Find the shortest distance between the lines given by
`(x-8)/(3)=(y+9)/(-1)=(z-10)/(7)and(x-1)/(3)=(y-2)/(8)=(z-5)/(-5)`

Options

A. `(68)/(sqrt514)` units
B. `(86)/(sqrt541)` units
C. `(206)/(sqrt514)` units
D. `(321)/(sqrt514)` units

Answer

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

### Step 1: Write the lines in vector form

The first line is given by:
\[ \frac{x-8}{3} = \frac{y+9}{-1} = \frac{z-10}{7} \]
This can be expressed in vector form as:
\[ \mathbf{r_1} = \begin{pmatrix} 8 \\ -9 \\ 10 \end{pmatrix} + \lambda \begin{pmatrix} 3 \\ -1 \\ 7 \end{pmatrix} \]
where \(\begin{pmatrix} 8 \\ -9 \\ 10 \end{pmatrix}\) is a point on the line and \(\begin{pmatrix} 3 \\ -1 \\ 7 \end{pmatrix}\) is the direction vector.

The second line is given by:
\[ \frac{x-1}{3} = \frac{y-2}{8} = \frac{z-5}{-5} \]
This can be expressed in vector form as:
\[ \mathbf{r_2} = \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ 8 \\ -5 \end{pmatrix} \]
where \(\begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix}\) is a point on the line and \(\begin{pmatrix} 3 \\ 8 \\ -5 \end{pmatrix}\) is the direction vector.

### Step 2: Identify the points and direction vectors

From the vector forms, we have:
- For Line 1:
- Point \(A_1 = \begin{pmatrix} 8 \\ -9 \\ 10 \end{pmatrix}\)
- Direction vector \(B_1 = \begin{pmatrix} 3 \\ -1 \\ 7 \end{pmatrix}\)

- For Line 2:
- Point \(A_2 = \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix}\)
- Direction vector \(B_2 = \begin{pmatrix} 3 \\ 8 \\ -5 \end{pmatrix}\)

### Step 3: Calculate \(A_2 - A_1\)

Now we compute:
\[ A_2 - A_1 = \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix} - \begin{pmatrix} 8 \\ -9 \\ 10 \end{pmatrix} = \begin{pmatrix} 1 - 8 \\ 2 - (-9) \\ 5 - 10 \end{pmatrix} = \begin{pmatrix} -7 \\ 11 \\ -5 \end{pmatrix} \]

### Step 4: Compute the cross product \(B_1 \times B_2\)

To find the cross product \(B_1 \times B_2\):
\[ B_1 = \begin{pmatrix} 3 \\ -1 \\ 7 \end{pmatrix}, \quad B_2 = \begin{pmatrix} 3 \\ 8 \\ -5 \end{pmatrix} \]
Using the determinant:
\[ B_1 \times B_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -1 & 7 \\ 3 & 8 & -5 \end{vmatrix} \]
Calculating the determinant:
\[ = \mathbf{i}((-1)(-5) - (7)(8)) - \mathbf{j}((3)(-5) - (7)(3)) + \mathbf{k}((3)(8) - (-1)(3)) \]
\[ = \mathbf{i}(5 - 56) - \mathbf{j}(-15 - 21) + \mathbf{k}(24 + 3) \]
\[ = \mathbf{i}(-51) + \mathbf{j}(36) + \mathbf{k}(27) = \begin{pmatrix} -51 \\ 36 \\ 27 \end{pmatrix} \]

### Step 5: Calculate the magnitude of \(B_1 \times B_2\)

Now, we find the magnitude:
\[ |B_1 \times B_2| = \sqrt{(-51)^2 + 36^2 + 27^2} = \sqrt{2601 + 1296 + 729} = \sqrt{4626} \]

### Step 6: Calculate the dot product \((A_2 - A_1) \cdot (B_1 \times B_2)\)

Now we compute the dot product:
\[ (A_2 - A_1) \cdot (B_1 \times B_2) = \begin{pmatrix} -7 \\ 11 \\ -5 \end{pmatrix} \cdot \begin{pmatrix} -51 \\ 36 \\ 27 \end{pmatrix} \]
\[ = (-7)(-51) + (11)(36) + (-5)(27) = 357 + 396 - 135 = 618 \]

### Step 7: Calculate the shortest distance

Finally, the shortest distance \(d\) between the two lines is given by:
\[ d = \frac{|(A_2 - A_1) \cdot (B_1 \times B_2)|}{|B_1 \times B_2|} = \frac{618}{\sqrt{4626}} \]

### Step 8: Simplify the expression

Calculating the numerical value:
\[ d \approx \frac{618}{68.0} \approx 9.1 \]

### Final Answer
The shortest distance between the two lines is approximately \(9.1\).