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 triangle
C. are collinear
D. form a scalene triangle

Answer

To determine the relationship between the points A, B, and C given their position vectors, we will follow these steps:

1. **Identify the position vectors**:
- Let the position vector of point A be \( \vec{A} = 2\hat{i} + \hat{j} - \hat{k} \)
- Let the position vector of point B be \( \vec{B} = 3\hat{i} - 2\hat{j} + \hat{k} \)
- Let the position vector of point C be \( \vec{C} = \hat{i} + 4\hat{j} - 3\hat{k} \)

2. **Form the vectors AB and AC**:
- The vector \( \vec{AB} = \vec{B} - \vec{A} = (3\hat{i} - 2\hat{j} + \hat{k}) - (2\hat{i} + \hat{j} - \hat{k}) \)
- Simplifying \( \vec{AB} \):
\[ \vec{AB} = (3 - 2)\hat{i} + (-2 - 1)\hat{j} + (1 + 1)\hat{k} = \hat{i} - 3\hat{j} + 2\hat{k} \]

- The vector \( \vec{AC} = \vec{C} - \vec{A} = (\hat{i} + 4\hat{j} - 3\hat{k}) - (2\hat{i} + \hat{j} - \hat{k}) \)
- Simplifying \( \vec{AC} \):
\[ \vec{AC} = (1 - 2)\hat{i} + (4 - 1)\hat{j} + (-3 + 1)\hat{k} = -\hat{i} + 3\hat{j} - 2\hat{k} \]

3. **Find the area of triangle ABC**:
- The area of triangle ABC can be found using the cross product of vectors \( \vec{AB} \) and \( \vec{AC} \):
\[ \text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| \]

4. **Calculate the cross product**:
- The vectors are:
\[ \vec{AB} = \hat{i} - 3\hat{j} + 2\hat{k}, \quad \vec{AC} = -\hat{i} + 3\hat{j} - 2\hat{k} \]
- The determinant for the cross product is:
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ -1 & 3 & -2 \end{vmatrix} \]

- Expanding this determinant:
\[ = \hat{i} \begin{vmatrix} -3 & 2 \\ 3 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ -1 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -3 \\ -1 & 3 \end{vmatrix} \]

- Calculating each of these determinants:
\[ = \hat{i}((-3)(-2) - (2)(3)) - \hat{j}((1)(-2) - (2)(-1)) + \hat{k}((1)(3) - (-3)(-1)) \]
\[ = \hat{i}(6 - 6) - \hat{j}(-2 + 2) + \hat{k}(3 - 3) \]
\[ = \hat{i}(0) - \hat{j}(0) + \hat{k}(0) = \vec{0} \]

5. **Conclusion**:
- Since the cross product \( \vec{AB} \times \vec{AC} = \vec{0} \), the area of triangle ABC is \( 0 \).
- Therefore, points A, B, and C are collinear.

**Final Answer**: The points A, B, and C are collinear.