The position vectors of A, B, C are `2hat(i)+hat(j)-hat(k), 3hat(i)-2hat(j)+hat(k)` and `hat(i)+4hat(j)-3hat(k)` respectively. Show that A, B and C are collinear.

Answer

To show that points A, B, and C are collinear, we need to demonstrate that the vectors \( \vec{AB} \) and \( \vec{BC} \) are scalar multiples of each other.

### Step-by-Step Solution

1. **Identify the Position Vectors**:
- The position vector of point A is given as:
\[ \vec{A} = 2\hat{i} + \hat{j} - \hat{k} \]
- The position vector of point B is given as:
\[ \vec{B} = 3\hat{i} - 2\hat{j} + \hat{k} \]
- The position vector of point C is given as:
\[ \vec{C} = \hat{i} + 4\hat{j} - 3\hat{k} \]

2. **Calculate the Vector \( \vec{AB} \)**:
- The vector \( \vec{AB} \) is calculated as:
\[ \vec{AB} = \vec{B} - \vec{A} \]
- Substituting the position vectors:
\[ \vec{AB} = (3\hat{i} - 2\hat{j} + \hat{k}) - (2\hat{i} + \hat{j} - \hat{k}) \]
- Simplifying this:
\[ \vec{AB} = (3 - 2)\hat{i} + (-2 - 1)\hat{j} + (1 + 1)\hat{k} = \hat{i} - 3\hat{j} + 2\hat{k} \]

3. **Calculate the Vector \( \vec{BC} \)**:
- The vector \( \vec{BC} \) is calculated as:
\[ \vec{BC} = \vec{C} - \vec{B} \]
- Substituting the position vectors:
\[ \vec{BC} = (\hat{i} + 4\hat{j} - 3\hat{k}) - (3\hat{i} - 2\hat{j} + \hat{k}) \]
- Simplifying this:
\[ \vec{BC} = (1 - 3)\hat{i} + (4 + 2)\hat{j} + (-3 - 1)\hat{k} = -2\hat{i} + 6\hat{j} - 4\hat{k} \]

4. **Check for Collinearity**:
- For points A, B, and C to be collinear, \( \vec{AB} \) must be a scalar multiple of \( \vec{BC} \). We need to find a scalar \( \lambda \) such that:
\[ \vec{AB} = \lambda \vec{BC} \]
- We have:
\[ \vec{AB} = \hat{i} - 3\hat{j} + 2\hat{k} \]
\[ \vec{BC} = -2\hat{i} + 6\hat{j} - 4\hat{k} \]
- Setting up the equations:
\[ \hat{i} = \lambda (-2\hat{i}), \quad -3\hat{j} = \lambda (6\hat{j}), \quad 2\hat{k} = \lambda (-4\hat{k}) \]
- Solving for \( \lambda \):
- From the first equation:
\[ 1 = -2\lambda \implies \lambda = -\frac{1}{2} \]
- From the second equation:
\[ -3 = 6\lambda \implies \lambda = -\frac{1}{2} \]
- From the third equation:
\[ 2 = -4\lambda \implies \lambda = -\frac{1}{2} \]
- Since \( \lambda \) is consistent across all equations, we conclude that:
\[ \vec{AB} = -\frac{1}{2} \vec{BC} \]

5. **Conclusion**:
- Since \( \vec{AB} \) is a scalar multiple of \( \vec{BC} \), points A, B, and C are collinear.