Show that the lines :
` vec(r) = 3 hati + 2 hatj - 4 hatk + lambda (hati + 2 hatj + 2 hatk)`
and ` vec(r) = 5 hati - 2 hatj + mu (3 hati + 2 hatj + 6 hatk)`
(ii) `vec(r) = (hati + hatj - hatk) + lambda (3 hati - hatj)`.
and `vec(r) = (4 hati - hatk) + mu (2 hati + 3 hatk)`
are intersecting. Hence, find their point of intersection.

Answer

To show that the given lines are intersecting and find their point of intersection, we will follow these steps:

### Step 1: Identify the lines and their vector forms
The first pair of lines is given by:
1. \( \vec{r_1} = (3\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(\hat{i} + 2\hat{j} + 2\hat{k}) \)
2. \( \vec{r_2} = (5\hat{i} - 2\hat{j}) + \mu(3\hat{i} + 2\hat{j} + 6\hat{k}) \)

The second pair of lines is given by:
1. \( \vec{r_3} = (1\hat{i} + 1\hat{j} - 1\hat{k}) + \lambda(3\hat{i} - \hat{j}) \)
2. \( \vec{r_4} = (4\hat{i} - 1\hat{k}) + \mu(2\hat{i} + 3\hat{k}) \)

### Step 2: Find the direction vectors and points for each line
For the first pair:
- Point \( A = (3, 2, -4) \)
- Direction vector \( \vec{b} = (1, 2, 2) \)
- Point \( C = (5, -2, 0) \)
- Direction vector \( \vec{d} = (3, 2, 6) \)

For the second pair:
- Point \( A' = (1, 1, -1) \)
- Direction vector \( \vec{b'} = (3, -1, 0) \)
- Point \( C' = (4, 0, -1) \)
- Direction vector \( \vec{d'} = (2, 0, 3) \)

### Step 3: Check if the lines intersect using the condition for intersection
Two lines intersect if the shortest distance between them is zero. This can be checked using the formula:
\[ d_{min} = \frac{|(\vec{C} - \vec{A}) \cdot (\vec{b} \times \vec{d})|}{|\vec{b} \times \vec{d}|} \]
If \( d_{min} = 0 \), the lines intersect.

#### For the first pair of lines:
1. Calculate \( \vec{C} - \vec{A} = (5 - 3, -2 - 2, 0 + 4) = (2, -4, 4) \)
2. Calculate \( \vec{b} \times \vec{d} \):
\[ \vec{b} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 3 & 2 & 6 \end{vmatrix} = \hat{i}(2 \cdot 6 - 2 \cdot 2) - \hat{j}(1 \cdot 6 - 2 \cdot 3) + \hat{k}(1 \cdot 2 - 2 \cdot 3) = \hat{i}(12 - 4) - \hat{j}(6 - 6) + \hat{k}(2 - 6) = 8\hat{i} + 0\hat{j} - 4\hat{k} \]

3. Calculate \( | \vec{b} \times \vec{d} | = \sqrt{8^2 + 0^2 + (-4)^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5} \)

4. Now calculate \( |(\vec{C} - \vec{A}) \cdot (\vec{b} \times \vec{d})| \):
\[ (2, -4, 4) \cdot (8, 0, -4) = 2 \cdot 8 + (-4) \cdot 0 + 4 \cdot (-4) = 16 + 0 - 16 = 0 \]

Since \( |(\vec{C} - \vec{A}) \cdot (\vec{b} \times \vec{d})| = 0 \), the lines intersect.

### Step 4: Find the point of intersection
To find the point of intersection, set the equations equal:
\[ 3 + \lambda = 5 + 3\mu \quad (1) \]
\[ 2 + 2\lambda = -2 + 2\mu \quad (2) \]

From equation (1):
\[ \lambda - 3\mu = 2 \quad (3) \]

From equation (2):
\[ 2\lambda - 2\mu = -4 \Rightarrow \lambda - \mu = -2 \quad (4) \]

Now solve equations (3) and (4):
From (4):
\[ \lambda = \mu - 2 \]
Substituting into (3):
\[ (\mu - 2) - 3\mu = 2 \Rightarrow -2\mu - 2 = 2 \Rightarrow -2\mu = 4 \Rightarrow \mu = -2 \]
Substituting \( \mu = -2 \) back into (4):
\[ \lambda = -2 - 2 = -4 \]

### Step 5: Substitute back to find the intersection point
Using \( \lambda = -4 \) in the first line:
\[ \vec{r_1} = (3 + (-4), 2 + 2(-4), -4 + 2(-4)) = (3 - 4, 2 - 8, -4 - 8) = (-1, -6, -12) \]

Thus, the point of intersection is:
\[ \text{Point of Intersection} = (-1, -6, -12) \]