If the position vectors of the vertices of a triangle be `2hat(i)+4hat(j)-hat(k), 4hat(i)+5hat(j)+hat(k) and 3hat(i)+6hat(j)-3hat(k)`, then the triangle is

Options

A. right angled
B. isosceles
C. equilateral
D. right angled isosceles

Answer

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

### Step 1: Identify the Position Vectors
Let the vertices of the triangle be:
- A: \( \vec{A} = 2\hat{i} + 4\hat{j} - \hat{k} \)
- B: \( \vec{B} = 4\hat{i} + 5\hat{j} + \hat{k} \)
- C: \( \vec{C} = 3\hat{i} + 6\hat{j} - 3\hat{k} \)

### Step 2: Calculate the Vectors for the Sides of the Triangle
We need to find the vectors representing the sides of the triangle:
- \( \vec{AB} = \vec{B} - \vec{A} \)
- \( \vec{BC} = \vec{C} - \vec{B} \)
- \( \vec{CA} = \vec{A} - \vec{C} \)

Calculating each:
1. **For \( \vec{AB} \)**:
\[ \vec{AB} = (4\hat{i} + 5\hat{j} + \hat{k}) - (2\hat{i} + 4\hat{j} - \hat{k}) = (4 - 2)\hat{i} + (5 - 4)\hat{j} + (1 + 1)\hat{k} = 2\hat{i} + 1\hat{j} + 2\hat{k} \]

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

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

### Step 3: Calculate the Lengths of the Sides
Now, we will calculate the lengths of these vectors:
1. **Length of \( AB \)**:
\[ |\vec{AB}| = \sqrt{(2)^2 + (1)^2 + (2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

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

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

### Step 4: Analyze the Lengths
Now we have:
- \( AB = 3 \)
- \( BC = 3\sqrt{2} \)
- \( CA = 3 \)

### Step 5: Determine the Type of Triangle
From the lengths:
- \( AB = CA \) (both are equal)
- \( BC = 3\sqrt{2} \)

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

Since \( AB^2 + CA^2 = BC^2 \), it confirms that triangle ABC is a right triangle.

### Conclusion
The triangle formed by the given position vectors is a right-angled isosceles triangle.

---