The shortest distance between the two lines `L_1:x=k_1, y=k_2 and L_2: x=k_3, y=k_4` is equal to

Options

A. `|sqrt(k_1^2+k_2^2)-sqrt(k_3^2+k_4^2)|`
B. `sqrt(k_1k_3+k_3k_4)`
C. `sqrt((k_1+k_3)^2+(k_2+k_4)^2)`
D. `sqrt((k_1-k_3)^2+(k_2-k_4)^2)`

Answer

To find the shortest distance between the two lines given by \( L_1: x = k_1, y = k_2 \) and \( L_2: x = k_3, y = k_4 \), we can follow these steps:

### Step 1: Understand the Lines
The lines \( L_1 \) and \( L_2 \) are both vertical lines in three-dimensional space, meaning they extend infinitely along the z-axis. The coordinates of points on these lines can be represented as:
- For \( L_1 \): \( (k_1, k_2, z) \) where \( z \) can take any value.
- For \( L_2 \): \( (k_3, k_4, z) \) where \( z \) can also take any value.

### Step 2: Identify the Direction Vectors
Since both lines are vertical, their direction vectors can be represented as:
- For \( L_1 \): \( \mathbf{b_1} = (0, 0, 1) \)
- For \( L_2 \): \( \mathbf{b_2} = (0, 0, 1) \)

### Step 3: Find Points on Each Line
Choose points on each line:
- Point on \( L_1 \): \( A = (k_1, k_2, 0) \)
- Point on \( L_2 \): \( B = (k_3, k_4, 0) \)

### Step 4: Calculate the Vector Between the Points
The vector \( \mathbf{A_2 - A_1} \) from point \( A \) to point \( B \) is given by:
\[ \mathbf{A_2 - A_1} = (k_3 - k_1, k_4 - k_2, 0) \]

### Step 5: Use the Formula for Distance Between Parallel Lines
The formula for the distance \( d \) between two parallel lines is given by:
\[ d = \frac{|\mathbf{b} \cdot (\mathbf{A_2 - A_1})|}{|\mathbf{b}|} \]
where \( \mathbf{b} \) is the direction vector of the lines.

### Step 6: Calculate the Cross Product
Since both lines are parallel, we can use the cross product to find the distance:
\[ \mathbf{b} = (0, 0, 1) \]
The cross product \( \mathbf{b} \times (\mathbf{A_2 - A_1}) \) is calculated as follows:
\[ \mathbf{b} \times (\mathbf{A_2 - A_1}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 0 & 1 \\ k_3 - k_1 & k_4 - k_2 & 0 \end{vmatrix} \]
Calculating the determinant gives:
\[ = (0 \cdot 0 - 1 \cdot (k_4 - k_2)) \mathbf{i} - (0 \cdot 0 - 1 \cdot (k_3 - k_1)) \mathbf{j} + (0 \cdot (k_4 - k_2) - 0 \cdot (k_3 - k_1)) \mathbf{k} \]
Thus, we have:
\[ \mathbf{b} \times (\mathbf{A_2 - A_1}) = (- (k_4 - k_2), (k_3 - k_1), 0) \]

### Step 7: Calculate the Magnitude of the Cross Product
The magnitude of the cross product is:
\[ |\mathbf{b} \times (\mathbf{A_2 - A_1})| = \sqrt{(k_4 - k_2)^2 + (k_3 - k_1)^2} \]

### Step 8: Calculate the Magnitude of the Direction Vector
The magnitude of the direction vector \( \mathbf{b} \) is:
\[ |\mathbf{b}| = \sqrt{0^2 + 0^2 + 1^2} = 1 \]

### Step 9: Substitute into the Distance Formula
Substituting into the distance formula gives:
\[ d = \frac{\sqrt{(k_4 - k_2)^2 + (k_3 - k_1)^2}}{1} = \sqrt{(k_4 - k_2)^2 + (k_3 - k_1)^2} \]

### Final Result
Thus, the shortest distance between the two lines \( L_1 \) and \( L_2 \) is:
\[ d = \sqrt{(k_4 - k_2)^2 + (k_3 - k_1)^2} \]

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