The shortest distance between the lines
`(x-5)/(1)=(y-2)/(2)=(z-4)/(-3)` and `(x+3)/(1)=(y+5)/(4)=(z-1)/(-5)` is

Options

A. `7sqrt(3)`
B. `6sqrt(3)`
C. `4sqrt(3)`
D. `5sqrt(3)`

Answer

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

### Step 1: Identify the lines in vector form
The two lines are given in symmetric form. We can convert them into vector form.

1. **Line 1**: \((x-5)/(1) = (y-2)/(2) = (z-4)/(-3)\)
- This can be expressed as:
\[ \mathbf{r_1} = (5, 2, 4) + \lambda_1 (1, 2, -3) \]
- Here, \( \mathbf{a_1} = (5, 2, 4) \) and \( \mathbf{b_1} = (1, 2, -3) \).

2. **Line 2**: \((x+3)/(1) = (y+5)/(4) = (z-1)/(-5)\)
- This can be expressed as:
\[ \mathbf{r_2} = (-3, -5, 1) + \lambda_2 (1, 4, -5) \]
- Here, \( \mathbf{a_2} = (-3, -5, 1) \) and \( \mathbf{b_2} = (1, 4, -5) \).

### Step 2: Calculate the cross product of direction vectors
To find the shortest distance between the two skew lines, we need to calculate the cross product of their direction vectors \( \mathbf{b_1} \) and \( \mathbf{b_2} \).

\[ \mathbf{b_1} \times \mathbf{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -3 \\ 1 & 4 & -5 \end{vmatrix} \]

Calculating this determinant, we have:

\[ \mathbf{b_1} \times \mathbf{b_2} = \mathbf{i} \begin{vmatrix} 2 & -3 \\ 4 & -5 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & -3 \\ 1 & -5 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2 \\ 1 & 4 \end{vmatrix} \]

Calculating each of these determinants:

1. For \( \mathbf{i} \):
\[ 2 \cdot (-5) - (-3) \cdot 4 = -10 + 12 = 2 \]

2. For \( \mathbf{j} \):
\[ 1 \cdot (-5) - (-3) \cdot 1 = -5 + 3 = -2 \quad \Rightarrow \quad -(-2) = 2 \]

3. For \( \mathbf{k} \):
\[ 1 \cdot 4 - 2 \cdot 1 = 4 - 2 = 2 \]

Thus,
\[ \mathbf{b_1} \times \mathbf{b_2} = (2, 2, 2) \]

### Step 3: Calculate the distance using the formula
The formula for the shortest distance \( d \) between two skew lines is given by:

\[ d = \frac{|(\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2})|}{|\mathbf{b_1} \times \mathbf{b_2}|} \]

First, calculate \( \mathbf{a_2} - \mathbf{a_1} \):
\[ \mathbf{a_2} - \mathbf{a_1} = (-3 - 5, -5 - 2, 1 - 4) = (-8, -7, -3) \]

Now calculate the dot product:
\[ (\mathbf{a_2} - \mathbf{a_1}) \cdot (\mathbf{b_1} \times \mathbf{b_2}) = (-8, -7, -3) \cdot (2, 2, 2) = -8 \cdot 2 + -7 \cdot 2 + -3 \cdot 2 = -16 - 14 - 6 = -36 \]

Now calculate the magnitude of \( \mathbf{b_1} \times \mathbf{b_2} \):
\[ |\mathbf{b_1} \times \mathbf{b_2}| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{4 + 4 + 4} = \sqrt{12} = 2\sqrt{3} \]

Finally, substitute into the distance formula:
\[ d = \frac{| -36 |}{2\sqrt{3}} = \frac{36}{2\sqrt{3}} = \frac{18}{\sqrt{3}} = 6\sqrt{3} \]

### Final Answer
The shortest distance between the lines is \( 6\sqrt{3} \).

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