If the position vectors of the vertices of a triangle be `2hati+4hatj-hatk,4hati+5hatj+hatk and 3 hati+6hatj-3hatk`, then the triangle is

Options

A. right angled
B. isosceles
C. equilateral
D. none of these

Explanation

To determine the type of triangle formed by the given position vectors of its vertices, we will follow these steps:

### Given Position Vectors:
- Vertex A: \( \vec{A} = 2\hat{i} + 4\hat{j} - \hat{k} \)
- Vertex B: \( \vec{B} = 4\hat{i} + 5\hat{j} + \hat{k} \)
- Vertex C: \( \vec{C} = 3\hat{i} + 6\hat{j} - 3\hat{k} \)

### Step 1: Calculate the vectors AB, BC, and CA

1. **Vector AB**:
\[ \vec{AB} = \vec{B} - \vec{A} = (4\hat{i} + 5\hat{j} + \hat{k}) - (2\hat{i} + 4\hat{j} - \hat{k}) \]
\[ = (4 - 2)\hat{i} + (5 - 4)\hat{j} + (1 + 1)\hat{k} = 2\hat{i} + 1\hat{j} + 2\hat{k} \]

2. **Vector BC**:
\[ \vec{BC} = \vec{C} - \vec{B} = (3\hat{i} + 6\hat{j} - 3\hat{k}) - (4\hat{i} + 5\hat{j} + \hat{k}) \]
\[ = (3 - 4)\hat{i} + (6 - 5)\hat{j} + (-3 - 1)\hat{k} = -1\hat{i} + 1\hat{j} - 4\hat{k} \]

3. **Vector CA**:
\[ \vec{CA} = \vec{A} - \vec{C} = (2\hat{i} + 4\hat{j} - \hat{k}) - (3\hat{i} + 6\hat{j} - 3\hat{k}) \]
\[ = (2 - 3)\hat{i} + (4 - 6)\hat{j} + (-1 + 3)\hat{k} = -1\hat{i} - 2\hat{j} + 2\hat{k} \]

### Step 2: Calculate the magnitudes of the vectors

1. **Magnitude of AB**:
\[ |\vec{AB}| = \sqrt{(2)^2 + (1)^2 + (2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

2. **Magnitude of BC**:
\[ |\vec{BC}| = \sqrt{(-1)^2 + (1)^2 + (-4)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2} \]

3. **Magnitude of CA**:
\[ |\vec{CA}| = \sqrt{(-1)^2 + (-2)^2 + (2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

### Step 3: Determine the type of triangle

Now we have:
- \( |\vec{AB}| = 3 \)
- \( |\vec{BC}| = 3\sqrt{2} \)
- \( |\vec{CA}| = 3 \)

Since \( |\vec{AB}| = |\vec{CA}| \), triangle ABC is an **isosceles triangle**.

### Step 4: Check if it is a right triangle

To check if triangle ABC is also a right triangle, we can apply the Pythagorean theorem:
\[ |\vec{BC}|^2 = |\vec{AB}|^2 + |\vec{CA}|^2 \]
Calculating:
\[ (3\sqrt{2})^2 = 3^2 + 3^2 \]
\[ 18 = 9 + 9 \]
\[ 18 = 18 \]

Since the equation holds true, triangle ABC is also a **right triangle**.

### Conclusion
The triangle formed by the given position vectors is both an **isosceles triangle** and a **right triangle**.

---