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

Options

A. equilateral
B. isosceles
C. right angled but not isosceles
D. right angled

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:
- \( A = 2\hat{i} - \hat{j} + \hat{k} \)
- \( B = \hat{i} - 3\hat{j} - 5\hat{k} \)
- \( C = 3\hat{i} - 4\hat{j} - 4\hat{k} \)

### Step 2: Calculate the lengths of the sides
We will use the distance formula for vectors to find the lengths of sides \( AB \), \( BC \), and \( AC \).

#### Length of \( AB \):
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Where:
- \( A(2, -1, 1) \)
- \( B(1, -3, -5) \)

Calculating:
\[ AB = \sqrt{(1 - 2)^2 + (-3 + 1)^2 + (-5 - 1)^2} \]
\[ = \sqrt{(-1)^2 + (-2)^2 + (-6)^2} \]
\[ = \sqrt{1 + 4 + 36} = \sqrt{41} \]

#### Length of \( BC \):
Using the same formula:
\[ BC = \sqrt{(3 - 1)^2 + (-4 + 3)^2 + (-4 + 5)^2} \]
\[ = \sqrt{(2)^2 + (-1)^2 + (1)^2} \]
\[ = \sqrt{4 + 1 + 1} = \sqrt{6} \]

#### Length of \( AC \):
Calculating:
\[ AC = \sqrt{(3 - 2)^2 + (-4 + 1)^2 + (-4 - 1)^2} \]
\[ = \sqrt{(1)^2 + (-3)^2 + (-5)^2} \]
\[ = \sqrt{1 + 9 + 25} = \sqrt{35} \]

### Step 3: Check the triangle type
Now we have:
- \( AB = \sqrt{41} \)
- \( BC = \sqrt{6} \)
- \( AC = \sqrt{35} \)

To determine if the triangle is a right triangle, we can check if:
\[ AB^2 = AC^2 + BC^2 \]
Calculating:
\[ AB^2 = 41, \quad AC^2 = 35, \quad BC^2 = 6 \]
\[ AC^2 + BC^2 = 35 + 6 = 41 \]
Since \( AB^2 = AC^2 + BC^2 \), the triangle is a right triangle.

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

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