k} \)

### Step 2: Calculate the vectors representing the sides of the triangle
1. **Vector \( \mathbf{AB} \)**:
\[ \mathbf{AB} = \mathbf{B} - \mathbf{A} = (\hat{i} - 3\hat{j} - 5\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) \]
\[ = (\hat{i} - 2\hat{i}) + (-3\hat{j} + \hat{j}) + (-5\hat{k} - \hat{k}) \]
\[ = -\hat{i} - 2\hat{j} - 6\hat{k} \]

2. **Vector \( \mathbf{BC} \)**:
\[ \mathbf{BC} = \mathbf{C} - \mathbf{B} = (3\hat{i} - 4\hat{j} - 4\hat{k}) - (\hat{i} - 3\hat{j} - 5\hat{k}) \]
\[ = (3\hat{i} - \hat{i}) + (-4\hat{j} + 3\hat{j}) + (-4\hat{k} + 5\hat{k}) \]
\[ = 2\hat{i} - \hat{j} + \hat{k} \]

3. **Vector \( \mathbf{CA} \)**:
\[ \mathbf{CA} = \mathbf{A} - \mathbf{C} = (2\hat{i} - \hat{j} + \hat{k}) - (3\hat{i} - 4\hat{j} - 4\hat{k}) \]
\[ = (2\hat{i} - 3\hat{i}) + (-\hat{j} + 4\hat{j}) + (\hat{k} + 4\hat{k}) \]
\[ = -\hat{i} + 3\hat{j} + 5\hat{k} \]

### Step 3: Calculate the magnitudes of the vectors
1. **Magnitude of \( \mathbf{AB} \)**:
\[ |\mathbf{AB}| = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} = \sqrt{1 + 4 + 36} = \sqrt{41} \]

2. **Magnitude of \( \mathbf{BC} \)**:
\[ |\mathbf{BC}| = \sqrt{(2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \]

3. **Magnitude of \( \mathbf{CA} \)**:
\[ |\mathbf{CA}| = \sqrt{(-1)^2 + (3)^2 + (5)^2} = \sqrt{1 + 9 + 25} = \sqrt{35} \]

### Step 4: Verify the Pythagorean theorem
To check if the triangle is right-angled, we verify if:
\[ |\mathbf{AB}|^2 = |\mathbf{BC}|^2 + |\mathbf{CA}|^2 \]
Calculating:
\[ |\mathbf{AB}|^2 = 41, \quad |\mathbf{BC}|^2 = 6, \quad |\mathbf{CA}|^2 = 35 \]
\[ |\mathbf{BC}|^2 + |\mathbf{CA}|^2 = 6 + 35 = 41 \]
Since \( |\mathbf{AB}|^2 = |\mathbf{BC}|^2 + |\mathbf{CA}|^2 \), the triangle is right-angled at point \( B \).

### Conclusion
Thus, the vectors \( \mathbf{A} \), \( \mathbf{B} \), and \( \mathbf{C} \) form a right-angled triangle.

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