Find the shortest distance between the lines `l_(1) and l_(2)` whose vector equation are `vecr =lambda (2hati+ 3hatj+ 4hatk) and vecr=(2hati+3hatj)+mu(2hati+3hatj+ 4hatk) `

Answer

To find the shortest distance between the two lines given by their vector equations, we can follow these steps:

### Step 1: Identify the vector equations of the lines
The vector equations of the lines are given as:
- Line \( l_1: \vec{r} = \lambda (2\hat{i} + 3\hat{j} + 4\hat{k}) \)
- Line \( l_2: \vec{r} = (2\hat{i} + 3\hat{j}) + \mu (2\hat{i} + 3\hat{j} + 4\hat{k}) \)

### Step 2: Determine direction vectors and points on each line
From the equations:
- For line \( l_1 \), the direction vector \( \vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k} \) and a point on the line (when \( \lambda = 0 \)) is \( \vec{a} = \vec{0} = 0\hat{i} + 0\hat{j} + 0\hat{k} \).
- For line \( l_2 \), the direction vector \( \vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k} \) and a point on the line is \( \vec{p} = 2\hat{i} + 3\hat{j} + 0\hat{k} \).

### Step 3: Check if the lines are parallel
Both lines have the same direction vector \( \vec{b} \), indicating that they are parallel.

### Step 4: Use the formula for the distance between two parallel lines
The formula for the distance \( d \) between two parallel lines is given by:
\[ d = \frac{|\vec{p} - \vec{a}|}{|\vec{b}|} \]
where \( \vec{p} \) is a point on line \( l_2 \), \( \vec{a} \) is a point on line \( l_1 \), and \( \vec{b} \) is the direction vector.

### Step 5: Calculate \( |\vec{p} - \vec{a}| \)
Calculate \( \vec{p} - \vec{a} \):
\[ \vec{p} - \vec{a} = (2\hat{i} + 3\hat{j} + 0\hat{k}) - (0\hat{i} + 0\hat{j} + 0\hat{k}) = 2\hat{i} + 3\hat{j} \]
Now, find the magnitude:
\[ |\vec{p} - \vec{a}| = \sqrt{(2)^2 + (3)^2} = \sqrt{4 + 9} = \sqrt{13} \]

### Step 6: Calculate \( |\vec{b}| \)
Calculate the magnitude of the direction vector \( \vec{b} \):
\[ |\vec{b}| = \sqrt{(2)^2 + (3)^2 + (4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \]

### Step 7: Substitute into the distance formula
Now substitute the values into the distance formula:
\[ d = \frac{|\vec{p} - \vec{a}|}{|\vec{b}|} = \frac{\sqrt{13}}{\sqrt{29}} = \sqrt{\frac{13}{29}} \]

### Final Answer
The shortest distance between the lines \( l_1 \) and \( l_2 \) is:
\[ d = \sqrt{\frac{13}{29}} \]

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