The position vectors of the points A, B, C are `2 hati + hatj - hatk , 3 hati - 2 hatj + hatk and hati + 4hatj - 3 hatk ` respectively . These points

Options

A. form an isosceles triangle
B. form a right angled triangle
C. are collinear
D. form a scalene triangle

Explanation

To solve the problem, we need to find the relationship between the points A, B, and C given their position vectors. The position vectors are:

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

We will find the vectors \( \vec{AB} \), \( \vec{BC} \), and \( \vec{CA} \), and then check if the points are collinear.

### Step 1: Find the vector \( \vec{AB} \)

The vector \( \vec{AB} \) is given by:

\[ \vec{AB} = \vec{B} - \vec{A} \]

Substituting the position vectors:

\[ \vec{AB} = (3 \hat{i} - 2 \hat{j} + \hat{k}) - (2 \hat{i} + \hat{j} - \hat{k}) \]

Calculating the components:

\[ \vec{AB} = (3 - 2) \hat{i} + (-2 - 1) \hat{j} + (1 + 1) \hat{k} \]

\[ \vec{AB} = 1 \hat{i} - 3 \hat{j} + 2 \hat{k} \]

### Step 2: Find the vector \( \vec{BC} \)

The vector \( \vec{BC} \) is given by:

\[ \vec{BC} = \vec{C} - \vec{B} \]

Substituting the position vectors:

\[ \vec{BC} = (\hat{i} + 4 \hat{j} - 3 \hat{k}) - (3 \hat{i} - 2 \hat{j} + \hat{k}) \]

Calculating the components:

\[ \vec{BC} = (1 - 3) \hat{i} + (4 + 2) \hat{j} + (-3 - 1) \hat{k} \]

\[ \vec{BC} = -2 \hat{i} + 6 \hat{j} - 4 \hat{k} \]

### Step 3: Find the vector \( \vec{CA} \)

The vector \( \vec{CA} \) is given by:

\[ \vec{CA} = \vec{A} - \vec{C} \]

Substituting the position vectors:

\[ \vec{CA} = (2 \hat{i} + \hat{j} - \hat{k}) - (\hat{i} + 4 \hat{j} - 3 \hat{k}) \]

Calculating the components:

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

\[ \vec{CA} = 1 \hat{i} - 3 \hat{j} + 2 \hat{k} \]

### Step 4: Check the magnitudes

Now we calculate the magnitudes of the vectors:

1. Magnitude of \( \vec{AB} \):

\[ |\vec{AB}| = \sqrt{(1)^2 + (-3)^2 + (2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \]

2. Magnitude of \( \vec{BC} \):

\[ |\vec{BC}| = \sqrt{(-2)^2 + (6)^2 + (-4)^2} = \sqrt{4 + 36 + 16} = \sqrt{56} = 2\sqrt{14} \]

3. Magnitude of \( \vec{CA} \):

\[ |\vec{CA}| = \sqrt{(1)^2 + (-3)^2 + (2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \]

### Step 5: Check the relationship

We observe that:

\[ |\vec{AB}| + |\vec{CA}| = \sqrt{14} + \sqrt{14} = 2\sqrt{14} = |\vec{BC}| \]

This shows that:

\[ \vec{AB} + \vec{CA} = \vec{BC} \]

### Conclusion

Since \( \vec{AB} + \vec{CA} = \vec{BC} \), and the vectors are in the same line, we conclude that points A, B, and C are collinear.