A line passes through the points whose position vectors are `hati+hatj-2hatk` and `hati-3hatj+hatk`. The position vector of a point on it at unit distance from the first point is (A)`(1)/(5)(5hati+hatj-7hatk)` (B)`(1)/(5)(5hati+9hatj-13hatk)` (C)`(hati-4hatj+3hatk)` (D)`(1)/(5)(hati-4hatj+3hatk)`

Options

A. A. `(1)/(5)(5hati+hatj-7hatk)`
B. `(1)/(5)(4hati+9hatj-15hatk)`
C. `(hati-4hatj+3hatk)`
D. `(1)/(5)(hati-4hatj+3hatk)`

Explanation

To solve the problem step by step, we need to find the position vector of a point on the line that passes through the given points at a unit distance from the first point.

### Step 1: Identify the position vectors of the given points
The position vectors of the points are:
- Point A: \(\mathbf{A} = \hat{i} + \hat{j} - 2\hat{k}\)
- Point B: \(\mathbf{B} = \hat{i} - 3\hat{j} + \hat{k}\)

### Step 2: Write the coordinates of points A and B
From the position vectors:
- Coordinates of A: \( (1, 1, -2) \)
- Coordinates of B: \( (1, -3, 1) \)

### Step 3: Find the direction ratios of the line AB
The direction ratios can be found by subtracting the coordinates of A from B:
\[ \text{Direction ratios} = (1 - 1, -3 - 1, 1 - (-2)) = (0, -4, 3) \]

### Step 4: Write the parametric equations of the line
Using point A and the direction ratios, the parametric equations of the line can be written as:
\[ x = 1 + 0t \quad (x = 1) \]
\[ y = 1 - 4t \]
\[ z = -2 + 3t \]

### Step 5: Find the general point on the line
The general point on the line can be expressed as:
\[ \mathbf{P}(t) = (1, 1 - 4t, -2 + 3t) \]

### Step 6: Calculate the distance from point A to point P
The distance \(PA\) should be equal to 1. The distance formula gives:
\[ PA = \sqrt{(1 - 1)^2 + (1 - (1 - 4t))^2 + (-2 - (-2 + 3t))^2} \]
This simplifies to:
\[ PA = \sqrt{0 + (4t)^2 + (3t)^2} = \sqrt{16t^2 + 9t^2} = \sqrt{25t^2} = 5|t| \]
Setting this equal to 1 gives:
\[ 5|t| = 1 \implies |t| = \frac{1}{5} \]
Thus, \(t = \frac{1}{5}\) or \(t = -\frac{1}{5}\).

### Step 7: Find the position vectors for both values of t
1. For \(t = \frac{1}{5}\):
\[ \mathbf{P}\left(\frac{1}{5}\right) = \left(1, 1 - 4\left(\frac{1}{5}\right), -2 + 3\left(\frac{1}{5}\right)\right) = \left(1, 1 - \frac{4}{5}, -2 + \frac{3}{5}\right) = \left(1, \frac{1}{5}, -\frac{7}{5}\right) \]
This can be expressed as:
\[ \mathbf{P}\left(\frac{1}{5}\right) = \frac{1}{5}(5\hat{i} + \hat{j} - 7\hat{k}) \]

2. For \(t = -\frac{1}{5}\):
\[ \mathbf{P}\left(-\frac{1}{5}\right) = \left(1, 1 - 4\left(-\frac{1}{5}\right), -2 + 3\left(-\frac{1}{5}\right)\right) = \left(1, 1 + \frac{4}{5}, -2 - \frac{3}{5}\right) = \left(1, \frac{9}{5}, -\frac{13}{5}\right) \]
This can be expressed as:
\[ \mathbf{P}\left(-\frac{1}{5}\right) = \frac{1}{5}(5\hat{i} + 9\hat{j} - 13\hat{k}) \]

### Step 8: Conclusion
Thus, the position vector of a point on the line at unit distance from the first point can be either:
- \(\frac{1}{5}(5\hat{i} + 9\hat{j} - 13\hat{k})\) (Option B)
- \(\frac{1}{5}(5\hat{i} + \hat{j} - 7\hat{k})\) (Option A)

Since the question asks for a point at unit distance, both options are valid, but the first one corresponds to the positive value of \(t\).