Find the shortest distance between lines ` vec r=( hat i+2 hat j+ hat k)+lambda(2 hat i+ hat j+2 hat k)a n d vec r=2 hat i- hat j- hat k+mu(2 hat i+ hat j+2 hat k)dot`

Answer

To find the shortest distance between the given lines, we can follow these steps:

### Step 1: Identify the lines and their direction vectors
The lines are given in vector form:
1. Line 1: \(\vec{r_1} = \hat{i} + 2\hat{j} + \hat{k} + \lambda(2\hat{i} + \hat{j} + 2\hat{k})\)
2. Line 2: \(\vec{r_2} = 2\hat{i} - \hat{j} - \hat{k} + \mu(2\hat{i} + \hat{j} + 2\hat{k})\)

From these equations, we can extract the direction vectors:
- Direction vector of Line 1, \(\vec{B} = 2\hat{i} + \hat{j} + 2\hat{k}\)
- Direction vector of Line 2, \(\vec{B'} = 2\hat{i} + \hat{j} + 2\hat{k}\)

### Step 2: Check if the lines are parallel or skew
Since both lines have the same direction vector \(\vec{B}\), they are parallel lines.

### Step 3: Identify the position vectors of the lines
The position vectors (points on the lines) are:
- Point on Line 1, \(\vec{A_1} = \hat{i} + 2\hat{j} + \hat{k}\)
- Point on Line 2, \(\vec{A_2} = 2\hat{i} - \hat{j} - \hat{k}\)

### Step 4: Calculate the vector between the two points
We find the vector \(\vec{A_2} - \vec{A_1}\):
\[ \vec{A_2} - \vec{A_1} = (2\hat{i} - \hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + \hat{k}) = (2 - 1)\hat{i} + (-1 - 2)\hat{j} + (-1 - 1)\hat{k} \]
\[ = \hat{i} - 3\hat{j} - 2\hat{k} \]

### Step 5: Use the formula for the distance between two parallel lines
The formula for the distance \(d\) between two parallel lines is given by:
\[ d = \frac{|\vec{B} \cdot (\vec{A_2} - \vec{A_1})|}{|\vec{B}|} \]
Where \(\vec{B}\) is the direction vector.

### Step 6: Calculate the dot product
Now, we compute the dot product \(\vec{B} \cdot (\vec{A_2} - \vec{A_1})\):
\[ \vec{B} = 2\hat{i} + \hat{j} + 2\hat{k} \]
\[ \vec{A_2} - \vec{A_1} = \hat{i} - 3\hat{j} - 2\hat{k} \]
\[ \vec{B} \cdot (\vec{A_2} - \vec{A_1}) = (2)(1) + (1)(-3) + (2)(-2) = 2 - 3 - 4 = -5 \]

### Step 7: Calculate the magnitude of the direction vector
Next, we calculate the magnitude of \(\vec{B}\):
\[ |\vec{B}| = \sqrt{(2^2) + (1^2) + (2^2)} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

### Step 8: Substitute into the distance formula
Now substituting into the distance formula:
\[ d = \frac{|-5|}{3} = \frac{5}{3} \]

### Final Answer
The shortest distance between the two parallel lines is \(\frac{5}{3}\) units.

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