Determine whether or not the following pairs of lines intersect :
`vec(r) = (2 lambda + 1) hati - (lambda + 1) hatj + (lambda + 1) hatk.`
and `vec(r) = (3 mu + 2)hati - (5mu + 5 ) hatj + (2 mu - 1) hatk. `

Answer

To determine whether the given pairs of lines intersect, we will follow these steps:

### Step 1: Write the equations of the lines in vector form

The first line is given as:
\[ \vec{r_1} = (2\lambda + 1) \hat{i} - (\lambda + 1) \hat{j} + (\lambda + 1) \hat{k} \]

The second line is given as:
\[ \vec{r_2} = (3\mu + 2) \hat{i} - (5\mu + 5) \hat{j} + (2\mu - 1) \hat{k} \]

### Step 2: Identify the components of the lines

From the first line, we can identify:
- Point \( A \) (constant part): \( \vec{A} = \hat{i} + (-1) \hat{j} + (1) \hat{k} = \hat{i} - \hat{j} + \hat{k} \)
- Direction vector \( \vec{B} = (2 \hat{i} - \hat{j} + \hat{k}) \)

From the second line, we can identify:
- Point \( C \) (constant part): \( \vec{C} = 2 \hat{i} - 5 \hat{j} - 1 \hat{k} \)
- Direction vector \( \vec{D} = (3 \hat{i} - 5 \hat{j} + 2 \hat{k}) \)

### Step 3: Calculate the shortest distance between the two lines

The formula for the shortest distance \( d_{\text{min}} \) between two skew lines is given by:
\[ d_{\text{min}} = \frac{|\vec{C} - \vec{A} \cdot (\vec{B} \times \vec{D})|}{|\vec{B} \times \vec{D}|} \]

### Step 4: Calculate \( \vec{C} - \vec{A} \)

Calculating \( \vec{C} - \vec{A} \):
\[ \vec{C} - \vec{A} = (2 \hat{i} - 5 \hat{j} - 1 \hat{k}) - (1 \hat{i} - 1 \hat{j} + 1 \hat{k}) = (2 - 1) \hat{i} + (-5 + 1) \hat{j} + (-1 - 1) \hat{k} = 1 \hat{i} - 4 \hat{j} - 2 \hat{k} \]

### Step 5: Calculate \( \vec{B} \times \vec{D} \)

Now we need to calculate the cross product \( \vec{B} \times \vec{D} \):
\[ \vec{B} = 2 \hat{i} - \hat{j} + \hat{k}, \quad \vec{D} = 3 \hat{i} - 5 \hat{j} + 2 \hat{k} \]

Using the determinant to calculate the cross product:
\[ \vec{B} \times \vec{D} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} -1 & 1 \\ -5 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 3 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -1 \\ 3 & -5 \end{vmatrix} \]
\[ = \hat{i}((-1)(2) - (1)(-5)) - \hat{j}((2)(2) - (1)(3)) + \hat{k}((2)(-5) - (-1)(3)) \]
\[ = \hat{i}(-2 + 5) - \hat{j}(4 - 3) + \hat{k}(-10 + 3) \]
\[ = 3\hat{i} - 1\hat{j} - 7\hat{k} \]

### Step 6: Calculate the magnitude of \( \vec{B} \times \vec{D} \)

Now we calculate the magnitude:
\[ |\vec{B} \times \vec{D}| = \sqrt{(3)^2 + (-1)^2 + (-7)^2} = \sqrt{9 + 1 + 49} = \sqrt{59} \]

### Step 7: Calculate the dot product \( \vec{C} - \vec{A} \cdot (\vec{B} \times \vec{D}) \)

Now we calculate the dot product:
\[ \vec{C} - \vec{A} = 1\hat{i} - 4\hat{j} - 2\hat{k} \]
\[ \vec{B} \times \vec{D} = 3\hat{i} - 1\hat{j} - 7\hat{k} \]
\[ (1)(3) + (-4)(-1) + (-2)(-7) = 3 + 4 + 14 = 21 \]

### Step 8: Calculate the shortest distance

Now substituting back into the distance formula:
\[ d_{\text{min}} = \frac{|21|}{\sqrt{59}} \neq 0 \]

### Conclusion

Since the shortest distance \( d_{\text{min}} \) is not zero, the two lines do not intersect.

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