Consider the equations of the straight lines given by :
`L_(1) : vec(r) = (hati + 2 hatj + hatk ) + lambda ( hati - hatj + hatk)`
`L_(2) : vec(r) = (2 hati - hatj - hatk) + mu ( 2 hati + hatj + 2 hatk)` .
If `vec(a_(1))= hati + 2 hatj + hatk, " " vec(b_(1)) = hati - hatj + hatk`,
`vec(a_(2)) = 2 hat(i) - hatj - hatk, vec(b_(2)) = 2 hati + hatj + 2 hatk `, then find :
(i) `vec(a_(2)) - vec(a_(1)) " " (ii) vec(b_(2)) - vec(b_(1))`
(iii) `vec(b_(1))xx vec(b_(2)) " " (iv) vec(a_(1)) xx vec(a_(2))`
(v) ` (vec(b_(1)) xx vec(b_(2))).(vec(a_(1)) xxvec(a_(2)))`
(vi) the shortest distance between `L_(1) and L_(2)` .

Answer

To solve the given problem step by step, we will follow the instructions provided in the question and perform the necessary vector operations.

### Given:
1. **Line L1**: \(\vec{r} = \hat{i} + 2\hat{j} + \hat{k} + \lambda(\hat{i} - \hat{j} + \hat{k})\)
- \(\vec{a_1} = \hat{i} + 2\hat{j} + \hat{k}\)
- \(\vec{b_1} = \hat{i} - \hat{j} + \hat{k}\)

2. **Line L2**: \(\vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k})\)
- \(\vec{a_2} = 2\hat{i} - \hat{j} - \hat{k}\)
- \(\vec{b_2} = 2\hat{i} + \hat{j} + 2\hat{k}\)

### Step 1: Calculate \(\vec{a_2} - \vec{a_1}\)

\[ \vec{a_2} - \vec{a_1} = (2\hat{i} - \hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + \hat{k}) \]
\[ = (2 - 1)\hat{i} + (-1 - 2)\hat{j} + (-1 - 1)\hat{k} \]
\[ = \hat{i} - 3\hat{j} - 2\hat{k} \]

### Step 2: Calculate \(\vec{b_2} - \vec{b_1}\)

\[ \vec{b_2} - \vec{b_1} = (2\hat{i} + \hat{j} + 2\hat{k}) - (\hat{i} - \hat{j} + \hat{k}) \]
\[ = (2 - 1)\hat{i} + (1 + 1)\hat{j} + (2 - 1)\hat{k} \]
\[ = \hat{i} + 2\hat{j} + \hat{k} \]

### Step 3: Calculate \(\vec{b_1} \times \vec{b_2}\)

Using the determinant method:
\[ \vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} \]

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

### Step 4: Calculate \(\vec{a_1} \times \vec{a_2}\)

Using the determinant method:
\[ \vec{a_1} \times \vec{a_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 2 & -1 & -1 \end{vmatrix} \]

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

### Step 5: Calculate the dot product \((\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_1} \times \vec{a_2})\)

\[ (-3\hat{i} + 3\hat{k}) \cdot (-\hat{i} + 3\hat{j} - 5\hat{k}) = (-3)(-1) + (0)(3) + (3)(-5) \]
\[ = 3 + 0 - 15 = -12 \]

### Step 6: Calculate the shortest distance between lines \(L_1\) and \(L_2\)

Using the formula for the shortest distance:
\[ d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} \]

First, calculate \(\vec{a_2} - \vec{a_1}\):
\[ \vec{a_2} - \vec{a_1} = \hat{i} - 3\hat{j} - 2\hat{k} \]

Now, calculate the dot product:
\[ (\hat{i} - 3\hat{j} - 2\hat{k}) \cdot (-3\hat{i} + 3\hat{k}) = (1)(-3) + (-3)(0) + (-2)(3) = -3 + 0 - 6 = -9 \]

Now, calculate the magnitude of \(\vec{b_1} \times \vec{b_2}\):
\[ |\vec{b_1} \times \vec{b_2}| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{9 + 0 + 9} = \sqrt{18} = 3\sqrt{2} \]

Finally, calculate the shortest distance:
\[ d = \frac{| -9 |}{3\sqrt{2}} = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \]

### Final Answers:
1. \(\vec{a_2} - \vec{a_1} = \hat{i} - 3\hat{j} - 2\hat{k}\)
2. \(\vec{b_2} - \vec{b_1} = \hat{i} + 2\hat{j} + \hat{k}\)
3. \(\vec{b_1} \times \vec{b_2} = -3\hat{i} + 3\hat{k}\)
4. \(\vec{a_1} \times \vec{a_2} = -\hat{i} + 3\hat{j} - 5\hat{k}\)
5. \((\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_1} \times \vec{a_2}) = -12\)
6. Shortest distance \(d = \frac{3\sqrt{2}}{2}\)