Let `bara=hati-hatj,b=hati+2hatj+2hatk,c=hati-hatj+hatk and bard = -2hati-hatj-hatk` then the shortest distance between the lines `barr=bara+tbarb and barr=barc+pbard` is k , then the value of `1/k^2` is

Answer

To find the shortest distance between the lines defined by the vectors \( \bar{a} = \hat{i} - \hat{j} \), \( \bar{b} = \hat{i} + 2\hat{j} + 2\hat{k} \), \( \bar{c} = \hat{i} - \hat{j} + \hat{k} \), and \( \bar{d} = -2\hat{i} - \hat{j} - \hat{k} \), we can follow these steps:

### Step 1: Define the Lines
The first line \( L_1 \) can be represented as:
\[ \vec{r_1} = \bar{a} + t \bar{b} = (\hat{i} - \hat{j}) + t(\hat{i} + 2\hat{j} + 2\hat{k}) \]
The second line \( L_2 \) can be represented as:
\[ \vec{r_2} = \bar{c} + p \bar{d} = (\hat{i} - \hat{j} + \hat{k}) + p(-2\hat{i} - \hat{j} - \hat{k}) \]

### Step 2: Find the Direction Vectors
The direction vector of line \( L_1 \) is:
\[ \bar{b} = \hat{i} + 2\hat{j} + 2\hat{k} \]
The direction vector of line \( L_2 \) is:
\[ \bar{d} = -2\hat{i} - \hat{j} - \hat{k} \]

### Step 3: Calculate the Vector Between Points on the Lines
To find the shortest distance, we need to calculate the vector from a point on line \( L_1 \) to a point on line \( L_2 \):
\[ \vec{C} - \vec{A} = \bar{c} - \bar{a} = (\hat{i} - \hat{j} + \hat{k}) - (\hat{i} - \hat{j}) = \hat{k} \]

### Step 4: Calculate the Cross Product of Direction Vectors
Next, we need to calculate the cross product \( \bar{b} \times \bar{d} \):
\[ \bar{b} \times \bar{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ -2 & -1 & -1 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 2 & 2 \\ -1 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ -2 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ -2 & -1 \end{vmatrix} \]
Calculating these determinants:
- For \( \hat{i} \): \( 2 \cdot (-1) - 2 \cdot (-1) = -2 + 2 = 0 \)
- For \( \hat{j} \): \( 1 \cdot (-1) - 2 \cdot (-2) = -1 + 4 = 3 \)
- For \( \hat{k} \): \( 1 \cdot (-1) - 2 \cdot (-2) = -1 + 4 = 3 \)

Thus:
\[ \bar{b} \times \bar{d} = 0\hat{i} - 3\hat{j} + 3\hat{k} = -3\hat{j} + 3\hat{k} \]

### Step 5: Calculate the Magnitude of the Cross Product
The magnitude of \( \bar{b} \times \bar{d} \) is:
\[ |\bar{b} \times \bar{d}| = \sqrt{0^2 + (-3)^2 + 3^2} = \sqrt{0 + 9 + 9} = \sqrt{18} = 3\sqrt{2} \]

### Step 6: Calculate the Shortest Distance
The shortest distance \( d \) between the two lines is given by:
\[ d = \frac{|\vec{C} - \vec{A} \cdot (\bar{b} \times \bar{d})|}{|\bar{b} \times \bar{d}|} \]
Calculating \( |\vec{C} - \vec{A}| \):
\[ |\hat{k}| = 1 \]
Thus:
\[ d = \frac{1}{3\sqrt{2}} \]
This means \( k = 3\sqrt{2} \).

### Step 7: Find \( \frac{1}{k^2} \)
Now, we need to find \( \frac{1}{k^2} \):
\[ k^2 = (3\sqrt{2})^2 = 9 \cdot 2 = 18 \]
Thus:
\[ \frac{1}{k^2} = \frac{1}{18} \]

### Final Answer
The value of \( \frac{1}{k^2} \) is \( \frac{1}{18} \).

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