(i) Find the vector equation of a line passing through a point with position vector 2` hat(i) - hat(j) + hat(k)` and parallel to the line joining the points with position vectors `- hat(i) + 4 hat(j) + hat(k) and hat(i) + 2 hat(j) + 2 hat(k).` Also, find the cartesian equivalent of the equation.
(ii) Find the vector equation of a line passing through the point with position vector `hat(i) - 2 hat(j) - 3 hat(k)` and parallel to the line joining the points with position vectors `hat(i) - hat(j) + 4 hat(k) and 2 hat(i) + hat(j) + 2 hat(k)`. Also, find the cartesian form of the equation.

Answer

To solve the given problem, we will break it down into two parts as stated in the question.

### Part (i)

**Step 1: Identify the fixed point and the points for the line.**

The fixed point has a position vector:
\[ \mathbf{A} = 2\hat{i} - \hat{j} + \hat{k} \]

The points with position vectors are:
\[ \mathbf{P} = -\hat{i} + 4\hat{j} + \hat{k} \]
\[ \mathbf{Q} = \hat{i} + 2\hat{j} + 2\hat{k} \]

**Step 2: Find the direction vector of the line joining points P and Q.**

The direction vector \(\mathbf{PQ}\) is given by:
\[ \mathbf{PQ} = \mathbf{Q} - \mathbf{P} = (\hat{i} + 2\hat{j} + 2\hat{k}) - (-\hat{i} + 4\hat{j} + \hat{k}) \]
Calculating this gives:
\[ \mathbf{PQ} = \hat{i} + 2\hat{j} + 2\hat{k} + \hat{i} - 4\hat{j} - \hat{k} = 2\hat{i} - 2\hat{j} + \hat{k} \]

**Step 3: Write the vector equation of the line.**

The vector equation of the line passing through point A and parallel to vector \(\mathbf{PQ}\) is:
\[ \mathbf{r} = \mathbf{A} + \lambda \mathbf{PQ} \]
Substituting the values:
\[ \mathbf{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda (2\hat{i} - 2\hat{j} + \hat{k}) \]
This simplifies to:
\[ \mathbf{r} = (2 + 2\lambda)\hat{i} + (-1 - 2\lambda)\hat{j} + (1 + \lambda)\hat{k} \]

**Step 4: Convert to Cartesian form.**

Let \(x = 2 + 2\lambda\), \(y = -1 - 2\lambda\), \(z = 1 + \lambda\).

From these equations, we can express \(\lambda\):
\[ \lambda = \frac{x - 2}{2} \quad (1) \]
\[ \lambda = -\frac{y + 1}{2} \quad (2) \]
\[ \lambda = z - 1 \quad (3) \]

Setting (1) = (2):
\[ \frac{x - 2}{2} = -\frac{y + 1}{2} \]
This simplifies to:
\[ x - 2 = -y - 1 \implies x + y = 1 \quad (4) \]

Setting (1) = (3):
\[ \frac{x - 2}{2} = z - 1 \]
This simplifies to:
\[ x - 2 = 2z - 2 \implies x - 2z = 0 \quad (5) \]

Setting (2) = (3):
\[ -\frac{y + 1}{2} = z - 1 \]
This simplifies to:
\[ -y - 1 = 2z - 2 \implies y + 2z = 1 \quad (6) \]

Thus, the Cartesian equations of the line are:
\[ x + y = 1, \quad x - 2z = 0, \quad y + 2z = 1 \]

### Part (ii)

**Step 1: Identify the fixed point and the points for the line.**

The fixed point has a position vector:
\[ \mathbf{A} = \hat{i} - 2\hat{j} - 3\hat{k} \]

The points with position vectors are:
\[ \mathbf{P} = \hat{i} - \hat{j} + 4\hat{k} \]
\[ \mathbf{Q} = 2\hat{i} + \hat{j} + 2\hat{k} \]

**Step 2: Find the direction vector of the line joining points P and Q.**

The direction vector \(\mathbf{PQ}\) is given by:
\[ \mathbf{PQ} = \mathbf{Q} - \mathbf{P} = (2\hat{i} + \hat{j} + 2\hat{k}) - (\hat{i} - \hat{j} + 4\hat{k}) \]
Calculating this gives:
\[ \mathbf{PQ} = 2\hat{i} + \hat{j} + 2\hat{k} - \hat{i} + \hat{j} - 4\hat{k} = \hat{i} + 2\hat{j} - 2\hat{k} \]

**Step 3: Write the vector equation of the line.**

The vector equation of the line passing through point A and parallel to vector \(\mathbf{PQ}\) is:
\[ \mathbf{r} = \mathbf{A} + \mu \mathbf{PQ} \]
Substituting the values:
\[ \mathbf{r} = (\hat{i} - 2\hat{j} - 3\hat{k}) + \mu (\hat{i} + 2\hat{j} - 2\hat{k}) \]
This simplifies to:
\[ \mathbf{r} = (1 + \mu)\hat{i} + (-2 + 2\mu)\hat{j} + (-3 - 2\mu)\hat{k} \]

**Step 4: Convert to Cartesian form.**

Let \(x = 1 + \mu\), \(y = -2 + 2\mu\), \(z = -3 - 2\mu\).

From these equations, we can express \(\mu\):
\[ \mu = x - 1 \quad (1) \]
\[ \mu = \frac{y + 2}{2} \quad (2) \]
\[ \mu = -\frac{z + 3}{2} \quad (3) \]

Setting (1) = (2):
\[ x - 1 = \frac{y + 2}{2} \]
This simplifies to:
\[ 2(x - 1) = y + 2 \implies 2x - y = 4 \quad (4) \]

Setting (1) = (3):
\[ x - 1 = -\frac{z + 3}{2} \]
This simplifies to:
\[ 2(x - 1) = -z - 3 \implies 2x + z = 1 \quad (5) \]

Setting (2) = (3):
\[ \frac{y + 2}{2} = -\frac{z + 3}{2} \]
This simplifies to:
\[ y + 2 = -z - 3 \implies y + z = -5 \quad (6) \]

Thus, the Cartesian equations of the line are:
\[ 2x - y = 4, \quad 2x + z = 1, \quad y + z = -5 \]